SOLUTIONS MANUAL for Game Theory Basics New Edition by Bernhard von Stengel

Game Theory Basics Solutions to Exercises © Bernhard von Stengel 2022 Solution to Exercise 1.1 (a) Let ≤ be defined by (1.7). To show that ≤ is transitive, consider 𝑥, 𝑦, 𝑧 with 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑧. If 𝑥 = 𝑦 then 𝑥 ≤ 𝑧, and if 𝑦 = 𝑧 then also 𝑥 ≤ 𝑧. So the only case left is 𝑥 < 𝑦 and 𝑦 < 𝑧, which implies 𝑥 < 𝑧 because < is transitive, and hence 𝑥 ≤ 𝑧.

Clearly, ≤ is reflexive because 𝑥 = 𝑥 and therefore 𝑥 ≤ 𝑥.

To show that ≤ is antisymmetric, consider 𝑥 and 𝑦 with 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥. If we had 𝑥 ≠ 𝑦 then 𝑥 < 𝑦 and 𝑦 < 𝑥, and by transitivity 𝑥 < 𝑥 which contradicts (1.38). Hence 𝑥 = 𝑦, as required. This shows that ≤ is a partial order.

Finally, we show (1.6), so we have to show that 𝑥 < 𝑦 implies 𝑥 ≤ 𝑦 and 𝑥 ≠ 𝑦 and vice versa. Let 𝑥 < 𝑦, which implies 𝑥 ≤ 𝑦 by (1.7). If we had 𝑥 = 𝑦 then 𝑥 < 𝑥, contradicting (1.38), so we also have 𝑥 ≠ 𝑦. Conversely, 𝑥 ≤ 𝑦 and 𝑥 ≠ 𝑦 imply by (1.7) 𝑥 < 𝑦 or 𝑥 = 𝑦 where the second case is excluded, hence 𝑥 < 𝑦, as required. (b) Consider a partial order ≤ and assume (1.6) as a definition of <. To show that < is transitive, suppose 𝑥 < 𝑦, that is, 𝑥 ≤ 𝑦 and 𝑥 ≠ 𝑦, and 𝑦 < 𝑧, that is, 𝑦 ≤ 𝑧 and 𝑦 ≠ 𝑧.

Because ≤ is transitive, 𝑥 ≤ 𝑧. If we had 𝑥 = 𝑧 then 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 and hence 𝑥 = 𝑦 by antisymmetry of ≤, which contradicts 𝑥 ≠ 𝑦, so we have 𝑥 ≤ 𝑧 and 𝑥 ≠ 𝑧, that is, 𝑥 < 𝑧 by (1.6), as required.

Also, < is irreflexive, because 𝑥 < 𝑥 would by definition mean 𝑥 ≤ 𝑥 and 𝑥 ≠ 𝑥, but the latter is not true.

Finally, we show (1.7), so we have to show that 𝑥 ≤ 𝑦 implies 𝑥 < 𝑦 or 𝑥 = 𝑦 and vice versa, given that < is defined by (1.6). Let 𝑥 ≤ 𝑦. Then if 𝑥 = 𝑦, we are done, otherwise 𝑥 ≠ 𝑦 and then by definition 𝑥 < 𝑦. Hence, 𝑥 ≤ 𝑦 implies 𝑥 < 𝑦 or 𝑥 = 𝑦.

Conversely, suppose 𝑥 < 𝑦 or 𝑥 = 𝑦. If 𝑥 < 𝑦 then 𝑥 ≤ 𝑦 by (1.6), and if 𝑥 = 𝑦 then 𝑥 ≤ 𝑦 because ≤ is reflexive. This completes the proof.

Solution to Exercise 1.2 (a) In analysing the games of three Nim heaps where one heap has size one, we first look at some examples, and then use mathematical induction to prove what we conjecture to be the losing positions. A losing position is one where every move is to a winning position, because then the opponent will win. The point of this exercise is to formulate a precise statement to be proved, and then to prove it.

First, if there are only two heaps recall that they are losing if and only if the heaps are of equal size. If they are of unequal size, then the winning move is to reduce the larger heap so that both heaps have equal size.

1

Consider three heaps of sizes 1, 𝑚, 𝑛, where 1 ≤ 𝑚 ≤ 𝑛. We observe the following: 1, 1, 𝑚 is winning, by moving to 1, 1, 0. Similarly, 1, 𝑚, 𝑚 is winning, by moving to 0, 𝑚, 𝑚. Next, 1, 2, 3 is losing (observed earlier in the lecture), and hence 1, 2, 𝑛 for 𝑛 ≥ 4 is winning. 1, 3, 𝑛 is winning for any 𝑛 ≥ 3 by moving to 1, 3, 2. For 1, 4, 5, reducing any heap produces a winning position, so this is losing.

The general pattern for the losing positions thus seems to be: 1, 𝑚, 𝑚 + 1, for even numbers 𝑚. This includes also the case 𝑚 = 0, which we can take as the base case for an induction. We now proceed to prove this formally.

First we show that if the positions of the form 1, 𝑚, 𝑛 with 𝑚 ≤ 𝑛 are losing when 𝑚 is even and 𝑛 = 𝑚 + 1, then these are the only losing positions because any other position 1, 𝑚, 𝑛 with 𝑚 ≤ 𝑛 is winning. Namely, if 𝑚 = 𝑛 then a winning move from 1, 𝑚, 𝑚 is to 0, 𝑚, 𝑚, so we can assume 𝑚 < 𝑛. If 𝑚 is even then 𝑛 > 𝑚 + 1 (otherwise we would be in the position 1, 𝑚, 𝑚 + 1) and so the winning move is to 1, 𝑚, 𝑚 + 1. If 𝑚 is odd then the winning move is to 1, 𝑚, 𝑚 − 1, the same as position 1, 𝑚 − 1, 𝑚 (this would also be a winning move from 1, 𝑚, 𝑚 so there the winning move is not unique).

Second, we show that any move from 1, 𝑚, 𝑚 + 1 with even 𝑚 is to a winning position, using as inductive hypothesis that 1, 𝑚 0 , 𝑚 0 + 1 for even 𝑚 0 and 𝑚 0 < 𝑚 is a losing position. The move to 0, 𝑚, 𝑚 + 1 produces a winning position with counter-move to 0, 𝑚, 𝑚. A move to 1, 𝑚 0 , 𝑚 + 1 for 𝑚 0 < 𝑚 is to a winning position with the counter-move to 1, 𝑚 0 , 𝑚 0 + 1 if 𝑚 0 is even and to 1, 𝑚 0 , 𝑚 0 − 1 if 𝑚 0 is odd. A move to 1, 𝑚, 𝑚 is to a winning position with counter-move to 0, 𝑚, 𝑚. A move to 1, 𝑚, 𝑚 0 with 𝑚 0 < 𝑚 is also to a winning position with the counter-move to 1, 𝑚 0 − 1, 𝑚 0 if 𝑚 0 is odd, and to 1, 𝑚 0 + 1, 𝑚 0 if 𝑚 0 is even (in which case 𝑚 0 + 1 < 𝑚 because 𝑚 is even).

This concludes the induction proof.

This result is in agreement with the theorem on Nim heap sizes represented as sums of powers of 2: ∗1+∗𝑚 +∗𝑛 is losing if and only if, except for 20 , the powers of 2 making up 𝑚 and 𝑛 come in pairs. So these must be the same powers of 2, except for 1 = 20 , which occurs in only 𝑚 or 𝑛, where we have assumed that 𝑛 is the larger number, so 1 appears in the representation of 𝑛: We have 𝑚 = 2𝑎 + 2𝑏 + 2𝑐 + · · · for 𝑎 > 𝑏 > 𝑐 > · · · ≥ 1, so 𝑚 is even, and, with the same 𝑎, 𝑏, 𝑐, . . ., 𝑛 = 2𝑎 + 2𝑏 + 2𝑐 + · · · + 1 = 𝑚 + 1. Then ∗1 + ∗𝑚 + ∗𝑛 ≡ ∗0. The following is an example using the bit representation where 𝑚 = 12 (which determines the bit pattern 1100, which of course depends on 𝑚): 1 12 13 Nim-sum 0

= = = =

0001 1100 1101 0000

(b) We use (a). Clearly, 1, 2, 3 is losing as shown in (1.2), and because the Nim-sum of the binary representations 01, 10, 11 is 00. Examples show that any other position is winning. The three numbers are 𝑛, 𝑛 + 1, 𝑛 + 2. If 𝑛 is even then reducing the heap of size 𝑛 + 2 to 1 creates the position 𝑛, 𝑛 + 1, 1 which is losing as shown in (a). If 𝑛 is odd, then 𝑛 + 1 is even and 𝑛 + 2 = (𝑛 + 1) + 1 so by the same argument, a winning move is to reduce the Nim heap of size 𝑛 to 1 (which only works if 𝑛 > 1).

2

Without (a), the following is a more complicated argument based entirely on the binary representation. Suppose the position 𝑛, 𝑛 + 1, 𝑛 + 2 was losing. First, if the first and last number is even then the three consecutive numbers have only one odd number among them, with a 1 in the final position of the “sum without carry” of the Nim sizes. Hence the Nim-sum is nonzero and the position is winning. So the three numbers must start and end with an odd number, with a single even number in the middle. If the three numbers have the same largest power of 2 in their binary representation, the leftmost binary column in the Nim-sum is odd, again a winning position. So the middle even number must be a power of 2, which is at least 4 because the case 1, 2, 3 is excluded. So the three numbers have binary representations of the form 01 𝑘 11 (with 𝑘 ≥ 0), 10 𝑘 00, and 10 𝑘 01. So the leftmost and rightmost column have an even number of 1’s (necessary to have a losing position), but the second-to-last column has only one 1 in it and hence the Nim-sum is again nonzero. (c) This is a winning position because the Nim-sum of the heaps is binary 1110 and thus nonzero, as shown in the following table. The table also shows the three winning moves obtained by changing the bits of the binary representation corresponding to the Nim-sum, which are one of the heap changes 8 → 6, 11 → 5, or 13 → 3: heapsize

move 1

move 2

8 = 1000

→ 0110

= 6

1000

= 8

1000

= 8

11 = 1011

1011

= 11

→ 0101

= 5

1011

= 11

13 = 1101

1101

= 13

1101

= 13

→ 0011

= 3

Nim-sum 1110

0000

0000

move 3

0000

Note that the number of removed tokens is different for each of the three heaps, because the three changed bits from the Nim-sum 1110 affect different patterns for the first three bits 100 (for heap size 8), 101 (for heap size 11), and 110 (for heap size 13).

Solution to Exercise 1.3 (a) In misère Nim, a single heap with 𝑛 tokens is losing if 𝑛 = 1, otherwise winning: If 𝑛 > 1 then the player wins by reducing the heap to size 1. If 𝑛 = 0 then the player cannot move and has won; although this is the case of “no heap” rather than “one heap”, this will be useful in (c).

Two heaps are a winning position if one of the heaps has size 1, where the winning move is to remove the entire other heap. Otherwise, both heaps have at least two tokens in them. Then this is a losing position if the two heaps are equal, because any move from there leads to a winning position as follows: If one heap is removed entirely, then the other player responds by reducing the remaining heap to size 1; if one heap is reduced to size 1, then the other player removes the entire other heap; finally, if one heap is reduced to size 2 or larger, then the other player equalizes the heaps again. Consequently, two unequal heaps are therefore a winning position. (b) In misère Nim, 1, 2, 3 is a losing position by the following counter-moves to the first move, in analogy to (1.2):

3

1, 2, 3

2, 3

1, 1, 3

1, 3

1, 2, 2

1, 2, 1

1, 2

2, 2

1, 1, 1

1

2, 2

1, 1, 1

1

(c) Losing positions in misère Nim are given by any odd number of heaps with a single token each, because then all move sequences finish with the current player taking the last token. Conversely, any even number of heaps with a single token each is a winning position. More generally, any even number of single-token heaps can be added (as a game sum) to a misère Nim position without affecting whether it is winning or losing, where the winning player removes a single-token heap if the other player has just done so, or otherwise makes a winning move in another Nim heap.

Consider now a general position and remove all pairs of single-token heaps. If at most two heaps remain, this is covered in (a). Otherwise there are 𝑘 ≥ 3 heaps of sizes 𝑎 1 ≤ 𝑎 2 ≤ · · · ≤ 𝑎 𝑘 with 𝑎 1 ≥ 1 and 𝑎 2 ≥ 2. We claim that this is a losing position if and only if it is a losing position in normal play, that is, if the binary Nim-sum of 𝑎 1 , 𝑎2 , . . . , 𝑎 𝑘 is zero (which we call, as before, a zero position). Suppose 𝑎 1 , 𝑎2 , . . . , 𝑎 𝑘 is a zero position, and consider any move, which creates a non-zero position. If the new position has only two heaps, which can only happen if 𝑘 = 3 and an entire heap has been removed, then the remaining two heaps are unequal, which by (a) is a winning position. If at least three heaps remain, then we have to consider the possibility (if 𝑘 = 3 and 𝑎1 = 1) that these are of the form 1, 1, 𝑎3 or 1, 𝑎2 , 1 where 𝑎 2 , 𝑎3 ≥ 2; in both cases, the winning counter-move is to 1, 1, 1. In all other cases, the winning counter-move is back to a zero position, as in Nim with normal play; this includes the case that this zero position has only two heaps, because then these two heaps are equal and at least of size two, which is also a losing position in misère Nim. In short, misère Nim is played optimally very similarly to Nim, except that an odd rather than even number of single-token heaps define a losing position, and that care has to be taken when only two heaps remain.

Solution to Exercise 1.4 (a) In 3 × 3 Cram, there are, up to symmetry, only two moves for player I, namely placing the domino such that it occupies a corner square or such that it occupies the center square. In either case, player II can respond by placing her domino alongside the first domino, such that a 2 × 2 square in one corner is occupied, leaving the L-shaped remaining 5 squares: II

II

→

→ 4

Then no matter what player I does, II will still be able to place the last domino, and player I loses. Hence, 3 × 3 Cram is a win for player II. (b) When both 𝑚 and 𝑛 are even, player II will win in 𝑚 × 𝑛 Cram by playing “copycat”, using the central symmetry of the board. That is, whenever player I places his domino, player II responds by placing her domino on the square obtained by point-reflection on the center of the board (shown as a dot in the following pictures). The following is an example move sequence on a 4 × 6 board: I

II

→ I

→ II

→

I

→

→

II

I

→ II

→ I

→

II

→

→

Then the domino pattern of the board will have that central symmetry whenever player I makes his move, so that the response move of player II will always be possible.

That is to say, the two adjacent squares in question are always empty because their symmetric counterparts have been empty when player I made his move.

Be careful: Any “copycat” strategy requires that the underlying symmetry of the situation is preserved. This fails when using the symmetry where the board is reflected along a line: here player I could place a domino on the line itself and II would not have a counter-move. So “by copycat” is not a satisfactory answer. (c) In 𝑚 × 𝑛 Cram when 𝑚 is odd and 𝑛 is even (or vice versa), player I will win by placing his domino on the center pair of squares (in the middle row or column), and then playing copycat as described before for player II in (b). The following shows how this could start on a 4 × 5 board: I

→

II

→

I

→

Solution to Exercise 1.5 (a) Player I always wins in these situations. We describe directly the winning move in the games with two rows of dots: For 2 × 𝑚 Chomp, the winning move is to remove the bottom right dot (2, 𝑚), leaving a pattern like the following when 𝑚 = 4:

5

Afterwards, player I can always re-create this pattern by removing the dot that is diagonally adjacent to the dot that player II removed. That is, any move of player II of the form (1, 𝑖) for 𝑖 > 1 can be countered by (2, 𝑖 − 1), and any move (2, 𝑖) by (1, 𝑖 + 1).

So player I has always a move left and wins.

For square games of size 𝑚 × 𝑚 with 𝑚 ≥ 2, the winning move is (2, 2). Then player I can respond to a move of type (𝑖, 1) by removing the dot (1, 𝑖) and vice versa, until player II is forced to take (1, 1) and loses. (b) Remove the “poisoned cookie”, that is, the dot on (1, 1). Then the last player loses by not being able to move any more, exactly when before she would have had to take the poisoned cookie. (c) If removing the bottom right dot (𝑚, 𝑛) is a winning move, then we are done. If not, there is a counter-move (𝑖, 𝑗) by player II that would create a losing situation for player I. But then player I could make (𝑖, 𝑗) to start with, creating the same losing situation for player II. So this game is a win for player I, even though we don’t know the winning move. This beautiful argument works only when considering the removal of a single dot as a first move, which is “absorbed” into the rectangle of removed dots with the next move; it would not work for any other first move. Note that in the case 𝑚 = 𝑛, (a) provides an explicit winning move.

Solution to Exercise 1.6 (a) The completed table for columns 5 and 6 is 0

1

2

3

4

5

6

0

∗0

∗1

∗2

∗3

∗4

∗5

∗6

1

∗1

∗2

∗0

∗4

∗5

∗3

∗7

2

∗2

∗0

∗1

∗5

∗3

∗4

∗8

3

∗3

∗4

∗5

∗6

∗2

∗0

∗1

(b) The queen is on a square with Nim value 2, so one winning move is to reduce the Nim heap to size 2 to make this a losing position. However, there are also two squares with Nim value 4 that the queen can reach, on row 3 column 1 and on row 0 column 4.

These are the only such squares. Moving the queen to either of these also gives a losing position because it is equivalent to ∗4 + ∗4. Any other move would produce a sum of two different Nim heaps, which is not a losing position, so there are no other winning moves.

6

Solution to Exercise 1.7 (a) Putting the domino anywhere on the board produces two independent boards of size 1 × 𝑘 and 1 × (𝑛 − 𝑘 − 2). The two board lengths add up to 𝑛 − 2 because two squares are taken away by the domino. The resulting position is a sum of two games because the player can only move in one of them, which is equivalent to the sum of Nim heaps ∗𝐷 𝑘 + ∗𝐷𝑛−𝑘−2 . By symmetry, we only have to consider 𝑘 ≤ 𝑛2 − 1. As an example, if 𝑛 = 10, then the largest 𝑘 to be considered (where the domino is placed in the middle of the strip) is 𝑘 = 4 = 10 2 − 1,

and if 𝑛 is odd, as when 𝑛 = 9, then the largest 𝑘 to be considered is 𝑘 = 3 ≤ 92 − 1, because for 𝑘 = 4 the domino is already past the middle:

Hence, 𝐷𝑛 = mex({𝐷 𝑘 ⊕ 𝐷𝑛−𝑘−2 | 0 ≤ 𝑘 ≤

𝑛 − 1}).

2

(b) We use the result in (a). Instead of mex({. . .}) we use the simpler notation mex(. . .) and obtain 𝐷0

=

𝐷1 = 0,

𝐷2

=

mex(𝐷0 ⊕ 𝐷0 ) = mex(0) = 1,

𝐷3

=

mex(𝐷0 ⊕ 𝐷1 ) = mex(0) = 1,

𝐷4

=

mex(𝐷0 ⊕ 𝐷2 , 𝐷1 ⊕ 𝐷1 ) = mex(1, 0) = 2,

𝐷5

=

mex(𝐷0 ⊕ 𝐷3 , 𝐷1 ⊕ 𝐷2 ) = mex(1, 1) = 0,

𝐷6

=

mex(𝐷0 ⊕ 𝐷4 , 𝐷1 ⊕ 𝐷3 , 𝐷2 ⊕ 𝐷2 ) = mex(2, 1, 0) = 3,

𝐷7

=

mex(𝐷0 ⊕ 𝐷5 , 𝐷1 ⊕ 𝐷4 , 𝐷2 ⊕ 𝐷3 ) = mex(0, 2, 0) = 1,

𝐷8

=

mex(𝐷0 ⊕ 𝐷6 , 𝐷1 ⊕ 𝐷5 , 𝐷2 ⊕ 𝐷4 , 𝐷3 ⊕ 𝐷3 ) = mex(3, 0, 3, 0) = 1,

𝐷9

=

mex(𝐷0 ⊕ 𝐷7 , 𝐷1 ⊕ 𝐷6 , 𝐷2 ⊕ 𝐷5 , 𝐷3 ⊕ 𝐷4 ) = mex(1, 3, 1, 3) = 0,

𝐷10

=

mex(𝐷0 ⊕ 𝐷8 , 𝐷1 ⊕ 𝐷7 , 𝐷2 ⊕ 𝐷6 , 𝐷3 ⊕ 𝐷5 , 𝐷4 ⊕ 𝐷4 )

=

mex(1, 1, 2, 1, 0) = 3.

For your possible interest, the sequence 𝐷0 , 𝐷1 , . . . is 0, 0, 1, 1, 2, 0, 3, 1, 1, 0, 3, 3, 2, 2, 4, 0, 5, 2, 2, 3, 3, 0, 1, 1, 3, 0, . . . and eventually repeats with period 34, the highest occurring value being 9 (which requires that 8 occurs – explain! But no 6 occurs – how can that be?). Recognised from those 𝑛 where 𝐷𝑛 = 0, Cram on a 1 × 𝑛 board is a losing game for 𝑛 = 0, 1, 5, 9, 15, and others.

Solution to Exercise 1.8 (a) Here, black will win, by simply closing the gap to the white counter whenever white makes a move. So black has always a move left and wins.

7

(b) The close relationship, hopefully revealed by (a), is to Poker Nim, where a move that widens the gap between the two counters amounts to adding tokens to the heap, whereas narrowing the gap amounts to reducing the heap size. That is, the gap between the two counters is the size of the Nim heap that corresponds to that row.

The equivalent Nim heaps are shown on the left: *2 *3 *1 *3 *0 *2 *6 *4

8 7 6 5 4 3 2 1 a b

c d e

f

g h

As an experienced Nim player, you may spot that the top three rows 8, 7, 6, equivalent to ∗2, ∗3, ∗1, sum to a losing position, and so do the bottom three rows 3, 2, 1 that are equivalent to ∗2, ∗6, ∗4, and of course row 4 which is equivalent to ∗0. So one winning move is to turn row 7 or row 5, either of which is equivalent to ∗3, to ∗0, by moving e5 to b5, or h7 to e7.

We now look systematically at all possible winning moves. First of all, the entire game, call it 𝐺, is a sum of eight independent Poker-Nim heaps. That is, 𝐺 is the game-sum 𝐺 = ∗2 + ∗3 + ∗1 + ∗3 + ∗0 + ∗2 + ∗6 + ∗4 where we saw that 𝐺 ≡ ∗3 and therefore 𝐺 + ∗3 ≡ 0. A winning move is therefore to add the Nim-sum ∗3 to 𝐺, which we do by adding ∗3 to any one of the eight Poker-Nim heaps above. We then have to check if the resulting new heap size is a feasible move: row 8 :

∗2 + ∗3 ≡ ∗1

row 7 :

∗3 + ∗3 ≡ ∗0

row 6 :

∗1 + ∗3 ≡ ∗2

row 5 :

∗3 + ∗3 ≡ ∗0

row 4 :

∗0 + ∗3 ≡ ∗3

row 3 :

∗2 + ∗3 ≡ ∗1

row 2 :

∗6 + ∗3 ≡ ∗5

row 1 :

∗4 + ∗3 ≡ ∗7

These equivalent Nim heap are computed by splitting the heaps into distinct powers of two. For example, in row 2 we have, in more detail, ∗6+∗3 ≡ ∗4+∗2+∗2+∗1 ≡ ∗4+∗1 ≡ ∗5.

The above moves are then indeed winning: For example, reducing in row 8 the gap of size 2 to size 1 (in terms of Nim heaps: ∗2, to which we added ∗3, becomes ∗1) reaches the overall position ∗1 + ∗3 + ∗1 + ∗3 + ∗0 + ∗2 + ∗6 + ∗4

8

which is equivalent to ∗0 and a losing position. Similarly, the above moves in rows 7, 5, 3, and 2 reduce the gap between the counters to obtain a smaller Nim-heap. In the other rows, the gap needs to be increased: •

The move in row 6 from ∗1 to ∗2 is a Poker-Nim move because it increases the gap from 1 to 2, where the white counter is moved from b6 to a6.

•

The move in row 4 from ∗0 to ∗3 increases the gap from 0 to 3, where the white counter is moved from c4 to f4. This move would not be possible for Black. Nevertheless, the game can be treated as an impartial game because the winning moves that reduce a gap between the counters (that is, reduce the respective Nim heap) are available to either player; the Poker-Nim moves are not needed in order to win (they never are, which is the whole point of the mex rule).

•

The required move in row 1 from ∗4 to ∗7 would need to increase the gap from 4 to 7, but this not possible for either player because neither counter can be moved three squares outwards. Hence, there is no winning move in this bottom row. With White to start this is the only row where White does not have a winning move.

This game is known as Northcott’s game. There are Java applets on the internet where the game could be played interactively, such as http://www.cut-the-knot.org/ recurrence/Northcott.shtml , but these no longer work on modern browsers.

Solution to Exercise 1.9 (a) This game fulfills the ending condition because no node can be visited again, because there are no “cycles” in the digraph. (b) Because the counters are moved independently, this is a sum of games, each corresponding to a counter. The nodes represent positions of a single game with one counter only. The options of a node are simply the nodes that the counter can move to. Every node is equivalent to a Nim heap. Nodes L and N have no successor node, so these are nodes from which a counter cannot be moved any more. Consequently, they are equivalent to the empty Nim heap ∗0. The nodes that have only L or N as successor are H, P, and M. They get Nim value 1 as they are equivalent to ∗1. Next, those nodes get Nim values that only have successors that already have Nim values, computed as the mex of those values. For example, D has only H as successor, which has Nim value 1, so D gets mex(1) = 0 as its Nim value. Node O gets mex(0, 1) = 2 as it Nim value, then K gets 3 which is mex(0, 1, 2). This continues in this fashion. The Nim values of all nodes are given as below.

In order to determine if the sum of games determined by the counters is winning or losing, we have to take the Nim sum of the respective Nim values. We can ignore the two counters on H because they cancel each other. The resulting Nim sum for the counters on A, F, G, K is 2 ⊕ 3 ⊕ 4 ⊕ 3 = 2 ⊕ 4 = 6. This is nonzero, so this is a winning position and the first player to move wins. There is one winning move, from G to J, because the counter on G has Nim value 4, the highest power of two in the Nim sum.

The only other potential alternative would be to obtain a second 4 in the Nim sum to

9

D

0

H

C

1 B

0

L

4

F

A

2

1 G

0 P

K

3 E

3

J

1 O

2

1 I

2 N

0

0 M

1

cancel the existing 4 of the counter on G by moving from F to G, but this would not give a Nim sum of 0, so moving from G to J is the only winning move. (c) When the arrow from J to K is reversed so that it points from K to J instead, the Nim value of J, which is 2, does not change, because 2 is still the mex of {0, 1}, which is the same as the mex of the larger set {0, 1, 3} when K is also a successor. Similarly, the Nim value of K, which is mex{0, 1, 2} = 3, does not change, because the additional successor J has value 2, which is already in the set {0, 1, 2}. Because all Nim values remain the same, the computation in (b) applies as well and the winning move is as before, so nothing changes.

Solution to Exercise 1.10 In order to represent Chomp in the normal play convention where the last player to move wins, we remove the top left dot, the “poisoned cookie”, indicated by a small circle in the picture below. Moreover, we replace the dots by squares. By the chomp rule, a move removes a square and with it all squares to the right and below it. The picture below describes all possible positions of the game. The starting position is A.

A

B

C

D

F

G

H

I

J

K

L

M

10

E

Positions B–E correspond to the options of A by removing a a square in the bottom row (and all squares to its right), and positions F, G, H to a move in the top row. From F, options J, K, and L are reached by removing a square in the bottom row; the other options of F are G and H. The options of G are positions H, I and M. The simpler games have obvious Nim values, partly because they represent a Nim heap, like B which is ∗3, or because they are sums of Nim heaps, like C which is ∗1 + ∗3.

In that way, the following Nim values are computed directly: B ≡ ∗3, C ≡ ∗1 + ∗3 ≡ ∗2, H ≡ I ≡ ∗1, J ≡ ∗2, K ≡ ∗1 + ∗2 ≡ ∗3, M ≡ ∗1 + ∗1 ≡ ∗0.

Now to the more complicated pictures. The options of G are H, I, and M, and mex(1, 1, 0) = 2, so G ≡ ∗2. The options of L are G, H, J, K, with mex(2, 1, 2, 3) = 0, so L ≡ ∗0, which agrees with the losing positions found in Exercise 1.5(a). The options of F are G, H, J, K, L with mex(2, 1, 2, 3, 0) = 4, so F ≡ ∗4. The options of D are B, C, G, H, L with mex(3, 2, 2, 1, 0) = 4, so D ≡ ∗4.

Position E should be losing according to Exercise 1.5(a), but let’s compute its Nim value. The options of E are B, C, D, F, G, H with mex(3, 2, 4, 4, 2, 1) = 0, so indeed E ≡ ∗0.

Finally, the options of A are B, C, D, E, F, G, H with mex(3, 2, 4, 0, 4, 2, 1) = 5, so A ≡ ∗5.

The Nim heap of size four cannot be increased to a heap of size five. So the only possible winning moves in the game A + ∗4 are to move to an option of A that has Nim value 4. The two options with that property are D and F. Solution to Exercise 1.11 (a) Cram on a 2 × 3 board

has, up to symmetry, the options

,

, and

. The first two options are clearly losing positions because exactly two more dominos will be placed subsequently. The last option has the options

and

with Nim values 0 and 1, respectively, so its Nim value is mex(0, 1) = 2. The Nim-value of the 2 × 3 board is therefore mex(0, 0, 2) = 1. (b) The Nim-value of 1 × 4 Cram is 2, like that of the 𝐿-shape in (a). So the game-sum + is equivalent to ∗1 + ∗2 (the sum of a Nim-heap of size 1 and of size 2) and is therefore winning. One obvious winning move is to reduce the second Nim heap to size 1, which means placing a domino at one end of the 1 × 4 board (not in the middle), which means moving to the position

+

. The other is to

increase the first Nim heap to size 2 because that was one of the options of the 2 × 3 board, which is the move to

+

.

Solution to Exercise 1.12 (a) A split-nim heap of size 1 cannot be split, so it is the same as an ordinary Nim heap of size 1, with Nim value 1.

A split-nim heap of size 2 has the additional option of being split into two equal-sized heaps of size 1, which is a losing position with Nim value 0. So the split-nim heap of

11

size 2 still has Nim value mex(0, 1, 0) = 2.

A split-nim heap of size 3 has the additional option of being split into two split-nim heaps of size 1 and 2, respectively. These have Nim-values 1 and 2 as seen before and Nim-sum 1 ⊕ 2 = 3. So because of this additional option, the split-nim heap of size 3 has Nim value mex(0, 1, 2, 3) = 4.

A split-nim heap of size 4 has two additional options of being split, namely into two split-nim heaps of size 1 and 3, and into two equal-sized heaps of size 2; the latter have Nim-sum zero and do not give any new option. The two split-nim heaps of size 1 and 3 have Nim values 1 and 4 (the latter is important!), so their Nim-sum is 1 ⊕ 4 = 5.

So the split-nim heap of size 4 has Nim value mex(0, 1, 2, 4, 5) = 3; the entry 4 here is the Nim value of the split-nim heap of size 3. (b) We compute the Nim-sum of the split-nim heaps of sizes 1, 2, 3, which is 1 ⊕ 2 ⊕ 4 = 7.

So this is a winning position. The move has to be made in the split-nim heap with Nim-value 4, because the heaps of size 1 and 2 do not have any options with higher Nim values. We have to obtain a Nim-value of 3, which is done by splitting the heap of size 3 into two heaps of sizes 1 and 2 (which creates an obvious losing position with heaps of sizes 1, 2, 1, 2). At this point, it should become clear that the split-nim heap of size 3 cannot have Nim value 3. (c) The Nim-sum of the split-nim heaps of sizes 1, 2, 4 is 1 ⊕ 2 ⊕ 3 = 0. So this is a losing position, and no winning moves exist. (At this point, some earlier mistakes in (a) have a chance of being corrected, since a “winning move” would possibly be seen as failing.) (d) This statement is true. Otherwise, if there were two split-nim heaps that are equivalent but have different size, their game sum would have to be losing. However, in such a sum of two different-sized split-heaps, the larger heap could be equalised with the smaller heap, which creates a sum of two identical games (the two equal heaps) which is always losing, and this would offer a winning move, a contradction.

Solution to Exercise 1.13 (a) By the rules of Gray Hackenbush, a single stalk of 𝑛 edges is just a Nim heap with 𝑛 tokens. So the leftmost picture is ∗3 + ∗2 + ∗1, which is equivalent to ∗0, a losing game.

In the middle picture, a player can either remove one of the edges directly connected to the ground, leaving a Nim heap ∗3, or one next to it, leaving ∗1 + ∗2, which is ∗3. Because mex(3, 3) = 0, this is a losing game with Nim value 0. In the rightmost picture, the options are ∗4, ∗1 + ∗3 ≡ ∗2, or ∗2 + ∗2 ≡ ∗0. Because mex(4, 2, 0) = 1, this is equivalent to ∗1. (b) The first picture is just ∗1 + ∗1, which is ∗0. Only edges are removed, not the dots connecting them, so it does not matter that the two edges are joined on the ground.

Similarly, the second picture is ∗1 + ∗2, which is ∗3. The options of the third picture are ∗2 by removing one of the top edges, or ∗0 by removing the bottom edge. Because mex(2, 0) = 1, that picture is equivalent to ∗1. In the last picture, the options are: ∗3 by removing the shorter branch, ∗1 (which is the third picture) by removing the

12

end of the longer branch, or ∗2 by removing the entire long branch, and finally ∗0 by removing the bottom edge. Because mex(3, 1, 2, 0) = 4, that picture is equivalent to ∗4. (c) In (b), two branches joined at the bottom with Nim value 0 put on top of a stem of length 1 gives a tree with Nim value 1. Two branches that together have Nim value 3 put on top of a stem of length 1 gives a tree with Nim value 4. Clearly, a single branch of length 𝑚 put on top of a stem of length 𝑛 produces a stalk of length 𝑛 + 𝑚, which is a Nim heap of size 𝑛 + 𝑚. This gives rise to the following conjecture: (*) If the Nim sum of the branches is 𝑚, and the stem has length 𝑛, then the tree obtained by putting the branches on top of the stem has Nim value 𝑛 + 𝑚.

In the example shown in (c), the Nim sum of the branches is 3 ⊕ 2 ⊕ 4 = 5, and the stem has length 2, so the resulting tree should have Nim value 7. This can be verified by examining the options of the tree, as was done in (b). We now give the general proof of (*), which proceeds exactly in that fashion. First, glue the branches together at the bottom, as in the following picture.

its *3 option:

By assumption in (*), the Nim sum of these branches is 𝑚, and hence all its options are equivalent to one of ∗0, ∗1, . . . , ∗(𝑚 − 1) (as far as relevant for applying the mex rule, there may also be options with Nim value 𝑘 for 𝑘 > 𝑚). The picture above shows an option with Nim value ∗3. Each such option ∗𝑘, for 0 ≤ 𝑘 < 𝑚, can be replaced by a single stalk of length 𝑘, as shown on the right above. When the branches are put on top of the stem (which has length 𝑛), these options, resulting from chopping off some edge in one of the branches, are equivalent to stalks of length 𝑛 + 𝑘 for 0 ≤ 𝑘 < 𝑚 because they are extended by the stem length. So the tree clearly has options equivalent to ∗𝑛, ∗(𝑛 + 1), . . . , ∗(𝑛 + 𝑚 − 1). It also does not have an option equivalent to ∗(𝑛 + 𝑚) because that can only result in chopping off some edge in one of the branches, not in the stem. Removing any stem edge results in a stalk of any length 0, 1, . . . , 𝑛 − 1. So the tree has all options of Nim values 0, 1, . . . , 𝑛 + 𝑚 − 1, but not 𝑛 + 𝑚, and hence is equivalent to a Nim heap of size 𝑛 + 𝑚, which proves (*).

The rightmost tree in (c) has four identical branches on top of a stem of length 3. The branches cancel out in pairs, so the Nim value of the tree is just that of the stem, which is 3.

Solution to Exercise 1.14 (a) In the position (1.32), Right should not move in the horizontal 1 × 2 board which cannot be taken away from him (with value −1) but at the top of one of the other Γ-shaped boards, and Left should reply similarly, as in

13

Right

→

Left

+

→

+

+

+

which results in the position with value 1 + (−1) which is the losing game 0. Hence, Right (as well as Left as already shown) loses as a starting player. (b) The negative −𝐺 of a Black-White Hackenbush game 𝐺 is the “White-Black” Hackenbush game that results from the original game by exchanging the colors. Example:

𝐺 =

= 𝑊, 𝐵

−𝐺 =

= 𝐵 . 𝑊

Then 𝐺 + (−𝐺) ≡ 0 because for any edge removed in 𝐺 in this game sum the corresponding edge of opposite color can be removed in −𝐺 and vice versa, which results again in a game sum of the form 𝐺 𝑖 + (−𝐺 𝑖 ) where 𝐺 𝑖 is an option of 𝐺. (c) A single black edge, written as 𝐵, has exactly one move for Left and none for Right, so it is given by the game { 0 | } which has value 1 (Left has exactly one safe move). A 𝐵 game with value 3 is a Hackenbush stalk 𝐵 that consists of three black edges. Then 𝐵 𝐵 the options of Left are 𝐵 , 𝐵, and 0, but the best option for Left is the first of these (removing the top black edge) which dominates the others. Right has no options.

That is, n o 𝐵 𝐵 = 𝐵 . 𝐵 𝐵 𝐵 𝐵 In turn, 𝐵 = {𝐵 | } = { 1 | } = 2 and therefore, as claimed, 𝐵 = { 2 | } = 3 . 𝐵 𝑊 (d) The first game 𝐵 has only one move per player and is a winning position for Left, so it should be a positive number but worth less than one move for Left because Right still has a move. This game is given by { 0 | 1 } which is the same as the Domineering position which has value 12 , by the same reasoning as for (1.32) or by the more general formula (1.34) for 𝑝 = 1.

The next position should be worth less than two moves to Left but more than one 𝑊 move, that is, be a number between 1 and 2. Indeed, 𝐵 = {1 | 2} because Left should 𝐵 clearly only take the topmost black edge, and {1 | 2} = 32 according to (1.36) for 𝑚 = 3, 𝑝 = 1.

The final position is better for Right than the first one, but still a win for Left and should therefore be a positive number. We have 𝑊 1 1 𝑊 = { 0 | 𝑊} = { 0 | } = 𝐵 2 4 𝐵 by (1.34) for 𝑝 = 1.

14

(e) By (1.36) we have 74 = stalk with options

3 2

𝑊 3 𝐵 . Now, a 2 and from part (d) we know = 4 2 2 𝐵 𝐵 𝑊 2 is given by 𝐵 . 𝐵

6

8 4

=

3

Solution to Exercise 1.15 (a) Left wins a game when moving second if and only if the outcome class of the game is ℒ or 𝒫. If 𝐺 ∈ 𝒫, then 𝐺 + 𝐻 has the same outcome class as 𝐻 where Left wins when moving second, and similarly if 𝐻 ∈ 𝒫, so we only have to consider the case that both 𝐺 and 𝐻 are in ℒ. This means that any right option 𝐺 𝑅 of 𝐺 is in ℒ or 𝒩 because Left wins in 𝐺 𝑅 after the move of Right to 𝐺 𝑅 . Consequently, Left has a counter-move to some game (𝐺 𝑅 )𝐿 which is in 𝒫 or ℒ. Hence, if Right moves in 𝐺 + 𝐻 to 𝐺 𝑅 + 𝐻, then Left counters by moving to (𝐺 𝑅 )𝐿 + 𝐻, which is equivalent to 𝐻 if (𝐺 𝑅 )𝐿 ∈ 𝒫, or again a game sum of two games in ℒ, which is a position where Left wins moving second by induction. The same applies if Right moves from 𝐺 + 𝐻 to 𝐺 + 𝐻 𝑅 for some right option 𝐻 𝑅 of 𝐻. (b) Left wins when moving first in a game if the game is in ℒ or 𝒩. If the games 𝐺 and 𝐻 were both numbers, then they would be positive numbers, and so would be their game sum 𝐺 + 𝐻. If the games 𝐺 and 𝐻 were both impartial, then so would be their game sum, which would be either in 𝒩 where Left wins by moving first, or in 𝒫 where Left wins by moving second.

Hence, it is a good idea to look for “hot” games that are won by the first player, but whose game sum is better for Right. Consider the Domineering positions 𝐺 = . In both 𝐺 and 𝐻, Left wins by moving first. However, 𝐺 + 𝐻 belongs

and 𝐻 =

to ℛ. Namely, if Left moves first from 𝐺 + 𝐻 to moving to +

+

, then Right counters by

, which is a win for Right. If Left moves first from 𝐺 + 𝐻 to

+

, then Right counters by moving to

+

, which is also a win for Right.

The latter position is also reached with Right moving first and Left moving second, where Right wins again. Another example is 𝐺 = 𝐻 = moves becomes

+

, where 𝐺 + 𝐻 after two

.

Solution to Exercise 2.1 (a) An equilibrium can be found by starting with some flow (distribution of users) and changing users who can improve their cost. A natural starting point is an equal split

15

with 𝑥 = 𝑦 = 5. Following the hint, 𝑧 of the 𝑦 users who arrive at node 𝑣 will use the bottom edge that starts at 𝑣, where the only possibilities are 𝑧 = 2 or 𝑧 = 3 because for 𝑧 ≤ 1 an extra user would have lower cost 1 or 2 rather than 3, and for 𝑧 ≥ 4 the top edge with constant cost 3 would be better.

If 𝑥 = 𝑦 = 5 and 𝑧 = 2 then the 𝑥 + 𝑧 = 7 users would incur a cost of 7 and the remaining 𝑦 − 𝑧 = 3 users a cost of 8, and neither could improve, so this is an equilibrium with average cost 7.3.

If 𝑥 = 𝑦 = 5 and 𝑧 = 3 then the 𝑥 users would incur a cost of 7 and all 𝑦 users a cost of 8, and again neither could improve, so this is an equilibrium with average cost 7.5.

If 𝑥 = 6 and 𝑦 = 4 and 𝑧 = 2, then the 𝑥 users have a cost of 8 and the 𝑧 users a cost of 6, but increasing 𝑦 and 𝑧 by 1 would raise that cost to 8. Hence, none of the 𝑥 users have an incentive to deviate. Similarly, the route of 𝑦 − 𝑧 = 2 users via the constant-cost edge has cost 7 and again does not offer an improvement for any of the 𝑥 users. The 1 resulting expected cost is 10 (6 · 8 + 2 · 6 + 2 · 7) = 7.4.

If 𝑥 = 6 and 𝑦 = 4 and 𝑧 = 3, then this is another equilibrium with 𝑦 = 4 users on the top route with cost 7 and 𝑥 = 6 on the bottom edge with cost 8 and expected cost 7.6.

For 𝑥 ≥ 7 the bottom edge has cost at least 9 but the top route at most cost 6 which would be increased to 7 with another user, and one of the 𝑥 users would change.

Similarly, 𝑥 ≤ 4 means a cost of at most 6 on the bottom edge and at least 𝑦+𝑧 ≥ 6+2 = 8 on the top route, with an incentive for any one of the 𝑦 users to move to the bottom edge. So these are not part of an equilibrium and only the described cases for 𝑥 = 5, 6 and 𝑧 = 2, 3 are equilibria. (b) With splittable flow, the edge with 𝑧 users will be filled to 𝑧 = 3 (if 𝑦 ≥ 3) because otherwise any fraction of users of the edge 𝑣𝑑 with constant cost 3 could improve their cost by switching, with a slight increase of 𝑧. So the top route has always cost 𝑦 + 3 and the bottom edge cost 𝑥 + 2 which have to be equal in a flow equilibrium. Because 𝑥 + 𝑦 = 10, this means 𝑥 = 5.5 and 𝑦 = 4.5 (and 𝑧 = 3). The resulting cost is 7.5 for all users. This is worse than in the two non-splittable equilibria in (a) where 𝑧 = 2.

Solution to Exercise 2.2 This is the example from the original paper by Braess (1968). (a) Assume that 𝑦 users take the bottom route from 𝑜 to 𝑑 via 𝑢, so that 6 − 𝑦 users take the top route via 𝑣. The cost for the bottom route is 10𝑦 + 50 + 𝑦, which is 11𝑦 + 50, and for the top route it is 50 + 6 − 𝑦 + 10(6 − 𝑦) = 116 − 11𝑦. For 𝑦 = 3 both costs are equal (to 83), which certainly defines an equilibrium because any user would only encounter higher costs by switching to a different route. Does 𝑦 = 2 also define equilibrium?

Then the top route has cost 116 − 11𝑦 = 94, so that one of the 4 users on that top route who switches to the bottom route would create a flow of 3 on the bottom route, which has the lower cost of 83 as already computed, which is less. So this is not equilibrium.

Neither would be any other value of 𝑦. So the equilibrium is unique. (b) As suggested, we consider the right network and first assume that there are 3 users each on the top and on the bottom route. As mentioned at the end of Section 2.5, we

16

let one user at a time improve her cost until no further improvement is possible, which will lead to an equilibrium. Suppose that there are 𝑧 users on the new, longer route 𝑜𝑢𝑣𝑑, currently 𝑧 = 0. When one of the 3 users on the top route, who currently has cost 83, switches to this new route, the flow on edge 𝑜𝑢 increases from 3 to 4, the flow on edge 𝑢𝑣 from 0 to 1, and the flow on the edge 𝑣𝑑 stays unchanged. That new route has cost 40 + 11 + 30 = 81, which is less than 83, so that user switches to the new route 𝑜𝑢𝑣𝑑. We then have 𝑧 = 1, which means one user on the route 𝑜𝑢𝑣𝑑, two users on the top route 𝑜𝑣𝑑, and three users on the bottom route 𝑜𝑢𝑑. At this point, the loads on the edges are as follows: edge

number of users

cost

𝑜𝑣

2

52

𝑜𝑢

4

40

𝑢𝑣

1

11

𝑣𝑑

3

30

𝑢𝑑

3

53

Next, another user who can improve is one of the 3 users of edge 𝑢𝑑 who currently pays 53 on that edge (plus the cost of 40 for using the edge 𝑜𝑢), by using the edge 𝑢𝑣 and then 𝑣𝑑 instead. Then 𝑢𝑣 has 2 users with cost 12 and 𝑣𝑑 has 4 users with cost 40.

That new cost is 52, less than 53, hence that user switches as well by changing her route from 𝑜𝑢𝑑 to 𝑜𝑢𝑣𝑑. Then 𝑧 = 2. We now have the following usage and costs: edge

number of users

cost

𝑜𝑣

2

52

𝑜𝑢

4

40

𝑢𝑣

2

12

𝑣𝑑

4

40

𝑢𝑑

2

52

so that all 3 routes have cost 92. This is an equilibrium because no user can find a better route. The equilibrium cost in the left network is 83 and in the right network is 92, which is more costly (by over 10 percent) despite the fact that the right network has higher capacity due to the extra edge 𝑢𝑣.

Solution to Exercise 2.3 We first consider the obvious choice that every user takes the direct route via one edge to their destination. These are exactly the four edges with cost function 𝑐(𝑥) = 𝑥, so every user has cost 1. Any other route any user may take has at least one edge with that cost function (nobody wants only to travel from 𝑣 to 𝑢 or from 𝑤 to 𝑢 which would be a zero-cost edge), so no user can ever pay less than 1, so this is both an equilibrium and the optimal flow.

17

Next, suppose every user takes their indirect route via two edges. This creates the following loads: edge

which users

cost

𝑢𝑣

2, 4

2

𝑣𝑢

3

0

𝑢𝑤

1, 3

2

𝑤𝑢

4

0

𝑣𝑤

2

1

𝑤𝑣

1

1

so that users 1 and 2 have cost 3, and users 3 and 4 have cost 2. The average cost per user is 2.5. This is an equilibrium for the following reason: User 1 currently has cost 3 and by switching to the direct route 𝑢𝑣 would become the third user on that edge 𝑢𝑣 and thus pay the same cost, without improvement. The same holds for user 2 who is in a symmetric situation to user 1. User 3 currently pays 2 and by switching to the direct route 𝑣𝑤 would become the second user on that edge and can also not improve. The same holds for user 4 (symmetrically to user 3).

Solution to Exercise 3.1 (a) Both 𝑇 and 𝐵 weakly dominate strategy 𝑀 of player I. For player II, strategy 𝑟 weakly dominates 𝑙. (b) The diagram in Figure 3.13 gives an overview of all possible orders of eliminating weakly dominated strategies. One iterated elimination of weakly dominated strategies is to first eliminate 𝑀, then 𝑟, then 𝑇, then 𝑙, which terminates at the strategy pair (𝐵, 𝑐).

In contrast, eliminating 𝑀, 𝑟, 𝑙 in that order leaves two strategies 𝑇 and 𝐵 with equal payoffs to player I, so neither of the two can be eliminated further. (c) Another possible sequence of iterated elimination of weakly dominated strategies is to first eliminate 𝑙, then 𝐵, then 𝑐, then 𝑀, which terminates at the strategy pair (𝑇, 𝑟). (d) The equilibria of the game are (𝑇, 𝑐), (𝑇, 𝑟), (𝐵, 𝑐). Note that (𝑇, 𝑐) is not among the strategy pairs found in (b) and (c) above.

Solution to Exercise 3.2 (a) Only 𝐵 strictly dominates 𝑇. Other dominance relations arise only after elimination of 𝐵 as in (b), and do not count. (b) After eliminating 𝑇, player III’s 𝐿 strictly dominates 𝑅. After eliminating 𝑅, player II’s 𝑙 strictly dominates 𝑟. The only remaining profile is (𝐵, 𝑙, 𝐿), which is the sole equilibrium of the game.

Solution to Exercise 3.3 The payoffs to the players are:

18

l

c

r

1

1

0

T 1 3 1 1 0 1 M 1 3 0 B 2 l 0

c

r

1

1

2

3

3

2

1

1

B 3

3

0

2

r

c

r

1

T 3 1 3 2 B 3 0

1

1

T 3 1 0 1 M 3 0

1

1

c 3

3

c B 3

r

c

0

B 2

c

c

T 1 3 2 3 B 2 3 l

2

T 3 1 0 1 M 3 0

T 1 3 1 2 3 2 B 2 3 0 l

0

3

r c

c

1

T 3

T 3 3 B 3

r 1

1

1

1

T 1 1 M 0 r T 1

1

Figure 3.13 All possible sequences of eliminating weakly dominated strategies in the game in Exercise 3.1.

payoff to I :

𝑎(𝑥, 𝑦) = 𝑥 · (12 − 𝑦 − 𝑥) − 𝑥 = 𝑥 · (11 − 𝑦 − 𝑥),

payoff to II :

𝑏(𝑥, 𝑦) = 𝑦 · (12 − 𝑥 − 𝑦) − 2𝑦 = 𝑦 · (10 − 𝑥 − 𝑦).

There are two ways of interpreting the question. Both are legitimate and answering either is fine, and answering both is excellent, in particular when all three equilibria of the second version of the game are found. The first, intended interpretation is that 𝑥 and 𝑦 are divisible quantities of the produced good (like mineral water, Cournot’s original example), that can be any real number. The second interpretation is that 𝑥 and 𝑦 are integer quantities where a unit of the produced good, like a bicycle, is not divisible. However, there is no reason to assume these integers are only 0, 3, 4, 6 as in the example in Figure 3.8, which are illustrative values that coincide with certain solutions to the continuous version of the game.

11−𝑦

First, let 𝑥 and 𝑦 be nonnegative reals. The best response of player I to 𝑦 is max 2 , 0 because negative amounts of production are not allowed. However, player II would not

19

choose 𝑦 > 11 because then her own payoff would be negative, and then the best response 𝑥

to 𝑦 is 𝑥 = 5.5 − 𝑦/2. Similarly, the best response of player II to 𝑥 is max 10−𝑥 2 , 0 . However, any best response 𝑥 to 𝑦 is at most 5.5, so then we can assume that the best response 𝑦 to 𝑥 is just 𝑦 = 10−𝑥 2 = 5 − 𝑥/2.

An equilibrium is a pair of mutual best responses, solved e.g. via 𝑥=

11 − 𝑦 11 − (5 − 𝑥/2) 𝑥 = =3+ , 2 2 4

3 4𝑥 = 3

and thus 𝑥 = 4 and therefore 𝑦 = 3, which defines the equilibrium. The payoffs are 4 · (11 − 3 − 4) = 16 to player I and 3 · (10 − 4 − 3) = 9 to player II. The intersecting best-response lines are shown in the following diagram on the left. 𝑦

𝑦

11

11 10 9 8 7 6 5 4 3 2 1 0

𝑥 = 5.5 − 𝑦/2

5 𝑦 = 5 − 𝑥/2

3

𝑥

0 0

4 5.5

10

𝑥 0 1 2 3 4 5 6 7 8 9 10

Second, suppose 𝑥 and 𝑦 can only take integer values. If 𝑥 ∈ 𝑋 = {0, 1, . . . , 10} and 𝑦 ∈ 𝑌 = {0, 1, . . . , 11}, then the possible pairs (𝑥, 𝑦) are gridpoints, shown in the right diagram above. The real-valued best responses 𝑥 = 5.5 − 𝑦/2 and 𝑦 = 5 − 𝑥/2 are certainly best responses if they are integral, which happens when 𝑦 is odd and 𝑥 is even. If 𝑦 is even, then 5.5 − 𝑦/2 is of the form 𝑘 + 12 for some integer 𝑘, and it can be verified that both 𝑘 and 𝑘 + 1 are best responses in 𝑋 to 𝑦, due to the quadratic payoff function with its apex 2 at 𝑘 + 12 : Suppose 𝑦 = 2𝑚, so that 𝑎(𝑥, 𝑦) = 𝑥(11 − 2𝑚 − 𝑥) = −(𝑥 − 11−2𝑚 2 ) + (11−2𝑚)2

(11−2𝑚)2 = 4

2 −(𝑥 − 11 . Because 𝑥 and 𝑚 are integers, this term is minimal if and only if 2 − 𝑚) + 4 11 1 𝑥 − 2 − 𝑚 = ± 2 , which are the described two choices for 𝑥. The best response would be unique if 11 2 was replaced by a slightly different value (due to a slightly different unit cost of production), so the non-uniqueness is a bit of a coincidence. The pairs (𝑥, 𝑦) in 𝑋 × 𝑌 where 𝑥 is such a best response to 𝑦 (with two choices for 𝑥 if 𝑦 is even) are indicated with diamonds in the right diagram above. Similarly, we consider player II’s best response to 𝑥 when 𝑥 is odd. Then 5 − 𝑥/2 is of the form 𝑘 + 21 for some integer 𝑘, and again both 𝑘 and 𝑘 + 1 are best responses in 𝑌 to 𝑥. The corresponding pairs (𝑥, 𝑦) are shown as black disks in the right diagram above. The entire diagram is shorthand for a strategic-form payoff table, which would be very big to draw up. This game with integer quantities 𝑥 and 𝑦 has three equilibria (𝑥, 𝑦), namely (4, 3) as found before, but also (3, 4) where both players get payoff 12, and (5, 2) with payoff 20 to player I and 6 to player II.

20

Solution to Exercise 3.4 The 𝑁 players have two strategies each, and the game is symmetric. (As we will show below, we can therefore apply the reasoning in Section 3.8.) Because the game is symmetric, it only matters for a strategy profile how many people contribute and how many do not, and to specify the corresponding payoff for either strategy. In order to formulate a strategic-form game, we need to specify payoffs for each strategy profile, where we could associate payoffs 4, 3, 2, 1 to the listed possibilities (i) to (iv). However, in the following, we only argue with what a player prefers.

When nobody contributes, this defines an equilibrium of the game: Then no single player can help provide the public good because 𝐾 ≥ 2, and therefore “not contributing” is a best response, because players prefer (iii) to (iv).

From the listed preferences, if a player’s action does not affect the outcome, then she strictly prefers not contributing to contributing. However, if it does affect the outcome, then this means the public good is provided if she contributes as in case (ii), and is not provided if she does not contribute as in case (iii) which she prefers less. This happens if and only if the player is one of exactly 𝐾 players who contribute. We claim that any equilibrium (other than “nobody contributing”) is of the form where exactly 𝐾 people contribute. First, this is an equilibrium: A player who does not contribute would not gain by contributing because she prefers (i) to (ii). Any of the 𝐾 players who do contribute would not gain by not contributing because she prefers (ii) to (iii).

Any other strategy profile is not an equilibrium: Suppose that more than 𝐾 people contribute. Then each of them (on her own) would gain by not contributing because with this unilateral deviation the good is still provided, and she prefers (i) to (ii). (This is enough to show that this is not an equilibrium.) Suppose that fewer than 𝐾 contribute, so the good is not provided. Then each person who contributes would gain by not contributing because she prefers (iii) to (iv). Hence, this is not an equilibrium unless there are no contributors at all, which has been described before.

We now apply Proposition 3.7 to this game. In fact, the above reasoning is rather close to the proof of the proposition. Let the two strategies be 𝐶 for “contribute” and 𝐷 for “don’t contribute”.

Above, we showed first that the profile (𝐷, 𝐷, . . . , 𝐷) is an equilibrium. However, 𝐷 does not dominate 𝐶, because players prefer (ii) to (iii). Consider the smallest number 𝑘 of people who play 𝐷, as in the profile in (3.12) where (𝐷, . . . , 𝐷 , 𝐶, . . . , 𝐶 )

| {z } | {z } 𝑘

𝑁−𝑘

and so that 𝐶 is a best response, which means the largest number 𝑁 − 𝑘 of people who contribute. The only possibility for that is 𝐾 = 𝑁 − 𝑘, i.e. exactly 𝐾 people contribute to the public good. The reasoning is as described above. (In applying the proposition, we could also switch the roles of the strategies and let 0, 1 stand for 𝐷, 𝐶; this would mean looking for the smallest number of contributors so that not contributing is a best response; this gives the equilibrium where nobody contributes. Note that in a different game there may be less extreme additional equilibria.)

21

In summary, apart from the “nobody contributes” equilibrium, any strategy profile where exactly 𝐾 out of the 𝑁 players contribute constitutes an equilibrium, and these are all equilibria.

Solution to Exercise 4.1 (a) Recall that a strategy in a game tree of a player specifies a move for every decision node of that player. In a game, we assume a certain order of the decision nodes of the player, and list a move for each of them, to define a strategy. In the game tree given here, player I has three decision nodes. His strategies are therefore 𝐿𝐴𝐷, 𝐿𝐴𝐸, 𝐿𝐵𝐷, 𝐿𝐵𝐸, 𝐿𝐶𝐷, 𝐿𝐶𝐸, 𝑅𝐴𝐷, 𝑅𝐴𝐸, 𝑅𝐵𝐷, 𝑅𝐵𝐸, 𝑅𝐶𝐷, 𝑅𝐶𝐸.

However, in order to count the number of strategies, it suffices to multiply the number of moves for all the decision nodes of the player. So number of strategies of player I is 2 × 3 × 2 = 12, and of player II it is 2 × 2 = 4. (b) The number of reduced strategies is found by listing them. Recall that a reduced strategy in a game tree of a player specifies a move for every decision node of that player, except for those moves that are unreachable due to an own earlier move. Given the order of the decision nodes, a reduced strategy is as before a list of moves, one for each decision node, except that any move unreachable due to an own earlier move, which is now unspecified, is replaced by a star ∗. In the game here, the third decision node of player I with moves 𝐷 and 𝐸 is unreachable when making move 𝐿, and the second decision node with moves 𝐴, 𝐵, 𝐶 is unreachable after making move 𝑅. So the reduced strategies of player I are 𝐿𝐴∗, 𝐿𝐵∗, 𝐿𝐶∗, 𝑅∗𝐷, 𝑅∗𝐸.

So player I has 5 reduced strategies, and player II has 4, since for her no reduction is possible. (If the reduction leaves the strategies unchanged, strategies and reduced strategies are the same.) (c) The reduced strategic form of the game is shown below, with response payoffs in boxes. (d) The equilibria of the game in reduced strategies are the pairs are (𝐿𝐴∗, 𝑟𝑏), (𝐿𝐵∗, 𝑙𝑏), (𝐿𝐵∗, 𝑟𝑏), (𝑅∗𝐸, 𝑙𝑎), (𝑅∗𝐸, 𝑟 𝑎). (e) Recall that subgame perfect equilibria are those obtained by backward induction, and have to be specified with a complete (not reduced) strategy for each player. Of the equilibria in (a), the last two are clearly not subgame perfect, even when specifying the missing move ∗ as 𝐴, 𝐵, or 𝐶, since they prescribe the move 𝐸 for player I which is not an optimal choice for player I when deciding between 𝐷 and 𝐸.

The equilibrium (𝐿𝐴∗, 𝑟𝑏) prescribes optimal moves for the players at their decision nodes after player I has chosen 𝐿. It looks as if we cannot decide whether move 𝑏 for player II is optimal since we have not specified if player I plays 𝐷 or 𝐸, since that is move is irrelevant after player I has chosen 𝐿. However, backward induction for player I chooses the move 𝐷 and then move 𝑏 of player II. So one subgame perfect equilibrium is (𝐿𝐴𝐷, 𝑟𝑏). By the same reasoning, (𝐿𝐵𝐷, 𝑙𝑏) and (𝐿𝐵𝐷, 𝑟𝑏) are the other two subgame perfect equilibria. Note that player II can choose to play 𝑙 by backward induction if (and only if) player I chooses 𝐵.

22

I

II

𝑙𝑎

𝑙𝑏

𝑟𝑎

2

𝐿𝐴∗

2

10

3

2

2

2

4

1

1

4

4

0

0

1

3

11

2

11

1

𝑅∗𝐸

8

2

11

4

8

10

𝑅∗𝐷

9

2

10

2

9

10

𝐿𝐶∗

9

2

10

3

9

10

𝐿𝐵∗

𝑟𝑏

3

11

Solution to Exercise 4.2 (a) Player I has 2 × 2 = 4 strategies, and player II has 2 × 2 = 4 strategies. The number of reduced strategies for player I is three, shown as rows in (b) below, and four (no reduction possible) for player II. (b) Best response payoffs are put in boxes: II

𝑎𝑐

I

𝑎𝑑 2

𝐿∗

1 1 0

3 2 1

4 1

3

4

1

1

2

4 𝑅𝑇

3

5 1

𝑏𝑑 1

2

5 𝑅𝑆

𝑏𝑐

4

3 0

(c) The equilibria in reduced strategies are (𝐿∗, 𝑎𝑐), (𝐿∗, 𝑎𝑑). Using backward induction, the only subgame perfect equilibrium of the game is (𝐿𝑆, 𝑎𝑑). Backward induction is done (easily) with the help of the game tree, and produces full, unreduced strategies, which are needed to verify subgame perfection. The corresponding pair of reduced strategies is an equilibrium in the reduced strategic form. (d) For player I, 𝑅𝑆 weakly dominates 𝑅𝑇.

23

For player II, 𝑎𝑐 weakly dominates 𝑏𝑐, 𝑎𝑑 weakly dominates 𝑎𝑐, 𝑎𝑑 strictly dominates 𝑏𝑐, 𝑎𝑑 weakly dominates 𝑏𝑑, and 𝑏𝑑 weakly dominates 𝑏𝑐.

Solution to Exercise 4.3 (a) The strategic form looks as follows: II

𝑙

I

𝑟 1

𝑇 2

1 2 0

3

𝐵 4

0

As shown by the boxed entries, the game has two equilibria (𝑇, 𝑟) and (𝐵, 𝑙). The equilibrium (𝐵, 𝑙) is obtained by backward induction and hence the unique subgame perfect equilibrium (SPE). (b) Of the stated conditions (i)–(iii), (i)

is true because every game tree has an SPE obtained by backward induction.

(ii) is false, for example by changing the payoff 1 in (a) to 4, which gives the equilibrium (𝑇, 𝑟) with higher payoff to player II. (iii) is true: If 𝑎 > 𝑐, then 𝑇 is strictly dominating 𝐵, and 𝑇 is always played by player I in equilibrium so that he gets payoff 𝑎, which is also the SPE payoff when player II plays 𝑙 (and player I plays 𝑇). If 𝑎 = 𝑐, then player I also gets 𝑎 in any equilibrium, and if 𝑎 < 𝑐, then the unique SPE is (𝐵, 𝑙) with payoff 𝑐 to player I, which is the highest payoff to player I in the game, and hence at least as high as any equilibrium payoff.

Solution to Exercise 4.4 (a) Player I has one decision node with five moves, and therefore five strategies. Player II has five decision nodes with two moves each, and therefore 25 = 32 strategies. These are also the numbers of reduced strategies because no decision node of a player is preceded by an own earlier move. (b) Here, player I has four decision nodes with two moves each, and therefore 24 = 16 strategies. Player II has two decision nodes with two moves each, and therefore 22 = 4 strategies. For player II, no reduction is possible, as no decision node is preceded by an own earlier move, so player II also has four reduced strategies. For player I, however, move 𝑋 precedes the two decision nodes on the left with moves 𝑃, 𝑄 and 𝑅, 𝑆 respectively, so the four combinations of these moves, preceded by 𝑋, form one set of reduced strategies (where no move on the right has to be specified). Similarly, the moves 𝑇 and 𝑈 are preceded by move 𝑌, which in turn makes the move combinations

24

with 𝑃 or 𝑄 and 𝑅 or 𝑆 irrelevant. In total, these are six reduced strategies for player I. The list of these reduced strategies as described is 𝑋𝑃𝑅∗, 𝑋𝑃𝑆∗, 𝑋𝑄𝑅∗, 𝑋𝑄𝑆∗, 𝑌∗∗𝑇, 𝑌∗∗𝑈.

Solution to Exercise 4.5 (a) Each player moves only once, so the number of a player’s moves is the number of strategies. The number of strategy profiles is 3 × 3 × 2 = 18. (b) Every player moves only once, so the players’ strategies are the same as their moves.

We first argue directly with the game tree, first for players II and III who move last.

For player II, move 𝑟 is best, but as a strategy it does not strictly dominate the other strategies, because player I could choose 𝑀, and then all strategies of player II give the same payoff. So we only have weak dominance, namely 𝑟 weakly dominates 𝑙 and 𝑐, and 𝑐 weakly dominates 𝑙. For the same reason, for player III 𝐿 weakly dominates 𝑅.

For player I, choosing 𝑇 could result in payoffs 0 or 2, which may be worse or better than the payoff 1 that results from choosing 𝑀 and possibly from choosing 𝑅. For player 1, only 𝐵 weakly dominates 𝑀.

The following figure shows the strategic form, for which an example has been given earlier in Exercise 3.2. It is not necessary to give the strategic form, but it gives the same answer, and also applies to (c). The boxes indicate best-response payoffs.

II

𝑙

𝑐

II

𝑟

I

𝑙

𝑐

𝑟

I 𝑇

2 , 1, 3

0, 2, 4

2, 3, 1

𝑇

2, 1, 3

0, 2, 4

2, 3 , 1

𝑀

1, 2 , 3

1, 2, 3

1, 2 , 3

𝑀

1, 2 , 3

1, 2 , 3

1, 2 , 3

𝐵

1, 4 , 2

1, 4, 2

1, 4 , 2

𝐵

3 , 5 ,0

3 , 5 ,0

3 , 5 ,0

III: 𝑅

III: 𝐿

(c) The table in (b) shows that the equilibria are (𝑇, 𝑟, 𝐿), (𝑀, 𝑐, 𝐿), (𝐵, 𝑐, 𝐿). Backward induction gives the only SPE, (𝑇, 𝑟, 𝐿).

Again, the game tree itself can also be used to find any other equilibrium. The SPE is obvious, so what would be equilibria that are not subgame perfect? If player I chooses 𝑇, then player II has to respond with 𝑟, but player III could also choose 𝑅 because player III is indifferent; but then the payoff to player I for 𝐵 would be higher than with 𝑇, which is then not an equilibrium choice. For the same reason, there is in general no equilibrium where player II chooses 𝑅, because the payoff 3 to player I

25

would always induce player I to play 𝐵, and then the only best response of player III is 𝐿 and not 𝑅. Hence, if player I chooses 𝑇, then this gives only the SPE (𝑇, 𝑟, 𝐿).

Next, if player I chooses 𝑀, then both other players are indifferent. Player III has to choose 𝐿, and player II can only choose 𝑐 with payoff 0 to player I, to make sure that 𝑀 is a best response of player I. This gives the equilibrium (𝑀, 𝑐, 𝐿). Note that we do not try to decide if it is “reasonable” for player II to choose 𝑐; we only identify the equilibria of the game. Finally, if player I chooses 𝐵, then player II is indifferent, and only her choice 𝑐 sustains 𝑅 as a best response of player I. This gives the equilibrium (𝐵, 𝑐, 𝐿). While these considerations that use the tree seem lengthy, they are less mechanical than constructing the strategic form, and tend to give a better understanding of the game.

Solution to Exercise 4.6 (a) The commitment game can be represented by a game tree where player I moves first, having 𝑚 choices, and player II moves second, having 𝑛 moves each. Thus, in the commitment game, player I has 𝑚 and player II has 𝑛 𝑚 many strategies (𝑛 moves for each of the 𝑚 commitments). (b) This is false. An example is the quality game in Figure 3.2(b), where in the commitment game player I commits to the strictly dominated strategy 𝑇. (c) This is true. In an SPE, player II’s chosen move must always be a best response, which is never true for a strategy in 𝐺 that is strictly dominated.

Solution to Exercise 4.7 (a) The payoff to player II is quadratic in 𝑦 and its derivative with respect to 𝑦 is 4 + 𝑥 − 2𝑦, which is zero if and only if 𝑦 = 𝑦(𝑥) = 2 + 𝑥/2. This is a maximum because the second derivative −2 is negative. By symmetry, player I’s best response is given by 𝑥 = 𝑥(𝑦) = 2 + 𝑦/2. An equilibrium is defined by 𝑥 = 𝑥(𝑦) and 𝑦 = 𝑦(𝑥), that is, 𝑥 = 𝑦 = 4, with payoff 16 to both players. (b) In the commitment game, player I chooses 𝑥 and player II a best response 𝑦(𝑥), so player I maximizes the function 𝑥 · (4 + 𝑦(𝑥) − 𝑥) = 𝑥 · (6 − 𝑥/2). Its derivative with respect to 𝑥 is 6 − 𝑥, which is zero for 𝑥 = 6, with best response 𝑦(6) = 5. The resulting payoffs to the two players are 18 for player I and 25 for player II. (c) Yes, they are unique, as unique solutions to two independent linear equations with two unknowns for the equilibrium and as a unique solution to a maximization problem for the commitment game. (d) In an SPE, player I solves the maximization problem of maximizing his payoff, say 𝑎(𝑥, 𝑦), over 𝑥 given that player II plays her best response 𝑦 = 𝑦(𝑥). In any equilibrium (𝑥 ∗ , 𝑦 ∗ ) of 𝐺, we have 𝑦 ∗ = 𝑦(𝑥 ∗ ) because the best response is unique, so the corresponding payoff to player I certainly fulfills 𝑎(𝑥 ∗ , 𝑦 ∗ ) = 𝑎(𝑥 ∗ , 𝑦(𝑥 ∗ )) ≤ max𝑥 𝑎(𝑥, 𝑦(𝑥)), where the last term is the SPE payoff of the commitment game.

26

In essence this says: Player I could always commit to his strategy in an equilibrium and thereby get the payoff in that equilibrium, so the commitment payoff cannot be smaller.

Solution to Exercise 5.1 To get an understanding of what is going on, it is useful to experiment with small values of 𝑤. The following table shows the value of 𝑐1 + 𝑤 · 𝑐 2 for 𝑤 = 1, 2, 3, 4 and also for 𝑤 = 2.2 and 𝑤 = 2.5. The utility for the most preferred candidate is shown with a box.

candi𝑐1 𝑐2 𝑤 = 1 𝑤 = 2 𝑤 = 3 𝑤 = 4 𝑤 = 2.2 𝑤 = 2.5 date weight 1 weight 𝑤 𝑥

121

5

126

131

136

141

132

133.5

𝑦

124

4

128

132

136

140

132.8

134

𝑧

128

2

130

132

134

136

132.4

133

Clearly, with increasing weight 𝑤 the more experienced candidate 𝑥 is preferred. The rankings are already different for 𝑤 = 1, 2, 3, 4, namely 𝑥 ≺ 𝑦 ≺ 𝑧 when 𝑤 = 1, and 𝑥 ≺ 𝑦 ∼ 𝑧 when 𝑤 = 2, and 𝑧 ≺ 𝑦 ∼ 𝑥 when 𝑤 = 3, and 𝑧 ≺ 𝑦 ≺ 𝑥 when 𝑤 = 4. However, for for 𝑤 = 2 and 𝑤 = 3 these rankings are not strict as asked for. When 𝑤 = 2 then candidate 𝑦 ties with 𝑧, so for slightly larger 𝑤 candidate 𝑦 should be ahead of 𝑧 and 𝑥, with 𝑧 still ahead of 𝑥, as indeed for 𝑤 = 2.2, which gives the ranking 𝑥 ≺ 𝑧 ≺ 𝑦. When 𝑤 = 3 then candidate 𝑦 ties with 𝑥 and both are ahead of 𝑧, so for slightly smaller 𝑤 we should have the ranking 𝑥 ≺ 𝑧 ≺ 𝑦, which is true for 𝑤 = 2.5, which gives the ranking 𝑧 ≺ 𝑥 ≺ 𝑦.

Figure 5.6 shows the affine functions 𝑐1 + 𝑤 · 𝑐 2 of 𝑤 ∈ [1, 4] for the different pairs (𝑐 1 , 𝑐2 ) given by the candidates 𝑥, 𝑦, 𝑧. It shows that there cannot be more than four different strict rankings between 𝑥, 𝑦, 𝑧 for any 𝑤 : The reason is that the ranking can only change when two of the lines cross (which creates a tie between the respective two candidates), and three lines can only cross in three places, shown by the dashed lines: for 𝑤 = 2 and 𝑤 = 3, and for 𝑤 = 37 which is the solution to 121 + 5𝑤 = 128 + 2𝑤 when 𝑥 and 𝑧 are tied.

Solution to Exercise 5.2 Because 𝑥 0 ≺ 𝑥1 ≺ 𝑧, we can find a lottery 𝑄 = 𝑥0 ∧𝛾 𝑧 so that 𝑄 ∼ 𝑥 1 , that is,

(1 − 𝛾)𝑢(𝑥0 ) + 𝛾𝑢(𝑧) = 𝑢(𝑥 1 ) .

By (5.18), this holds if and only if 𝛾 = 1/𝑢(𝑧) and 𝑢(𝑧) = 1/𝛾 where 0 < 𝛾 < 1 because 𝑢(𝑧) > 𝑢(𝑥 1 ) = 1. Because 𝑄 ∼ 𝑥 1 , we have

27

141 140

136

𝑧 130

𝑦

128

𝑥

126 𝑤 1

2

Figure 5.6

2.2 73

2.5

3

4

Utility 𝑐1 + 𝑤 · 𝑐2 for 𝑥, 𝑦, 𝑧 as a function of 𝑤 in Exercise 5.1.

E(𝑣, 𝑄)

=

𝑣(𝑥 1 )

(1 − 𝛾)𝑣(𝑥 0 ) + 𝛾𝑣(𝑧)

=

𝑣(𝑥 1 )

(1 − 𝛾)(𝑎 𝑢(𝑥 0 ) + 𝑏) + 𝛾𝑣(𝑧)

=

𝑎 𝑢(𝑥 1 ) + 𝑏

(1 − 𝛾)𝑏 + 𝛾𝑣(𝑧)

=

𝑎+𝑏

𝛾𝑣(𝑧)

=

𝑎 + 𝛾𝑏

𝑣(𝑧)

=

1 𝛾 𝑎+𝑏

𝐵 = (1 − 𝛼)𝐴 + 𝛼𝐶

⇔

𝐵 = 𝐴 + 𝛼(𝐶 − 𝐴)

⇔

𝐵−𝐴 𝛼 = 𝐶−𝐴

= 𝑎 𝑢(𝑧) + 𝑏 .

Solution to Exercise 5.3 (a) Because 𝐶 − 𝐴 > 0,

28

which shows that the two equations are equivalent. Therefore, 𝐴 < 𝐵 < 𝐶, which is equivalent to 0 < 𝐵 − 𝐴 < 𝐶 − 𝐴, is equivalent to 0 < 𝛼 < 1. (b) Let 𝑃 choose the outcome 𝑥 𝑖 with probability 𝑝 𝑖 and let 𝑅 choose 𝑥 𝑖 with probability 𝑟 𝑖 , for 1 ≤ 𝑖 ≤ 𝑛. Then the lottery (1 − 𝛼)𝑃 + 𝛼𝑅 chooses 𝑥 𝑖 with probability (1 − 𝛼)𝑝 𝑖 + 𝛼𝑟 𝑖 .

Therefore,

E(𝑢, (1 − 𝛼)𝑃 + 𝛼𝑅)

=

𝑛 Õ

𝑢(𝑥 𝑖 ) (1 − 𝛼)𝑝 𝑖 + 𝛼𝑟 𝑖

𝑖=1

=

(1 − 𝛼)

𝑛 Õ 𝑖=1

=

𝑢(𝑥 𝑖 )𝑝 𝑖 + 𝛼

𝑛 Õ

𝑢(𝑥 𝑖 )𝑟 𝑖

𝑖=1

(1 − 𝛼)E(𝑢, 𝑃) + 𝛼E(𝑢, 𝑅).

Solution to Exercise 6.1 Considered as matrices, the dimensions of 𝑥, 𝑥>, 𝑦, and 𝑦> are 𝑚 × 1, 1 × 𝑚, 𝑛 × 1, and 1 × 𝑛, respectively. Hence 𝑥>𝑦 is a 1 × 1 matrix (if 𝑚 = 𝑛, otherwise 𝑥>𝑦 is not defined), the scalar product of 𝑥 and 𝑦. In contrast, 𝑥 𝑦> is an 𝑚 × 𝑛 matrix (even if 𝑚 ≠ 𝑛), with entries 𝑥 𝑖 𝑦 𝑗 in row 𝑖 and column 𝑗. This is a matrix of rank at most 1, because every column is a scalar multiple of 𝑥 (multiplied with 𝑦 𝑗 ), or equally every row is a multiple of 𝑦> (multiplied with 𝑥 𝑖 ), and every matrix of rank 1 can be written in this way. If 𝑥 or 𝑦 is the zero vector, then this is the all-zero matrix which has rank 0.

Solution to Exercise 6.2 First, it is easy to see that the game has no equilibrium in pure strategies. We first find the mixed equilibria without using the upper-envelope method, to demonstrate the method described in Section 6.6.

Because the best response to each pure strategy is unique and there is no pure-strategy equilibrium, both players must mix at least two strategies in any equilibrium. Hence, in equilibrium, player II has to mix between 𝑙 and 𝑟, and therefore has to be indifferent between between 𝑙 and 𝑟 by the best-response condition. Let (1 − 𝑞, 𝑞) be the mixed strategy of player II, 0 < 𝑞 < 1.

Let (𝑎, 𝑏, 𝑐) be the mixed strategy of player I, where he plays 𝑇, 𝑀, 𝐵 with probabilities 𝑎, 𝑏, 𝑐, respectively. Suppose player I uses only 𝑇 and 𝑀 with positive probability (𝑐 = 0), with the mixed strategy (𝑎, 𝑏, 0). Making player II indifferent means 1 · 𝑎 + 0 · 𝑏 = 0 · 𝑎 + 2 · 𝑏, or 𝑎 = 2𝑏 = 2(1 − 𝑎), that is, 𝑎 = 23 , 𝑏 = 13 . Then player II gets an expected payoff of 23 for each of her pure strategies. In turn, she chooses (1 − 𝑞, 𝑞) so as to make player I indifferent between 𝑇 and 𝑀 because both have to be pure best responses to (1 − 𝑞, 𝑞) and hence must have equal payoff. This means 6𝑞 = 2(1 − 𝑞) + 5𝑞 or 𝑞 = 2 − 2𝑞 or 3𝑞 = 2, that is, the mixed strategy is ( 13 , 23 ). The expected payoff is 4 for both 𝑇 and 𝑀, which is higher than the payoff 3 player I would get when playing 𝐵. So indeed, we have the equilibrium (( 23 , 13 , 0), ( 13 , 32 )).

Next, we try a mixed strategy (𝑎, 0, 𝑐) of player I that only uses 𝑇 and 𝐵. Equating the expected payoffs to player II gives the equation 𝑎 + 3𝑐 = 4𝑐 or 𝑎 = 𝑐, that is, the

29

mixed strategy ( 12 , 0, 12 ) where player II gets expected payoff 2. Then player II has to play (1 − 𝑞, 𝑞) so that player I is indifferent between 𝑇 and 𝐵, that is, 6𝑞 = 3 or 𝑞 = 21 , with expected payoff 3 to player I. This looks like an equilibrium but in fact is not, because the expected payoff for 𝑀 is 3.5, which is higher than 3. So 𝑇 and 𝐵 have equal payoff but are nevertheless not best responses to ( 12 , 12 ), which means we don’t have an equilibrium.

Finally, consider the case that player I plays with a mixed strategy (0, 𝑏, 𝑐) where only 𝑀 and 𝐵 have positive probability. Equal payoff to the columns of player II means 3𝑐 = 2𝑏 + 4𝑐 or 0 = 2𝑏 + 𝑐 which is clearly not possible for probabilities 𝑏, 𝑐 that sum to one. This is due to the fact that against 𝑀 and 𝐵, the best pure response is always to play 𝑟, so the indifference cannot be achieved.

There is no strategy where player I can play all three pure strategies with positive probability: In that case he would have to be indifferent between 𝑇 and 𝑀, which already determines the mixed strategy of player II uniquely as ( 31 , 23 ), but then 𝐵 gets a different payoff. So it is not possible to make player I indifferent between all three pure strategies.

The game therefore has only one mixed equilibrium, given above.

We can also argue with the following upper-envelope diagram where we plot the expected payoff to player I as a function of (1 − 𝑞, 𝑞) with 𝑞 = prob(𝑟).

payoff to I

6

𝑇

5 𝑀 𝐵 3

3

2

prob(𝑟)

0 0

1 3

2 3

1

There are only two choices for 𝑞 where player I has more than one pure best response to (1 − 𝑞, 𝑞), namely 𝑞 = 13 with best responses 𝐵 and 𝑀, and 𝑞 = 32 with best responses 𝑀 and 𝑇; the case 𝑞 = 21 with equal payoffs for 𝐵 and 𝑇 can also be seen from the diagram, but this intersection point is not on the upper envelope. We can then proceed as before and identify the possible mixtures of 𝐵 and 𝑀, or of 𝑀 and 𝑇, respectively, to make player II indifferent between 𝑙 and 𝑟. This succeeds only for 𝑀 and 𝑇, and the corresponding point for 𝑞 = 23 is marked with a white dot on the upper envelope, which defines the only equilibrium of the game.

Solution to Exercise 6.3 (a) This is true. For player I, arbitrary mixing between 𝑀 and 𝐵 is part of an equilibrium if II and III choose 𝑐 and 𝐿, respectively. For player II, mixing so that 𝑐 has probability

30

at least 12 (so the expected payoff to player I via 𝑇 is at most 1) is part of an equilibrium if III chooses 𝐿 and player I chooses 𝑀 or 𝐵 or any mixture of 𝑀 and 𝐵. For player III, mixing so that 𝐿 has probability at least 12 (so the expected payoff to player I via 𝐵 is at most 2) is part of an equilibrium if I and II choose 𝑇 and 𝑟, respectively. (b) This is true. If 𝑇 has positive probability in an equilibrium, then player II chooses 𝑟 for sure (a pure strategy). If 𝐵 has positive probability in an equilibrium, then player III chooses 𝐿 for sure (a pure strategy). Thus, if both II and III mix properly in equilibrium, this can only happen if player I chooses the pure strategy 𝑀. (It does not matter if there actually is an equilibrium where player I plays the pure strategy 𝑀.)

Solution to Exercise 6.4 Suppose we draw arrows alongside the sides of the game to indicate a player’s best response, as an alternative to the best-response boxes as in the following example where 𝐴 > 𝐶, 𝐵 < 𝐷, 𝑎 < 𝑏, 𝑐 > 𝑑. Equality like 𝐴 = 𝐶 is excluded by assumption.

II

left → right

I

𝑎 Top

𝑏 𝐵

𝐴

↑

𝑐

Bottom

𝑑

𝐶

(∗)

↓

𝐷 ←

(a) If the arrows of one player point in the same direction, that player will choose only that pure strategy in equilibrium because it strictly dominates the other strategy, and the other player will have a unique best response. This is shown in the following arrow patterns (where a double arrow stands for an arrow in either direction): ↔ ↑

↔ ↑

↓

↔

← ↓

l

↔

→ l

l

←

l →

Hence, the arrows of each player must point in opposite directions to get a mixed equilibrium. This means 𝐴 < 𝐶 and 𝐵 > 𝐷 for player I, or the other way round, 𝐴 > 𝐶 and 𝐵 < 𝐷. Independently, player II’s payoffs must fulfill either 𝑎 < 𝑏 and 𝑐 > 𝑑, or 𝑎 > 𝑏 and 𝑐 < 𝑑. The actual probabilities in the mixed equilibrium will be shown in (c) below. These give four possibilities for orderings between the payoffs in total, as in the following arrow patterns. → ↑

← ↓

←

↓

← ↑

→

↑

↓ →

31

→ ↓

↑ ←

Note that nothing is said about the relative magnitude of player I’s payoffs across columns, like 𝐴 in comparison with 𝐵 or 𝐷, or about player II’s payoffs across rows, like 𝑏 versus 𝑑. (b) Arrows in opposite direction for each player could still imply the existence of two pure-strategy equilibria as in the Battle of the Sexes game or in the game of Chicken.

In order to have the mixed-strategy equilibrium as the only equilibrium, the arrows must be circular. This is so in the two cases of clockwise pointing arrows, as in (∗) above: 𝐴 > 𝐶, 𝑏 > 𝑎, 𝐷 > 𝐵, 𝑐 > 𝑑. and counter-clockwise pointing arrows: 𝐴 < 𝐶,

𝑏 < 𝑎,

𝐷 < 𝐵,

𝑐 < 𝑑.

(c) Assume the first case in (b) of clockwise arrows as in (∗). Let 1 − 𝑝 and 𝑝 be the probabilities for Top and Bottom and 1 − 𝑞 and 𝑞 be the probabilities for left and right, respectively. Player II is indifferent between his pure strategies if and only if (1 − 𝑝)𝑎 + 𝑝𝑐 = (1 − 𝑝)𝑏 + 𝑝𝑑 or equivalently 𝑝(𝑐 − 𝑑) = (1 − 𝑝)(𝑏 − 𝑎), that is, 𝑝(𝑐 − 𝑑 + 𝑏 − 𝑎) = 𝑏 − 𝑎, so that 𝑝=

𝑏−𝑎 , 𝑐−𝑑+𝑏−𝑎

1−𝑝 =

𝑐−𝑑 . 𝑐−𝑑+𝑏−𝑎

By assumption, 𝑏 − 𝑎 and 𝑐 − 𝑑 are both positive, so that 𝑝 and 1 − 𝑝 are strictly between 0 and 1.

Similarly, player I is indifferent between Top and Bottom if and only if (1 − 𝑞)𝐴 + 𝑞𝐵 = (1 − 𝑞)𝐶 + 𝑞𝐷 or equivalently (1 − 𝑞)(𝐴 − 𝐶) = 𝑞(𝐷 − 𝐵), that is, 𝐴 − 𝐶 = 𝑞(𝐷 − 𝐵 + 𝐴 − 𝐶), so that 𝑞 = (𝐴 − 𝐶)/(𝐷 − 𝐵 + 𝐴 − 𝐶), and correspondingly 1 − 𝑞 = (𝐷 − 𝐵)/(𝐷 − 𝐵 + 𝐴 − 𝐶).

Again, note that 𝐴 − 𝐶 and 𝐷 − 𝐵, and hence 𝑞 and 1 − 𝑞, are strictly positive. (d) We have (1 − 𝑝)/𝑝 = (𝑐 − 𝑑)/(𝑏 − 𝑎), which is a quotient of positive terms by assumption.

This means that the probabilities for playing Top and Bottom are in inverse proportion to the differences of player II’s payoffs in the respective rows. The differences in each row must be taken in opposite directions, right minus left versus left minus right (otherwise player I couldn’t make player II indifferent).

Analogously, (1 − 𝑞)/𝑞 = (𝐷 − 𝐵)/(𝐴 − 𝐶), again a quotient of positive terms. This means that the probabilities for left and right must be in inverse proportion to the differences of player I’s payoffs in the left and right column (choosing signs so as to make the differences positive). (e) Adding a constant 𝐸 to the payoffs of player I in the first column does not change the difference 𝐴 − 𝐶 which is used to determine the probabilities for player II above (in fact, this holds for any column – it even holds for larger games as long as the

32

same payoff is added to all entries in a column of player I payoffs). Similarly, adding a constant to the row (or any row) of the payoffs of player II does not change the equilibrium strategy of player I. Solution to Exercise 6.5 (a) We mark the best response payoffs in this game with boxes: II

𝑎

I

𝑏

2

𝑇

0

𝑐

4

3 1

2 7

𝐵

𝑒

5 4

0

5 0

4 0

1

𝑑

3 0

1

8 0

This shows that the game has no pure-strategy equilibria. Hence, in equilibrium, player I must assign positive probability to both 𝑇 and 𝐵.

The quickest way to find the mixed strategy equilibria is to look at the upper envelope of payoffs to player II as a function of the probability 𝑝 = prob(𝐵) that player I plays 𝐵.

This is shown in the picture below.

payoff to II

5

8

𝑎

𝑑

𝑒

7

𝑐

5

𝑏

4

4

3 2

prob(𝐵)

0 0

1 5

2 5

2 3

1

(b) The diagram shows that the only cases where player II has two best responses are when these best responses are 𝑑 and 𝑏, or 𝑏 and 𝑎, or 𝑎 and 𝑒. We consider these in turn.

When player II plays only 𝑑 or 𝑏, then 𝐵 is a best response to 𝑑 and 𝑇 is a best response to 𝑏 (in other words, the 2 × 2 game where player II is restricted to the columns 𝑏 and 𝑑

33

does not have dominated strategies), so player II can indeed make player I indifferent.

She does so by playing the mixed strategy (0, 13 , 0, 23 , 0) with resulting payoff 23 to player I (using the “difference trick”, for example). Conversely, player II is indifferent between 𝑏 and 𝑑 when player I plays the mixed strategy ( 45 , 15 ). The resulting payoff is 4 for both 𝑏 and 𝑑. This gives the first mixed equilibrium.

When player II plays only 𝑎 or 𝑏, it is again possible to make player I indifferent, with the mixed strategy ( 23 , 31 , 0, 0, 0). With player I playing ( 53 , 25 ) and expected payoff 4 for both 𝑎 and 𝑏, this gives the second mixed equilibrium.

Finally, when player II plays only 𝑎 or 𝑒, player I can be made indifferent, with the mixed strategy ( 34 , 0, 0, 0, 14 ), resulting in payoff 34 for both 𝑇 and 𝐵. Then player I can play ( 13 , 23 ) with expected payoff 16 3 to player II for 𝑎 and 𝑒. This defines the third mixed equilibrium. As the upper-envelope diagram shows, there are no other possibilities that player II can play more than one pure strategy as a best response.

Note: this game is not degenerate, despite the fact that against the mixed strategy ( 21 , 12 ) of player I there are three pure strategies 𝑏, 𝑐, 𝑒 of player II that give the same payoff 4, as this is not the best payoff (𝑎 gives the higher payoff 4.5), so these three pure strategies are not best responses.

Solution to Exercise 6.6 (a) In the diagrams below, the left picture shows the upper envelope of payoffs for player I, the right picture for player II. The circle marks each player’s equilibrium strategy.

payoff to I 3

payoff to II 𝐵

2

1

𝑟

1 prob(𝑟)

0 0

𝑙

2

𝑇

1 prob(𝐵)

0

1

0

1 2

1

(b) In the left picture, 𝐵 dominates 𝑇. (c) The left endpoint of the line of expected payoffs when player I uses 𝐵 is above the left endpoint of the line when he plays 𝑇. The same holds for the right endpoints because 𝐵 dominates 𝑇. So the line for 𝑇 is always below the line for 𝐵 because this holds for the endpoints, and 𝑇 is therefore never a best response. So if this holds for any pure strategy of player II (because of the dominance), it also holds for any mixed strategy of player II. Solution to Exercise 6.7 The strategic form is as follows. Best response payoffs are put in boxes.

34

II

𝑙

I

𝑟

0

0 𝑇

−10

0

−90

0 𝐵

−1

−6

The game has one pure-strategy equilibrium (𝑇, 𝑙). Moreover, the game is degenerate, because the pure strategy 𝑇 has two pure best responses, 𝑙 and 𝑟. Because player II is indifferent between 𝑙 and 𝑟 when player I plays 𝑇, she can mix between and 𝑙 and 𝑟. This defines an equilibrium as long as 𝑇 stays a best response. Let 𝑞 be the probability that player II plays 𝑟. Then 𝑇 is a best response to the mixed strategy (1 − 𝑞, 𝑞) if and only if −10𝑞 ≥ −(1 − 𝑞) − 6𝑞 (because −10𝑞 is the expected payoff for 𝑇 and −(1 − 𝑞) − 6𝑞 = −1 − 5𝑞 is the expected payoff for 𝐵), that is, 1 ≥ 5𝑞. Obviously, 𝑞 ≥ 0. We therefore obtain the following set of equilibria: (𝑇, (1 − 𝑞, 𝑞)) where 0 ≤ 𝑞 ≤ 51 .

Furthermore, suppose there is a mixed strategy equilibrium where 𝐵 has positive probability 𝑝. Then player II’s best response to that mixed strategy (1 − 𝑝, 𝑝) of player I is always 𝑙, which gives expected payoff zero, which is larger than the negative payoff −90𝑝 when playing 𝑟. But player I’s unique best response to 𝑙 is 𝑇, so 𝐵 cannot have positive probability in any equilibrium.

It is interesting that player II’s lack of any gain in cheating reduces the inspection probability to zero (player I always plays 𝑇), but there are still equilibria where player II cheats with probability 𝑞 up to 15 .

Solution to Exercise 6.8 Let (𝑥, 𝑦) be an equilibrium in mixed strategies of the given nondegenerate game, and define the sets supp(𝑥) = { 𝑖 | 𝑥 𝑖 > 0} and supp(𝑦) = { 𝑗 | 𝑦 𝑗 > 0}, also called the supports of the mixed strategies 𝑥 and 𝑦. Moreover, let BR(𝑥) = { 𝑗 | 𝑗 is a pure best response to 𝑥} and BR(𝑦) = { 𝑖 | 𝑖 is a pure best response to 𝑦}. The best response condition (Proposition 6.1) can be written as the set inclusions supp(𝑥) ⊆ BR(𝑦) and supp(𝑦) ⊆ BR(𝑥). The nondegeneracy condition can be stated as |BR(𝑥)| ≤ |supp(𝑥)| and |BR(𝑦)| ≤ |supp(𝑦)|. This implies

|BR(𝑥)| ≤ |supp(𝑥)| ≤ |BR(𝑦)| ≤ |supp(𝑦)| ≤ |BR(𝑥)|,

so all these inequalities hold as equalities, and |BR(𝑥)| = |BR(𝑦)| as claimed. In addition, this shows that every pure best response is played with positive probability, that is, there are no unused best responses.

35

Solution to Exercise 7.1 The set 𝐶 is the unit disk, and the set 𝐺 is the upper semi-circle of the unit circle. The function 𝑔 projects every point (𝑥, 𝑦) upwards to the point on 𝐺 with the same 𝑥-coordinate.

The function ℎ rotates every point on the disk left (counter-clockwise) by a quarter turn.

After applying 𝑔, the function 𝑓 turns 𝐺 by a quarter turn left. Because 𝐶 is compact and convex and 𝑔 and ℎ and therefore 𝑓 are continuous, each function has at least one fixed point by Brouwer’s fixed point theorem.

𝑦 apply 𝑔

𝑦 then ℎ

𝐺

ℎ(𝐺) 𝑥

𝑥

𝐶 −1 √ (√ , −1 ) 2

2

The set of fixed points of 𝑔 is 𝐺. The only fixed point of ℎ is (0, 0), because (𝑥, 𝑦) = ℎ(𝑥, 𝑦) = (−𝑦, 𝑥) implies 𝑥 = −𝑦 and 𝑦 = 𝑥 and thus 𝑥 = −𝑥 and therefore 𝑥 = 𝑦 √ = 0. There 2 is exactly one fixed point of 𝑓 , which √ √ is found as follows: (𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) = ℎ(𝑥, 1 − 𝑥 ) = 2 (− 1 − 𝑥 2 , 𝑥) is equivalent to 𝑥 = − 1 − 𝑥 2 and 𝑦 = 𝑥. The first equation implies 𝑥 = 1− 𝑥 2 −1 −1 √ and thus 𝑥 = ± √1 and thus 𝑥 = √ because 𝑥 < 0. Indeed, ( √ , −1 ) is the unique fixed 2 2 2 2 point of 𝑓 .

Solution to Exercise 7.2 (a) Suppose 𝑣 𝑖 was equal to the convex combination of some other points 𝑝 1 , . . . , 𝑝 𝑘 of 𝑆.

Each of these is a convex combination of 𝑣 1 , . . . , 𝑣 𝑑 (including 𝑣 𝑖 ) and therefore 𝑣 𝑖 is a convex combination of 𝑣 1 , . . . , 𝑣 𝑑 , 𝑣𝑖 =

𝑑 Õ

𝑣 𝑗𝜆𝑗

(∗)

𝑗=1

where 𝜆 𝑗 ≥ 0 and 𝑑𝑗=1 𝜆 𝑗 = 1, and 𝜆 𝑖 < 1 because otherwise each of the points 𝑝 1 , . . . , 𝑝 𝑘 is equal to 𝑣 𝑖 . Subtracting 𝑣 𝑖 𝜆 𝑖 from both sides in (∗) and multiplication with 1/(1 − 𝜆 𝑖 ) gives 𝑑 Õ 𝜆𝑗 𝑣𝑖 = 𝑣𝑗 1 − 𝜆𝑖

Í

𝑗=1, 𝑗≠𝑖

which shows that 𝑣 𝑖 is a convex and hence affine combination of 𝑣 1 , . . . , 𝑣 𝑖−1 , 𝑣 𝑖+1 , . . . , 𝑣 𝑑 which contradicts the affine independence of 𝑣 1 , . . . , 𝑣 𝑑 . (b) Let 𝑆 be a simplex of dimension 𝑑 − 1, given as the convex hull of two sets 𝐴 = {𝑣 1 , . . . , 𝑣 𝑑 } and 𝐵 = {𝑤 1 , . . . , 𝑤 𝑑 } of affinely independent points (their cardinality 𝑑

36

is determined by the dimension). If 𝑣 𝑖 ∉ 𝐵, then 𝑣 𝑖 is nevertheless a convex combination of 𝑤 1 , . . . , 𝑣 𝑑 and therefore not an extreme point of 𝑆, which contradicts (a). This shows 𝐴 ⊆ 𝐵, and analogously 𝐵 ⊆ 𝐴, and therefore 𝐴 = 𝐵.

Solution to Exercise 7.3 (a) An example is given by 𝑆, 𝑇 ⊂ R2 where 𝑆 has vertices (0, 0), (3, 0), and (0, 1) and 𝑇 has vertices (0, 0), (2, 0), and (2, 1). Then 𝑆 ∩ 𝑇 is a quadrilateral and not a triangle:

𝑇

𝑆

(b) The set R is convex but has no extreme points: any 𝑥 ∈ R can be represented as the convex combination (𝑥 − 𝜀) 12 + (𝑥 + 𝜀) 21 for some 𝜀 > 0. The same holds for any open interval in R. (c) The half-open interval [0, 1) is bounded and has the only extreme point 0, but is not the convex hull of its extreme points.

Solution to Exercise 7.4 Clearly, the Sperner condition has to be violated, which requires 𝑑 ≥ 3, and more than one sub-simplex in the triangulation. The following shows a corresponding simple counterexample for 𝑑 = 3, and an even simpler one on the right.

3

3

1

3

3

3

2

1

3

2

Solution to Exercise 8.1 (a) The following best responses show that the game does not have an equilibrium in pure strategies:

37

payoff to Max, = cost to min min Max 𝐿

𝑅

𝑙

𝑤

𝑟

0.6

0.7

1.0

1.0

0.8

1.0

1.0 𝑤

0.7 0.6

0.8 0.7

𝑟

𝑙

0.7

𝑝

0 0

𝑝ˆ = 34

1

An equilibrium strategy of player I is a max-min strategy, with maximum worst-case expected payoff to player I, given by the respective best response of player II. The above goalpost diagram shows the expected payoff to player I if the strategy of player II is 𝑙, 𝑤, or 𝑟. For a max-min strategy of player I, one has to consider the minimum of his expected payoffs (the lower envelope), which is shown as the bold line.

The picture shows that the max-min payoff is 0.775 when 𝑝 = prob(𝑅) = 34 , namely when player I receives equal expected payoff when player II plays 𝑟 or 𝑤. This equality, namely the equation 1(1 − 𝑝) + 0.7𝑝 = 0.7(1 − 𝑝) + 0.8𝑝, determines 𝑝 = 43 .

A quick way of finding the probabilities for 𝐿 and 𝑅 is given by the difference trick: In row 𝐿, the difference of the matrix entries for 𝑤 and 𝑟 is 0.3. In row 𝑅, the difference is 0.1. Hence, 𝐿 and 𝑅 have to be played with inversely proportional probabilities, namely ratio 1 to 3, that is, 𝐿 has probability 14 and 𝑅 has probability 34 . The equal resulting cost of 0.775 for 𝑤 and 𝑟 is easily verified.

Furthermore, the strategy 𝑙 of player II has expected cost 0.9, and thus will not be played by II as a best response.

In order to get an equilibrium, player II has to make player I indifferent between 𝐿 and 𝑅 by assigning suitable probabilities to 𝑤 and 𝑟. Again, we use the difference trick: The difference among the payoffs for column 𝑤 is 0.1, whereas for column 𝑟 it is 0.3. Hence 𝑤 and 𝑟 have probabilities 34 and 14 , respectively, giving player I expected payoff 0.775 in each case. Because only best responses have positive probability, this is an equilibrium.

This is a zero-sum game, so the equilibrium payoff is the unique value of the game. (b) Now player I has 𝑀 as an extra strategy. We first try the solution obtained so far.

It works if player II can still keep his min-max cost of 0.775 by using the above probabilities (0, 43 , 14 ) even when player I has the extra possible response. Indeed, by playing 𝑀 player I would only get 0.25 and this is not a best response for player I, so he does not play 𝑀 and the 2 × 3 solution carries over to the 3 × 3 game. This solves the game, because it is a zero-sum game and one equilibrium is sufficient for us because any other equilibrium would have the same payoff. (It is in fact the only equilibrium.)

38

Solution to Exercise 8.2 (a) Following the hint, here are two examples for 𝑛 = 2: min

1

2

1

2

3

2

1

0

Max

min

1

2

1

2

3

2

1

4

Max

In the first case, the top row dominates the bottom row, and in the second case, the left column dominates the right column. The only freedom is in the choice of the bottom-right matrix entry. This leads to the following general reasoning. Let the payoffs to the maximizing row player in this zero-sum game be 𝑎 𝑖𝑗 for 𝑖 = 1, 2 and 1 ≤ 𝑗 ≤ 𝑛. Because the top row played against the left column is an equilibrium and the best responses against these strategies are unique, we have 𝑎 11 > 𝑎 21 ,

𝑎11 < 𝑎 1𝑗

for 𝑗 = 2, . . . , 𝑛.

Therefore, any column 𝑗 with 𝑎21 < 𝑎2𝑗 is strictly dominated by column 1 and can be eliminated. After that, any column 𝑗 ≠ 1 fulfills 𝑎 2𝑗 ≤ 𝑎 21 < 𝑎 11 < 𝑎 1𝑗 , which means that row 1 strictly dominates row 2. So then row 2 can be eliminated, and then clearly all columns other than 1 for the single row 1 can also be eliminated as being strictly dominated. (b) For a degenerate game the claim does not hold, as in the following example, which is not dominance solvable: min

1

2

1

2

2

2

1

0

Max

(c) One has to make sure that the two cases in (a) apply to rows 2 and 3 in different columns. The following is an example, where each row and column is sometimes a best response, so none can be dominated.

39

min

1

2

3

1

5

6

7

2

4

8

1

3

3

2

9

Max

Solution to Exercise 8.3 (a) The game matrix looks as follows, with payoffs to player I. II

1

2

3

4

5

1

0

1

−1

−1

−1

2

−1

0

1

−1

−1

3

1

−1

0

1

−1

4

1

1

−1

0

1

5

1

1

1

−1

0

I

(b) There is no strict domination. Both 4 and 5 weakly dominate 1 (strategy 1 only beats 2 which is also beaten by 4 and 5), and 5 weakly dominates 2 (because 2 only beats 3 which is also beaten by 5). (c) A zero-sum game has a unique value. The game is symmetric, so when changing the players and changing the signs of the payoffs its value changes sign but must be the same, which is only possible if the value is zero. In more detail: The value is the max-min payoff to player I and min-max cost to player II. If the value was positive, then player I would get a positive payoff and player II could adopt player I’s strategy and also gain a positive amount, which is not possible. If the value was negative, then player II would get a positive amount (in terms of her own payoff), which could similarly be obtained by player I, also a contradiction. So the value is zero. (d) By Proposition 8.7, eliminating weakly dominated strategies does not change the value of a zero-sum game, so we can omit strategies 1 and 2 for both players when we look for the value of the game. The resulting 3 × 3 matrix game is the rock–scissors–paper game (with 3 for rock, 4 for scissors, 5 for paper) where the optimal mixed strategy is

40

to randomize uniformly. Hence, in the original 5 × 5 game an equilibrium strategy for each player is to play the strategies 1, 2, 3, 4, 5 with probabilities 0, 0, 31 , 13 , 13 . Then the unplayed strategies get negative expected payoff (−1 for 1 and − 13 for 2) and the three used strategies get payoff 0, which is optimal. Although not asked, this is in fact the unique equilibrium because strategies 1 and 2 are not best responses and cannot be played in any equilibrium of this zero-sum game, and the remaining strategies 3, 4, and 5 are uniquely played with probabilities 13 to give payoff zero for each. There is no equilibrium in rock–scissors–paper where one of the three strategies is not played with positive probability. For example, if the column player never uses paper, then for the row player rock strictly dominates scissors (tying instead of losing against rock, and winning instead of tying against scissors), and the mixed equilibrium of the resulting 2 × 2 game does not extend to the 3 × 3 game.

Solution to Exercise 8.4 (a) The best-response boxes show that the game has no equilibrium in pure strategies.

The upper envelope of payoffs to player II is shown on the right. payoff to II II

𝑙

I

𝑐

0

𝑇

2

𝐵

0

3

4

4

𝑐

𝑟

3

1 4

5

𝑙

𝑟

2

2

5

2

2

2

3

prob(𝐵)

0 0

1 3

2 3

1

The diagram shows that there are only two cases where player II can mix. The first case is when prob(𝐵) = 13 and 𝑟 and 𝑐 are best responses (black dot). However, in this case 𝐵 is always the best reply of player I who can therefore not be made indifferent.

The second case is when prob(𝐵) = 32 and 𝑐 and 𝑙 are best responses (white dot).

This gives the only equilibrium of the game (( 13 , 23 ), ( 12 , 12 , 0)) where player I is made indifferent between 𝑇 and 𝐵 when player II plays 𝑙 and 𝑐 with probability 21 each. (b) We consider the lower envelope of the payoffs to player I: payoff to I

4

𝑐

3 2

𝑙

𝑟

1

prob(𝐵)

0 0

1 4

1

41

The minimum payoff to player I is maximized for prob(𝐵) = 14 , which gives him max-min payoff 32 . (c) Finding a max-min strategy for player II involves probabilities for three strategies 𝑙, 𝑐, 𝑟. However, considering the zero-sum game with payoffs to player II, where player I is the minimizer, shows that we can find the max-min payoff to player II as the min-max cost to player I in that zero-sum game. Then the upper envelope of (best response) payoffs to player II, shown in (a), is minimized by player I as her opponent, which happens when prob(𝐵) = 13 , shown as a black dot. The resulting min-max cost to player I is 83 . By the minimax theorem, this is player II’s max-min payoff, where her best responses are the columns 𝑐 and 𝑟. The difference trick shows that she has to play these columns with probabilities 13 and 23 to guarantee herself that payoff in both rows. So player II’s max-min payoff is 83 , obtained with the max-min strategy (0, 13 , 23 ).

This demonstrates the power of the minimax theorem! (d) The game has a unique equilibrium (( 13 , 23 ), ( 12 , 12 , 0)) with payoffs 2 and 10 3 to players 3 I and II, respectively. The max-min payoff to player I is 2 , and the max-min payoff to player II is 83 . Both payoffs are lower than in the equilibrium. (In other games, a player’s max-min payoff may be the same as their equilibrium payoff, even though their max-min strategy is not an equilibrium strategy.) In general, an equilibrium payoff cannot be lower than the player’s max-min payoff, because otherwise the player could profitably deviate to their max-min strategy.

Solution to Exercise 9.1 (a) This game is symmetric, that is, it stays the same when exchanging the players. The payoff matrix 𝐴 to player I is given by (9.12), so the subdivision of 𝑌 into best response regions is the same as in Figure 9.14. Moreover, because of the symmetry of the game, the set 𝑋 is subdivided in the same way as 𝑌, except that labels 1 2 3 have to be exchanged with 4 5 6 , respectively. This results in the subdivision into best response regions shown below, which we directly extend with the zero vectors 0 for the application of the Lemke–Howson algorithm in (c), similar to Figure 9.16. (b) We investigate the points in 𝑋 that have three labels and check if there is a point in 𝑌 that has the remaining 3 labels. 𝑎 with labels 1 2 6 in 𝑋 – matches 𝑎 with labels 3 4 5 in 𝑌, symbol ; 𝑏 with labels 2 5 6 in 𝑋 – no match in 𝑌; 𝑐 with labels 2 3 5 in 𝑋 – no match in 𝑌; 𝑑 with labels 3 4 5 in 𝑋 – matches 𝑑 with labels 1 2 6 in 𝑌, symbol ; 𝑒 with labels 4 5 6 in 𝑋 – matches 𝑒 with labels 1 2 3 in 𝑌, symbol ; 𝑓 with labels 3 4 6 in 𝑋 – no match in 𝑌; 𝑔 with labels 1 3 6 in 𝑋 – no match in 𝑌.

The three equilibria found are given by

(𝑎, 𝑎) = (0, 0, 1), (0, 0, 1) ;

(𝑑, 𝑑) = ( 34 , 41 , 0), ( 43 , 41 , 0) ; and

(𝑒 , 𝑒) = ( 12 , 41 , 14 ), ( 21 , 41 , 14 ) .

The three equilibria of this symmetric game are symmetric strategy pairs (𝑥, 𝑦) with

42

0

0

3

6

𝑎

𝑎

𝑋˜

𝑌˜ 2

𝑐

2 3

3

𝑒

4

𝑑

4

𝑏

6

𝑒 5

5

1

𝑏

𝑓

𝑔

𝑐

1

𝑑

6

𝑓

𝑔

𝑥 = 𝑦, but in general a symmetric game may have nonsymmetric equilibria (𝑥, 𝑦) with 𝑥 ≠ 𝑦 (for which (𝑦, 𝑥) is then, due to the symmetry of the game, another nonsymmetric equilibrium); the game of Chicken is an example. (c) The computations for the missing labels 1 , 2 , 3 , 4 , 5 , 6 are as follows: For missing label 1 , the path is symbol

(𝑥, 𝑦)

labels

drop label

0, 0

123, 456

1 in 𝑋˜

𝑐, 0

235, 456

5 in 𝑌˜

𝑐, 𝑔

235, 346

3 in 𝑋˜

𝑏, 𝑔

256, 346

6 in 𝑌˜

𝑏, 𝑎

256, 345

5 in 𝑋˜

𝑎, 𝑎

126, 345

˜ (found 1 in 𝑋)

This is the longest path for all the possible missing labels and shown in the diagram in (a) above. Although not asked, we also show the second LH path for missing label 1 that connects the other two equilibria: symbol

(𝑥, 𝑦)

labels

drop label

𝑒, 𝑒

456, 123

1 in 𝑌˜

𝑒, 𝑏

456, 235

5 in 𝑋˜

𝑓,𝑏

346, 235

3 in 𝑌˜

𝑓,𝑐

346, 256

6 in 𝑋˜

𝑑, 𝑐

345, 256

5 in 𝑌˜

𝑑, 𝑑

345, 126

˜ (found 1 in 𝑌)

43

For missing label 2 , the path is symbol

(𝑥, 𝑦)

labels

drop label

0, 0

123, 456

2 in 𝑋˜

𝑔, 0

136, 456

6 in 𝑌˜

𝑔, 𝑎

136, 345

3 in 𝑋˜

𝑎, 𝑎

126, 345

˜ (found 2 in 𝑋)

(𝑥, 𝑦)

labels

drop label

0, 0

123, 456

3 in 𝑋˜

𝑎, 0

136, 456

6 in 𝑌˜

𝑎, 𝑎

126, 345

˜ (found 3 in 𝑌)

For missing label 3 , the path is symbol

For missing labels 4 , 5 , 6 , the paths are exactly the same as for 1 , 2 , 3 , due to the symmetry of the game, except for exchanging 𝑋˜ and 𝑌˜ and the labels 1 , 2 , 3 with labels 4 , 5 , 6 , respectively.

This means that the Lemke–Howson algorithm will always find the pure strategy equilibrium (𝑎, 𝑎) and never any of the two mixed equilibria in (b).

Solution to Exercise 9.2 First, we replace the strategy names 𝑇, 𝐵, 𝑙, 𝑐, 𝑟 with labels 1 , 2 , 3 , 4 , 5 . As in Figure 6.12, we see that the game has no pure-strategy equilibria.

II

3

I 1

2

1

2 0

3 0

1

5

4

0 2

1 2

2 0

The best response regions for 𝑋 are most easily constructed with the help of the upper envelope diagram, as shown below. The strategy ( 21 , 12 ) of player I has all columns as best responses and has therefore three labels, which shows the degeneracy.

The picture shows on the right also the best response regions that subdivide 𝑌. In order to make player I indifferent between 1 and 2 , player II can either mix 3 and 4 , where the difference trick shows that she should use the mixed strategy ( 12 , 12 , 0), or mix 3 and 5 , with the mixed strategy ( 23 , 0, 13 ); mixing 4 and 5 will only have the best response 1 .

The subdivision of the triangle 𝑌 is shown above.

We identify the sets of all equilibria with the help of the labels as follows: The only opportunity to have all five labels for a strategy pair (𝑥, 𝑦) is when 𝑥 = ( 12 , 21 ), but then

44

payoff to II 2

3

2

5

4

1

1

𝑌

1 2

0

1

1

𝑋 2

3

3

4

prob( 2 )

0

2 4

1

5

5

𝑥 already has the three labels 3 , 4 , 5 , so only 1 and 2 are missing. The line in 𝑌 defined by these two labels is the segment connecting ( 12 , 12 , 0) and ( 23 , 0, 13 ), so any point 𝑦 on that line segment (that is, any convex combination of these two strategies) together with 𝑥 = ( 21 , 12 ) defines an equilibrium (𝑥, 𝑦). Note that a slightly lower payoff to player II in column 4 would result in 4 never being a best response and not appearing as a label for the best-response regions in 𝑋, and the game would no longer be degenerate. In that case 4 has to have probability zero, so the only equilibrium is (𝑥, 𝑦) with 𝑦 = ( 23 , 0, 13 ), as the only pair (𝑥, 𝑦) that has all labels.

Solution to Exercise 9.3 (a) The best response payoffs of the game are indicated by boxes.

II

1

2

3

5

4

I

4

6

2 2

6

5 0

3

4 4

4 4

0 4 1

6

0

6

5

Hence, the only pure strategy equilibrium of the game is 3 , 5 . (b) The left triangle is 𝑌 and the right triangle is 𝑋, and they have the following labels:

45

1 p 0 0

a

Y

0,0,1

q

2

r

s0 0 1

6

g

5 v

5 3

t

X

1 e 1,0,0

1

3

2

f

4

4

d

c

b 0,1,0

u

0 1 0

6

The reason is that 3 is the only strategy that appears twice as a best response to a pure strategy (to 5 and 6 ), and the only best response region which includes two pure strategies appears in the left triangle, which is therefore 𝑌, with 𝑎 and 𝑏 representing the two pure strategies 5 and 6 , end 𝑒 for 4 . In order to determine if 𝑎 is 5 or 6 , consider the expected payoffs to player I against the mixed strategy 𝑦 = ( 12 , 21 , 0) of player II, which are 4, 4, 3, so player II is indifferent between 1 and 2 . So 𝑦 can only be the point 𝑑, which means the pure strategies (1, 0, 0), (0, 1, 0) and (0, 0, 1) are 𝑒 , 𝑏, 𝑎.

For 𝑋 the pure strategies (1, 0, 0), (0, 1, 0) and (0, 0, 1) are 𝑝, 𝑢, 𝑠 with best-response labels as indicated. The reason is that for 𝑥 = (𝑥 1 , 0, 𝑥3 ) we have indifference between 4 and 6 when 𝑥 1 = 34 and between 4 and 5 when 𝑥 1 = 12 , which is only possible at that side of the triangle.

The game is not degenerate because every point in 𝑋 or 𝑌 has at most three labels. (c) The pure strategy is (𝑠, 𝑏) (white circle), and two other equilibria (𝑣, 𝑓 ) (black circle) and (𝑡, 𝑐) (black square). (d) Starting from the artificial equilibrium (0𝑥 , 0 𝑦 ), the sequence of computations for missing label 1 is (0𝑥 , 0 𝑦 ) → (𝑝, 0 𝑦 ) → (𝑝, 𝑎) → (𝑞, 𝑎) → (𝑞, 𝑔) → (𝑣, 𝑔) → (𝑣, 𝑓 ) which is the equilibrium found. Starting from the pure equilibrium (𝑠, 𝑏)), the sequence of computations for missing label 1 is (𝑠, 𝑏) → (𝑟, 𝑏) → (𝑟, 𝑐) → (𝑡, 𝑐) so this is the third equilibrium.

Solution to Exercise 9.4 Suppose the pure strategy equilibrium is (𝑖, 𝑗) (with pure strategies 𝑖 and 𝑗 also seen as mixed strategies). Then the missing label 𝑖 found from the artificial equilibrium (0, 0𝑦) will lead to (𝑖, 0) and then to (𝑖, 𝑗) because 𝑗 is the duplicate label as a best response to 𝑖. This finds the pure equilibrium. If another pure equilibrium (𝑘, 𝑙) is found, then missing label 𝑖 from (𝑘, 𝑙) will not lead back to the artificial equilibrium (0, 0) because that is connected to (𝑖, 𝑗), but necessarily to a third equilibrium, as claimed.

46

Solution to Exercise 9.5

1 −2 The answer is provided by Lemma 9.9 and Figure 9.18, where we can take 𝐴 = . −2 1 Then (𝑦1 , 0) and (0, 𝑦2 ) as elements of 𝑄 are bounded because of the positive entries in both columns of 𝐴, but (𝑦1 , 𝑦1 ) ∈ 𝑄 for any 𝑦1 ≥ 0, so 𝑄 is unbounded and not a polytope.

Solution to Exercise 9.6 Suppose the 𝑚 × 𝑛 bimatrix game is (𝐴, 𝐵) and the pure equilibrium is (𝑖, 𝑗) with row 𝑖 and column 𝑗. We simply add a sufficiently large constant 𝑐 to column 𝑗 of 𝐴 and row 𝑖 of 𝐵, which defines a column shift, and thus produces an equivalent game. For that purpose, consider the maximum 𝑀 of all entries of 𝐴 and 𝐵 and let 𝑐 = max{𝑀 − 𝑎 𝑖𝑗 , 𝑀 − 𝑏 𝑖𝑗 } + 1, and consider the payoff matrices 𝐴0 and 𝐵0 where 𝑎 0𝑘 𝑗 = 𝑎 𝑘 𝑗 + 𝑐 for 1 ≤ 𝑘 ≤ 𝑚 and 𝑏 0𝑖𝑙 = 𝑏 𝑖𝑙 + 𝑐 for 1 ≤ 𝑙 ≤ 𝑛, and 𝑎 0𝑘𝑙 = 𝑎 𝑘𝑙 for 𝑙 ≠ 𝑗 and 𝑏 0𝑘𝑙 = 𝑏 𝑘𝑙 for 𝑘 ≠ 𝑖. Because the game is nondegenerate, row 𝑖 is the unique best response to column 𝑗 in (𝐴, 𝐵), that is, 𝑎 𝑖𝑗 > 𝑎 𝑘 𝑗 and therefore also 𝑎 0𝑖𝑗 > 𝑎 0𝑘 𝑗 for all 𝑘 ≠ 𝑖. Similarly, column 𝑗 is the best response to row 𝑖 in (𝐴, 𝐵), that is, 𝑏 𝑖𝑗 > 𝑏 𝑖𝑙 and therefore also 𝑏 0𝑖𝑗 > 𝑏 0𝑖𝑙 for all 𝑙 ≠ 𝑗. By construction, 𝑎 0𝑖𝑗 = 𝑎 𝑖𝑗 + 𝑐 ≥ 𝑀 + 1 > 𝑎 𝑘𝑙 = 𝑎 0𝑘𝑙 for all 𝑘 and 𝑙 ≠ 𝑗, and similarly 𝑏 0𝑖𝑗 > 𝑏 0𝑘𝑙 for all 𝑘 ≠ 𝑖 and all 𝑙. Hence, the equilibrium payoffs 𝑎 0𝑖𝑗 and 𝑏 0𝑖𝑗 are the largest payoffs in the new game (𝐴0 , 𝐵0), as required.

In the Prisoner’s Dilemma in Figure 3.1, the equilibrium (𝐷, 𝐷) has a lower payoff for both players than the strategy pair (𝐶, 𝐶). Adding 3, say, to the payoffs of player I in column 𝐷 and to the payoffs of player II in row 𝐷 gives a strategically equivalent game where (𝐷, 𝐷) has the highest payoffs 4, 4 in the game, so it is no longer a “dilemma” as to why the players cannot get anything better by cooperating.

Nondegeneracy is needed to get strictly higher payoffs: In the threat game in Figure 4.3 the equilibrium (𝑇, 𝑙) can be treated as described by adding a constant to column 𝑙 for player I and to row 𝑇 for player II to create a strategically equivalent game, but player II’s payoffs against 𝑇 stay equal for 𝑙 and 𝑟; the indifference of player II between 𝑙 and 𝑟 when player I plays 𝑇 is required for strategic equivalence.

Solution to Exercise 10.1 At the end of Section 10.4, this question has already been answered for the case that player I moves first. So we consider here only the case that player II moves first. That player has as many moves as she has strategies in the given strategic form, that is, 𝑛. Then player I moves second, but is not informed about the move of the first player, for otherwise this would not represent simultaneous action. For that reason, an information set is necessary.

This is a single information set, where the moves of player I are exactly his 𝑚 strategies in the strategic form. There is one decision node for player II, with 𝑛 moves for player II, then 𝑛 decision nodes for player I in the information set, followed by 𝑚 moves of player I, resulting in 𝑚 · 𝑛 terminal nodes, each of which has a payoff pair that corresponds to the respective cell in the strategic form.

47

I

II

𝑎

I

𝑏

0

𝑇

0

1 1

3

𝐵

3

𝑐

T

B

c b

2 II

2 4

4

II a

5

a b

I

c

a b

c

T

B

T

B

T

B

0 0

3 3

1 1

4 4

2 2

5 5

5 0 0

1 1

2 2

3 3

4 4

5 5

The above shows an example, for a 2 × 3 game with particularly simple payoffs so that they are easily identified in the respective extensive games. The extensive games show the two possibilities of games that have this strategic form, namely where player I moves first and player II moves second, and where player II moves first and player I moves second.

Solution to Exercise 10.2 It is best to use the formal Definition 10.1 of perfect recall to solve this exercise, on which we elaborate one more time: Look at a player (say player II), consider any information set of that player which has at least two nodes 𝑢 and 𝑣 (for information sets with a single node there will never be a problem). Then both 𝑢 and 𝑣 have a history of previous moves.

Along that history, consider only the sequence of moves of player II (which may be none).

If these earlier own moves are the same for both 𝑢 and 𝑣, then player II cannot distinguish between the two nodes, which is exactly what the information set is meant to express.

So this has to hold for any two nodes 𝑢 and 𝑣 in an information set. The player is then said to have perfect recall. If that is not the case, then there is an information set with two nodes 𝑢 and 𝑣 that have different sequences of earlier own moves. But then the player could distinguish 𝑢 and 𝑣, by remembering her earlier moves, contradicting the intention of an information set that she does not know if she is at 𝑢 or at 𝑣. So if the information set really means she cannot distinguish between 𝑢 and 𝑣, she must have forgotten what she did earlier. Therefore, the player is then said to have imperfect recall. Obviously, it is somewhat difficult to interpret the situation of imperfect recall, because why should a player forget her own earlier moves? At least it is difficult to consider “strategies” in a consistent manner. Therefore, we declare games without perfect recall as “illegal” and do not analyse them further.

In the game on the right, player II does not have perfect recall because the two nodes in her second information set are preceded by the different sequences of own moves empty sequence and move 𝑐, respectively. So player II has there forgotten if she made the move 𝑐 or not.

The game on the left does have perfect recall. We show its reduced strategic form, where we give only the payoffs to the maximizing player I. Player I’s best-response payoffs are surrounded by a box, the minimizing column player’s best-response costs are surrounded by a circle.

The game has a unique pure-strategy equilibrium (𝑅𝑆, 𝑏𝑐) with payoff 2 to player I and payoff −2 to player II. It is unique because 𝑅𝑆 is the unique best response to 𝑏𝑐 and

48

II

𝑎∗

𝑏𝑐

𝑏𝑑

𝐿∗

5

1

1

𝑅𝑆

4

2

3

𝑅𝑇

1

1

1

I

vice versa, which makes them the unique max-min and min-max strategy, respectively, in this zero-sum game.

If one does not want to use these properties of zero-sum games, observe that we can first eliminate the strictly dominated strategy 𝑅𝑇, then the strictly dominated strategy 𝑎∗ of the minimizing player II, then 𝐿∗, then 𝑏𝑑, so that the pure strategy pair (𝑅𝑆, 𝑏𝑐) remains.

Solution to Exercise 10.3 In the left game, player II does not have perfect recall because the two nodes in her second information set are preceded by the different sequences of own moves, namely the empty sequence and move 𝑐, respectively. (This is the same reason as in Exercise 10.2, right picture.) That is, player II has forgotten if she made the move 𝑐 or not.

The right game has perfect recall, because the only information with more than one node belongs to player II who only moves once. The strategic form looks as follows.

The upper envelope shows the expected payoffs to player I against the mixed strategy of player II. payoff to I II

𝑎

I

𝑏

0

1

𝐿∗

5

𝐿∗

2

0

𝑅𝑇

3

𝑅𝑇 3

0

6

𝑅𝑆

5

2 0

𝑅𝑆

6

3 2

0 3

prob(𝑏)

0 0

1 3

2 3

1

The game has one pure-strategy equilibrium (𝑅𝑇, 𝑏). Against 𝑅𝑇, both 𝑎 and 𝑏 are best responses. The upper envelope of the payoffs to player I shows that 𝑅𝑇 is a best response whenever 𝑏 is played with probability 23 or more. Then both 𝑎 and 𝑏 are best responses, so

49

player II can mix in equilibrium and this defines the mixed equilibria ((0, 0, 1), (1 − 𝑞, 𝑞)) for 𝑞 ∈ [ 23 , 1]. This includes the pure-strategy equilibrium (𝑅𝑇, 𝑏).

When player II plays 𝑎 and 𝑏 with ( 32 , 13 ), then both 𝐿∗ and 𝑅𝑆 are best responses of player I. If he plays ( 23 , 13 , 0), this defines an equilibrium. There is no other mixed strategy of player II where player I has more than one best response, and there are no further pure-strategy equilibria, so these are all equilibria.

Solution to Exercise 10.4 Let the cards be called 𝐻 (High), 𝑀 (Middle) and 𝐿 (Low) with 𝐻 beating 𝑀 and 𝐿, and 𝑀 beating 𝐿. There are six possible deals, each having probability 16 , namely 𝐻 𝑀, 𝐻𝐿, 𝑀𝐻, 𝑀𝐿, 𝐿𝐻, 𝐿𝑀, where, for example, 𝐻 𝑀 means that player I gets the H and player II the M card. The deal is modeled by an initial chance move, as shown in the extensive form game in Figure 10.17, with payoffs to the maximizing player I. Figure 10.18 shows an alternative but equivalent extensive game.

chance 1/6

1/6

1/6 1/6 HM

RH

HL

MH

I

1/6

FH RH

FH RM

−1

−1

ML

I

1/6

LH

LM

I

FM RM

FM RL

FL RL

FL

−1

−1

−1

−1

II mL

pL

mL

2 1

pL

2 1

II mH

pH

−2 1

mH

pH

−2 1

II mM

pM

mM

2 1

pM

−2 1 Figure 10.17 One possible answer to Exercise 10.4(a).

(a) After the chance move, three information sets for player I describe that he knows his own card 𝐻, 𝑀, or 𝐿, but not the card of the other player, which is why these information sets have two decision nodes each. In each case, player I can choose to Raise (𝑅) or Fold (𝐹), which we write with the subscripts as 𝑅 𝐻 for “Raise on seeing 𝐻”, 𝐹𝐻 for “Fold on seeing 𝐻”, and so on. The move “Fold” terminates the game

50

chance 1/3

1/3

I

1/3

I

I

RH chance

FH

RM chance

FM

RL chance

FL

1/2

−1

1/2

−1

1/2

−1

1/2

1/2

1/2

II mH

pH

mH

pH

−2

1

−2

1

II mL

pL

mL

pL

2

1

2

1

II mM

pM

mM

pM

2

1

−2

1

Figure 10.18 A different version of the game in Figure 10.17, where first the card for player I is chosen by chance, and only afterwards (if player I has not folded) the second card for player II.

with payoff −1 to player I, and payoff 1 to player II which we do not show because this a zero-sum game. The move “Raise” in each case continues the game, and these moves lead to three information sets for player II that model that player II sees her card, but not the card of player I. The information set when player gets an 𝑀 card contains the two decision nodes where player II is informed about her card but not the card of player I. The other two information sets correspond to player II getting the 𝐻 and the 𝐿 card, respectively. Player II’s moves are then, for example, 𝑚 𝑀 for “meet on seeing 𝑀”, 𝑝 𝑀 for “pass on seeing 𝑀”, and so on. A pass by player II means she has to pay one unit to player II for certain, a meet means the payoff is determined by a comparison of the hands, with possible payoffs 2 or −2 for player I. (b) In the extensive form above, each player has three information sets with two moves each, so a strategy for the player is a triple of moves, like 𝑅 𝐻 𝑅 𝑀 𝐹𝐿 for player I that states “Raise on seeing 𝐻 and 𝑀, and Fold on seeing 𝐿”, so player I has 2 × 2 × 2 = 8 strategies. Similarly, player II has eight strategies, an example of which is 𝑚𝐻 𝑝 𝑀 𝑝 𝐿 which states “meet on seeing 𝐻, and pass on seeing M and L”. Without simplifying the game, we would therefore obtain an 8 × 8 matrix game (shown in Figure 10.19) which would be rather laborious to analyze. The game is simplified as follows, starting from the information sets closest to the payoffs, which are those of player II. If player II gets the 𝐻 or the 𝐿 card, he knows that she will win or lose, respectively, upon choosing

51

Figure 10.19 The (computer-generated) strategic form of the game in Figure 10.17, which is way too large.

“meet”, and therefore will for sure choose the moves 𝑚𝐻 and 𝑝 𝐿 , which produces a unique choice at the two information sets of player II for the cards 𝐻 and 𝐿. A similar consideration applies to player I: Upon seeing 𝐻, player I knows he will win and will therefore choose 𝑅 𝐻 which gives him payoff 2 or 1 (depending on whether player II meets or passes) rather than −1 when player I chose 𝐹𝐻 , which is therefore eliminated as a move by player I. Concerning the move when player I sees the middle card, we take into account that player II will behave in a forseeable way. Given that player I sees 𝑀, there is an equal probability that player II either has 𝐻 and will raise (payoff −2 to player I) or has 𝐿 and will pass (payoff 1 to player I), with conditional expected payoff 12 × (−2) + 12 × 1 = − 12 to player I. When player I sees 𝑀, the alternative move to 𝑅 𝑀 is to Fold, choosing 𝐹𝑀 , and then he will for sure get payoff −1, which is worse.

Hence player I will definitely choose 𝑅 𝑀 rather 𝐹𝑀 . If player II does not behave as expected, this is even better for player I, and he should choose 𝑅 𝑀 as well.

Note: instead of conditional probabilites, one could also consider the absolute contribution to expected payoffs, where the conditional probability 12 above is replaced by the absolute probability 16 . Then these contributions to expected payoffs would

52

be 16 × (−2) + 16 × 1 = − 16 for 𝑅 𝑀 and 16 × (−1) + 16 × (−1) = − 31 for 𝐹𝑀 , which (because these are just the same numbers except multiplied by 13 ) shows that 𝑅 𝑀 is the better move. The only open question is whether player I should Raise or Fold on seeing a Low card 𝐿. This decision is not unique, because player II will possibly pass on seeing 𝑀, in which case the “bluff” 𝑅 𝐿 indeed is good for player I.

chance 1/6

1/6

1/6 1/6 HM

HL

I RH

MH

ML

1/6

RH

I

RM

1/6

LH

LM

I FL RL

FL

−1

−1

RL

RM

II pL

pL

1

1

II mH

mH

−2

−2

II mM

pM

mM

2 1

pM

−2 1 Figure 10.20

Simplified version of the game in Figure 10.17.

(c) The simpler game is essentially the same as the original extensive form game, where only the strategies 𝑚𝐻 𝑚 𝑀 𝑝 𝐿 and 𝑚𝐻 𝑝 𝑀 𝑝 𝐿 of player II are considered, and the strategies 𝑅 𝐻 𝑅 𝑀 𝑅 𝐿 and 𝑅 𝐻 𝑅 𝑀 𝐹𝐿 of player I. Any other strategy of player II, for example, would specify a nonoptimal move (for example 𝑚𝐿 ) which would result in a worse payoff to player II in certain cases (namely if player I does not Fold when seeing 𝐻 or 𝑀), and at best is the same payoff to player II. That is, any other strategy of player II is weakly dominated by one of the two remaining strategies. (The same applies to the strategies of player I.) These can be ignored by Proposition 8.7. (d) In the simplified game, the only choice for player I is between the strategies 𝑅 𝐻 𝑅 𝑀 𝑅 𝐿 and 𝑅 𝐻 𝑅 𝑀 𝐹𝐿 (in effect, between his moves 𝑅 𝐿 and 𝐹𝐿 ). Similarly, player II only chooses between the strategies 𝑚𝐻 𝑚 𝑀 𝑝 𝐿 and 𝑚𝐻 𝑝 𝑀 𝑝 𝐿 (in effect, between her moves 𝑚 𝑀 and 𝑝 𝑀 ). We now determine the strategic form for the simplified game. Given any pair of choices (strategies) of the two players, we compute expected payoffs for the six possible hands that are dealt. Each relevant leaf of the tree (given by the player’s choices) has a probability of 16 of being reached. Hence these expected payoffs for the

53

possible pairs of choices are 𝑅 𝐻 𝑅 𝑀 𝑅 𝐿 , 𝑚𝐻 𝑚 𝑀 𝑝 𝐿

:

1 2 1 2 2 2 6 + 6 − 6 + 6 − 6 − 6 =

− 13

𝑅 𝐻 𝑅 𝑀 𝑅 𝐿 , 𝑚𝐻 𝑝 𝑀 𝑝 𝐿

:

1 1 2 1 2 1 6 + 6 − 6 + 6 − 6 + 6 =

0

𝑅 𝐻 𝑅 𝑀 𝐹𝐿 , 𝑚 𝐻 𝑚 𝑀 𝑝 𝐿

:

2 1 2 1 1 1 6 + 6 − 6 + 6 − 6 − 6 =

0

𝑅 𝐻 𝑅 𝑀 𝐹𝐿 , 𝑚 𝐻 𝑝 𝑀 𝑝 𝐿

:

1 1 2 1 1 1 6 + 6 − 6 + 6 − 6 − 6 =

− 16 ,

so the strategic form looks as follows: min 𝑚𝐻 𝑚 𝑀 𝑝 𝐿

𝑚𝐻 𝑝 𝑀 𝑝 𝐿

𝑅𝐻 𝑅 𝑀 𝑅𝐿

− 13

0

𝑅 𝐻 𝑅 𝑀 𝐹𝐿

0

− 16

Max

The best responses show that this 2 × 2 matrix game has circular preferences and hence no pure-strategy equilibrium. In equilibrium, player I plays 𝑅 𝐿 with probability 13 and 𝐹𝐿 with probability 23 , so that player II will in both cases pay expected cost − 19 , and player II plays 𝑚 𝑀 with probability 13 and 𝑝 𝑀 with probability 23 , so as to make player I indifferent. The value of the game is − 19 , so the game is slightly advantageous for player II. The optimal strategies are in fact behavior strategies because they prescribe move probabilities when player I sees 𝐿 (and chooses “Raise” with probability 1 when he sees 𝐻 or 𝑀), and similarly for player II where her only randomization is when she gets the 𝑀 card. Note that player I has an advantage of “bluffing” with probability 13 when seeing a low card. If he always chose to Fold, that is, chose 𝐹𝐿 with certainty, then player II would know that when she sees 𝑀 (and player I has raised) that player I has the High card 𝐻. Hence, she would respond with 𝑝 𝑀 , which results in the lower expected payoff − 16 to player I, which is not his optimal payoff.

Solution to Exercise 10.5 This question tests your understanding of the definition of a subgame. By definition, a subgame is a subtree of the game tree where any information set that contains some node of the subtree must be fully contained in the subtree. Consider the information set that contains the root of a subgame. If that information set contained any other node, it would have to be contained in the subtree and hence be a descendant of the root of the subtree.

However, this violates Proposition 10.2.

Solution to Exercise 10.6 Note that this game has perfect recall, because the only non-singleton information belongs to player I, and it is the only time he makes a move. We consider the reduced strategic

54

form and the upper envelope of payoffs to player II as a function of the mixed strategy of player I. payoff to II II

𝑎∗∗

I

𝑏𝑐∗

1

𝑇

1

5

4

1 1

𝑏𝑐∗

5

4

𝑎∗∗

𝑏𝑑 𝑓

5

4

4

6

1 2

2

𝑏𝑑 𝑓

1

5

5

𝐵

𝑏𝑑𝑒

4 1

6

2

𝑏𝑑𝑒

1 prob(𝐵)

0 0

1 3

3 4

1

There is one pure equilibrium (𝐵, 𝑎∗∗) in reduced strategies with payoffs (4, 5), and another, (𝑇, 𝑏𝑐∗) with payoffs (5, 5), shown with dots in the upper envelope diagram.

From the upper envelope we see that there are two points where player II is indifferent between two of his reduced strategies: For 𝑝 = prob(𝐵) = 13 player II is indifferent between 𝑏𝑐∗ and 𝑏𝑑 𝑓 . Then player II, in turn, can make player I indifferent between 𝑇 and 𝐵 only by mixing between 𝑏𝑐∗ and 𝑏𝑑 𝑓 with probabilities 0 and 1. Because only 𝑏𝑑 𝑓 has to be a best response in that case, player I (see the upper envelope diagram) can achieve this by mixing with 13 ≤ 𝑝 ≤ 43 . This gives the set of mixed equilibria ((1 − 𝑝, 𝑝), 𝑏𝑑 𝑓 ) for 31 ≤ 𝑝 ≤ 34 , with payoffs (6, 4).

For 𝑝 = prob(𝐵) = 34 player II is indifferent between 𝑎∗∗ and 𝑏𝑑 𝑓 . Again, player I can be made indifferent between 𝑇 and 𝐵 when player II plays 𝑏𝑑 𝑓 with certainty, which is part of the set of equilibria that have just been described. As soon as player II chooses 𝑎∗∗ with positive probability relative to 𝑏𝑑 𝑓 , player I’s unique best response is 𝐵, against which 𝑏𝑑𝑓 is not a best response.

In a subgame perfect equilibrium, player II clearly has to play move 𝑓 at her last decision node. There are no further subgames in the game, which means any equilibrium where player II plays 𝑓 is subgame perfect. The infinite equilibrium set is given by the strategy pairs ((1 − 𝑝, 𝑝), 𝑏𝑑 𝑓 ) for 13 ≤ 𝑝 ≤ 34 . For any 𝑝, this is a pair of behavior strategies: The pure strategy 𝑏𝑑 𝑓 of player II is a behavior strategy (which indeed specifies move 𝑓 ), and player I has only one information set with two moves 𝑇 and 𝐵, so his mixed strategy (1 − 𝑝, 𝑝) is also a behavior strategy.

As an SPE, the pure equilibrium (𝑇, 𝑏𝑐∗) has to be written as (𝑇, 𝑏𝑐 𝑓 ) so that both players’ strategies are fully specified. To describe the pure equilibrium (𝐵, 𝑎∗∗) as an SPE, the reduced strategy 𝑎∗∗ has to be written as 𝑎∗ 𝑓 and ∗ can in fact be any behavior strategy 𝛽 that mixes between 𝑐 and 𝑑, so if written as (𝐵, 𝑎𝛽 𝑓 ) this defines another SPE. (Note that after move 𝑎, the behavior specified by 𝛽 is never used because that decision node of player II is not reached, but is necessary to define the SPE property.)

55

Solution to Exercise 10.7 This game has two subgames that start with the respective moves of player 2. In the left subgame, player 2’s unique optimal move is 𝑙. The right subgame is a 2 × 2 game with the following payoffs to players 2 and 3 (which are the players involved): 3

𝐴

2

𝐵

3

𝑎

3 1

3 2

𝑏

2

3 4

This subgame has two pure equilibria (𝑎, 𝐴) and (𝑏, 𝐵). This is a degenerate game which has also a mixed equilibrium (𝑎, (1 − 𝑞, 𝑞)) where player 2 plays 𝑎 and player 3 plays 𝐵 with probability 𝑞 so that 𝑎 stays a best response of player 2, which requires that 𝑞 ≤ 41 because for 𝑞 = 14 player 2 is exactly indifferent between 𝑎 and 𝑏, and 𝐴 needs to be played with probability at least 34 so that 𝑎 is a best response.

The payoffs resulting from these equilibria have now to be substituted for the second decision node of player 2 to determine the best response of player 1. Then if players 2 and 3 play (𝑏, 𝐵) in the subgame, player 1 chooses 𝐿 in response, and for the other equilibra where player 2 chooses 𝑎, player 1 chooses 𝑅. So the resulting subgame perfect equilibria are the following: (𝐿, (𝑙, 𝑏), 𝐵), (𝑅, (𝑙, 𝑎), 𝐴), and the mixed equilibria (𝑅, (𝑙, 𝑎), (1 − 𝑞, 𝑞)) for 𝑞 ∈ [0, 41 ].

Solution to Exercise 10.8

1 𝐿

2 𝑅

𝑙

2 𝑙 2 3 1

𝑟 2 0 2

0

3 𝑙

3 1−𝑞 𝐴

𝑟

1

𝐿

𝑟

𝑞 𝐵

2

3 𝐶

𝐷

2 3 − 2𝑞

4

𝑅 3 3 3

4 1 3

1 2 2

1 4 3

3+𝑞

1

This game has two subgames that start with the respective moves of player 3. In the left subgame, player 3 is indifferent between 𝐴 and 𝐵 and can mix them with (1 − 𝑞, 𝑞) for an

56

arbitrary 𝑞 ∈ [0, 1]. In the right subgame, player 3 will play 𝐷. The game tree with these optimal moves is shown on the left. Substituting the resulting payoffs, parameterized by 𝑞, for the decision nodes of player 3 gives the 2 × 2 game between players 1 and 2 shown on the right.

This game has a unique mixed equilibrium, where player 1 and 2 will mix as follows.

The difference trick shows that the differences for player 2’s payoffs for rows 𝐿 and 𝑅 are 3 and 4 − (3 − 2𝑞) = 1 + 2𝑞, respectively, so player 1’s equilibrium mixed strategy is 1+2𝑞 3 ( 4+2𝑞 , 4+2𝑞 ). The differences for player 1’s payoffs for columns 𝑙 and 𝑟 are 3 + 𝑞 − 2 = 1 + 𝑞 1

1+𝑞

and 1, respectively, so player 2’s equilibrium mixed strategy is ( 2+𝑞 , 2+𝑞 ). These are to be combined with the mixture (1 − 𝑞, 𝑞) of player 3 mixing between 𝐴 and 𝐵, and choosing 𝐷 with certainty.

Solution to Exercise 11.1 The pair (𝑈 , 𝑉) maximizes the Nash product (𝑢 − 0)(𝑣 − 0) if 𝑈𝑉 ≥ 𝑢𝑣 for all (𝑢, 𝑣) ∈ 𝑆.

This inequality, multiplied with 𝑎 and 𝑏, is equivalent to: 𝑎𝑈𝑏𝑉 ≥ 𝑎𝑢𝑏𝑣 for all (𝑢, 𝑣) ∈ 𝑆, that is, 𝑎𝑈𝑏𝑉 ≥ 𝑢 0 𝑣 0 for all (𝑢 0 , 𝑣 0) ∈ 𝑆0. This shows that the maximum Nash product in 𝑆0 is indeed obtained for (𝑈 0 , 𝑉 0) = (𝑎𝑈 , 𝑏𝑉), as claimed.

Solution to Exercise 11.2 We first consider the game in (i). (a) In order to find max-min strategies, it is best to consider the payoff matrix of the respective player as a zero-sum game where that player is the maximizer. For player I and player II, this defines the following zero-sum games, where we have indicated the pure best responses of the maximizer with a box and of the minimizer with a circle. min

𝑙

𝑟

𝑇

1

3

𝐵

0

3

max

max

𝑙

𝑟

𝑇

−2

3

𝐵

6

1

min

The left game has a unique pure-strategy equilibrium, which shows that the max-min strategy of player I is 𝑇, with max-min payoff 𝑢0 = 1. This also easy to see because the minimizing column against any row is 𝑙, which determines player I’s max-min strategy as the pure strategy 𝑇.

In contrast, the right game does not have a pure equilibrium, and as a 2 × 2 game therefore has a unique mixed equilibrium. The difference trick shows that the differences in payoffs for the left and right column are 8 and 2, respectively, so that these columns should be played with probabilities (0.2, 0.8), which defines player II’s ˆ Her max-min payoff is 𝑣 0 = 2, which is the expected payoff in max-min strategy 𝑦. both rows, given by −0.4 + 2.4 for 𝑇 and 1.2 + 0.8 for 𝐵.

57

𝑣 𝐵, 𝑙 : (0, 6)

(𝑈 , 𝑉) = (2.5, 3.5) 𝑆

𝑇, 𝑟 : (3, 3)

(𝑢0 , 𝑣0 ) = (1, 2) 𝐵, 𝑟 : (3, 1) 𝑢

𝑇, 𝑙 : (1, −2)

(b) The picture shows the convex hull of the payoff pairs with threat point (𝑢0 , 𝑣0 ) = (1, 2) and the bargaining set 𝑆, given by all points in the convex hull to the right and above that threat point. The Pareto-frontier is the shown thick line segment that connects (1, 5) (which is the highest point of 𝑆) and (3, 3). The vertical part of the right boundary of 𝑆 between (3, 2) and (3, 3) is not part of the Pareto-frontier, apart from its endpoint (3, 3), because any point (3, 𝑣) on that line with 𝑣 < 3 can be improved for player II without hurting player I. (c) The Nash bargaining solution (𝑈 , 𝑉) is found on the Pareto-frontier by finding an ascending slope from (𝑢0 , 𝑣0 ) to (𝑈 , 𝑉) which is the same as the descending slope for a tangent to the bargaining set, as described in Figure 11.6(c). Here that tangent is unique with slope −1, for the line that connects the points (0, 6) and (3, 3) (three steps to the right and three steps down). This is met by an ascending line of slope 1 starting from (𝑢0 , 𝑣0 ) = (1, 2), shown in the above picture, with (𝑈 , 𝑉) = (2.5, 3.5). This point is the convex combination 16 (0, 6) + 56 (3, 3).

Alternatively, one can maximize the Nash product (𝑈 − 𝑢0 )(𝑉 − 𝑣 0 ) for (𝑈 , 𝑉) on the line (1 − 𝑝)(0, 6) + 𝑝(3, 3), i.e. 𝑈 = 3𝑝, 𝑉 = 6 − 3𝑝. This Nash product is given by by (3𝑝 − 1)(4 − 3𝑝) = 12𝑝 − 4 − 9𝑝 2 + 3𝑝. Its derivative with respect to 𝑝 is 15𝑝 − 18, which is zero for 𝑝 = 56 as above. (d) The bargaining solution in (c) is implemented with the lottery that chooses (𝐵, 𝑙) and (𝑇, 𝑟) with probabilities 16 and 56 , respectively.

58

Solution to Exercise 11.2 This is the solution to the game in (ii). (a) The max-min strategy of player I is 𝑇, with max-min payoff 𝑢0 = 2, and the max-min strategy of player II is 𝑙, with max-min payoff 𝑣 0 = 3. Note that these max-min strategies are unique pure strategies of the respective player, because they have a unique best response when considering the other player as a minimizer in a zero-sum game that is defined by the player’s own payoffs. This defines the threat point (𝑢0 , 𝑣0 ) = (2, 3). (b) Here, the threat point (𝑢0 , 𝑣0 ) = (2, 3) is not part of the convex hull of the possible payoff pairs, as shown here (this can happen when the two max-min strategies are not played against each other): 𝑣 (𝑈 , 𝑉) = (3, 4)

𝑇, 𝑙 𝑆 (𝑢0 , 𝑣0 )

𝐵, 𝑟 : (1, −2)

𝐵, 𝑙 : (4, 3)

𝑇, 𝑟 : (3, 1) 𝑢

The bargaining set 𝑆 consists of those points (𝑢, 𝑣) in the convex hull with 𝑢 ≥ 𝑢0 , 𝑣 ≥ 𝑣 0 . The Pareto-frontier is the bold line that joins (4, 3) to (3, 4). (c) The bargaining solution (𝑈 , 𝑉) is the point (3, 4) which results from the strategy pair (𝑇, 𝑙). One way to see this is via Proposition 11.5, another is to maximize the Nash product on the Pareto-frontier. The latter is the line consisting of the points (1−𝑝)(4, 3)+𝑝(3, 4) for 𝑝 ∈ [0, 1], which are of the form (4−𝑝, 3+𝑝). The corresponding Nash product is (4 − 𝑝 − 2)(3 + 𝑝 − 3), that is, (2 − 𝑝)𝑝 or 2𝑝 − 𝑝 2 , which has its maximum for 𝑝 = 1. This shows (𝑈 , 𝑉) = (3, 4). (d) In the Nash bargaining solution, only (𝑇, 𝑙) is played with certainty.

Solution to Exercise 11.3 (a) The game tree looks as below, with the two SPE in (c) shown with white and black arrows, respectively. (b) Player I has 𝑀 + 1 strategies, given y any demand 𝑥 ∈ {0, 1, . . . , 𝑀}. Player II has 2𝑀+1 strategies, given by any combination of 𝐴 and 𝑅 at each of her 𝑀 + 1 decision nodes, starting from 𝐴𝐴 · · · 𝐴𝐴, 𝐴𝐴 · · · 𝐴𝑅, to 𝑅𝑅 · · · 𝑅𝑅. For 𝑀 = 3 as in (a), these are 16 strategies for player II. The numbers of reduced strategies are the same because no reduction is possible. (c) Some case distinctions necessary here. By backward induction, player II chooses to accept in an SPE whenever she is offered a positive amount. Hence, the only question

59

I 0 II

1

2

II

3 II

II

𝐴

𝑅

𝐴

𝑅

𝐴

𝑅

𝐴

𝑅

0 3

0 0

1 2

0 0

2 1

0 0

3 0

0 0

is what player II does when she is offered nothing (𝑥 = 𝑀), and can then either accept or reject. When player II accepts, then player I’s best response is to demand everything (𝑥 = 𝑀), which defines one SPE, shown with black arrows in (a), and the white arrows for player II for 𝑥 = 0, 1, 2. When player II rejects when offered nothing, then player I’s best response is 𝑥 = 𝑀 − 1, which defines the other SPE shown with the all-white arrows in (a). (d) Here, it is important to understand what a strategy of player II is. An arbitrary strategy of player II prescribes how for each demand 𝑥 = 0, 1, . . . , 𝑀, player II chooses either 𝐴 or 𝑅. Clearly, the best response of player I is to demand the maximum that player II will accept. Because that demand, say 𝑥, is accepted by player II, player II cannot improve her payoff, and so this defines an equilibrium. Consequently, any strategy of player II is part of an equilibrium.

There is are two cases where this reasoning does not apply, namely when player I gets zero for every possible demand. The first case where this happens is is when player II rejects everything. In that case, player I is indifferent between all demands because all give payoff zero. However, the only demand that defines an equilibrium is when player I demands everything, because rejection 𝑅 is only in that case a best response of player II (any positive amount she would have to accept on the equilibrium path).

The second case where this occurs is for the strategy 𝐴𝑅 · · · 𝑅𝑅 of player II where she rejects every offer except when she is offered the whole pie, that is, 𝑥 = 0. Then there are two possible best responses of player I which lead to an equilibrium. The first is if player I gives in to this demand and indeed offers the whole pie to player II, 𝑥 = 0, which player II according to her strategy accepts (this is a special case of the initially described equilibrium where player I demands the maximum that player II accepts).

The second strategy of player I is 𝑥 = 𝑀, against which 𝑅 is indeed a best response of player I, and on the equilibrium path both players get nothing. This can be described as the “no-deal” scenario of mutually maximal demands, which are rejected. (e) (This requires the concept of a behavior strategy from Chapter 10.) In addition to (c), player II can mix between 𝐴 and 𝑅 when offered nothing, which is her more general behavior strategy in equilibrium. Then the best response of player I depends on his

60

expected payoff against that behavior. If the probability for acceptance 𝐴 for demand 𝑀 is larger than 𝑀−1 𝑀 , then making demand 𝑀 is the unique equilibrium choice of player I, because then his expected payoff is larger than 𝑀 − 1. If the probability for acceptance 𝐴 for demand 𝑀 is less than 𝑀−1 𝑀 , then making demand 𝑀 − 1 is the unique equilibrium choice of player I. If the probability for acceptance 𝐴 for demand 𝑀 is exactly 𝑀−1 𝑀 , then player I is indifferent between making demand 𝑀 − 1 and making demand 𝑀, and can mix arbitrarily between the two in an SPE. Solution to Exercise 11.4 (a) The following picture shows the bargaining set, which is the shaded area between the threat point (0, 0) and the curve. 𝑣 1

𝑢

0 0

1

The Pareto-frontier of the bargaining set is the curve itself (the bold line). It is given by the set of pairs (𝑢(𝑥), 𝑣(1 − 𝑥)) for 𝑥 ∈ [0, 1]. Because 𝑢 = 𝑢(𝑥) = 𝑥, we have 𝑣(1 − 𝑥) = 1 − 𝑥 2 , so this curve is simply the graph of the function 1 − 𝑢 2 for 𝑢 ∈ [0, 1]. (b) The Nash bargaining solution is found on the Pareto-frontier, by maximizing the Nash product 𝑢(𝑥)𝑣(1 − 𝑥) = 𝑥(1 − 𝑥 2 ) = 𝑥 − 𝑥 3 for 𝑥 ∈ [0, 1]. The derivative of 𝑢(𝑥)𝑣(1 − 𝑥) is 1 − 3𝑥 2 √ √ which is zero (for 0 < 𝑥 < 1) if and only if 𝑥 = 1/ 3. The pie will be split into 𝑥 = 1/ 3 for player I and 1 − 𝑥 for player II, with utilities 𝑥 and 1 − 𝑥 2 = 2/3, respectively.

Solution to Exercise 11.5 (a) In the last round 2, player II can demand the whole pie, getting 𝛿 and leaving nothing to player I (point A in Figure 11.11). So in round 1, player I, when making a demand 𝑥, has to offer at least a utility of 𝑣(1 − 𝑥) = 𝛿 to player II (point B in Figure 11.11). The √ 2 resulting equation is 𝛿 = 𝑣(1 − 𝑥) = 1 − 𝑥 , that is, 𝑥 = 1 − 𝛿 because 𝑥 > 0. This is player I’s demand in round 1.

When 𝛿 = 34 we have 𝑥 =

q

1 1 4 = 2.

The players’ strategies √ in the subgame perfect equilibrium are as follows: Player I makes demand 𝑥 = 1 − 𝛿, and player II will accept this offer, which terminates the

61

game. Player II will accept any larger offer, and reject any lower offer, then countering by demanding the whole pie, and player I accepts any offer in the final round. (Strictly speaking, this describes a reduced strategy because no counter-demand of player II is specified when she accepts; for an SPE it is necessary to specify complete strategies, which means specifying a counter-demand for player II even if she accepts the offer in the first round, but that counter-demand would be exactly the same.) At any rate, be careful to describe more than just the equilibrium path. (b) Backward induction shows the following: In the final round 3, player I can demand everything, and receive utility 𝛿2 (point A in the first picture of Figure 11.13). In the round before, round 2, player II can therefore only demand an amount 𝑦 for herself such that the utility for player I is equal to 𝛿2 (shown at point B in the picture), that is, 𝛿𝑢(1 − 𝑦) = 𝛿2 or equivalently 1 − 𝑦 = 𝛿, that is, 𝑦 = 1 − 𝛿. In round 1, therefore, player I can make a demand 𝑥 such that player II’s payoff 𝑣(1 − 𝑥) equals the utility 𝛿𝑣(𝑦) that she can get in round 2 (shown at point C in the picture). That is, 𝑣(1 − 𝑥) = 1 − 𝑥 2 = 𝛿𝑣(𝑦) = 𝛿𝑣(1 − 𝛿) = 𝛿(1 − 𝛿2 ) = 𝛿 − 𝛿3 √ or 𝑥 2 = 1 − 𝛿 + 𝛿3 and thus 𝑥 = 1 − 𝛿 + 𝛿3 . This is demanded by player I in round 1 and accepted by player II. For 𝛿 = 14 we have 𝑥 =

q

1 1 − 14 + 64 =

q

64−16+1) , so 𝑥 = 64

q

49 7 64 = 8 .

(c) The stationary solution in an infinite number of rounds is that player I makes demand 𝑥 whenever it is his turn, and player II makes demand 𝑦 whenever it is her turn, such that subsequently the other player is indifferent between accepting and rejecting.

This is expressed by the equations 𝑣(1 − 𝑥) = 𝛿𝑣(𝑦) (line B–C in the last picture of Figure 11.15), and 𝛿𝑢(1 − 𝑦) = 𝛿2 𝑢(𝑥) (line A–B in the picture), the latter being equivalent to 𝑢(1 − 𝑦) = 𝛿𝑢(𝑥). Here, these equations are equivalent to 1 − 𝑥 2 = 𝛿(1 − (1 − 𝑦)2 ),

1 − 𝑦 = 𝛿𝑥.

Substituting the second equation into the first gives 1 − 𝑥 2 = 𝛿(1 − (𝛿𝑥)2 ) = 𝛿 − 𝛿3 𝑥 2 , so 1 − 𝛿 = 𝑥 2 (1 − 𝛿3 ), and thus 𝑥2 =

q

For 𝛿 = 53 we have 𝑥 = 1/

1−𝛿 1 = .

1 − 𝛿3 1 + 𝛿 + 𝛿2

q

9 1 + 35 + 25 = 1/

25+15+9 = 75 .

25

When 𝛿 → 1, then 𝑥 → √1 , which is the same as the Nash bargaining solution 3 computed in Exercise 11.4(b).

Solution to Exercise 12.1 This is the game in Figure 10.16, with its Nash equilibrium ( 35 , 25 ), ( 15 , 45 ) . In order to find its correlated equilibria, assume that the correlation device 𝑧 is given by the probabilities

62

𝑎, 𝑏, 𝑐, 𝑑 as in

II

𝑙

𝑟

𝑇

𝑎

𝑏

𝐵

𝑐

𝑑

I

Then the incentive constraints for the players are (with their simplified form shown on the right) 4𝑎 ≥ 𝑏,

4𝑎 + 2𝑏 ≥ 3𝑏, 3𝑑 ≥ 4𝑐 + 2𝑑,

𝑑 ≥ 4𝑐,

−4𝑎 ≥ −2𝑎 − 3𝑐,

3𝑐 ≥ 2𝑎,

−2𝑏 − 3𝑑 ≥ −4𝑏,

2𝑏 ≥ 3𝑑.

These imply 4𝑎 ≥ 𝑏 ≥ 23 𝑑 ≥ 32 4𝑐 = 6𝑐 ≥ 4𝑎, so all these inequalities hold as equalities. The smallest of these numbers is 𝑐, and setting 𝑐 = 2 to obtain integers gives 𝑐 = 2,

𝑎 = 3,

𝑏 = 12,

𝑑=8

2 𝑐 = 25 ,

8 𝑑 = 25 ,

which sum to 25. Normalization gives 3 , 𝑎 = 25

12 𝑏 = 25 ,

so the only correlated equilibrium is the Nash equilibrium.

Solution to Exercise 12.2 (a) As the symmetry of the game in players and strategies suggests, an equilibrium of the game is (𝑥, 𝑦) with 𝑥 = 𝑦 = ( 13 , 13 , 13 ), with payoff − 13 in each row and column. For a correlated equilibrium 𝑧, the incentive constraints (12.4) for 𝑖 = 1 and 𝑘 = 2, 3 state − 𝑧 11 − 2𝑧 12 + 3𝑧 13 ≥ 0, 2𝑧 11 − 3𝑧 12 + 𝑧 13 ≥ 0, and similarly for 𝑖 = 2, 3 and all 𝑘 = 1, 2, 3. For player II, the incentive constraints (12.5) are the same with 𝑧 𝑗𝑖 instead of 𝑧 𝑖𝑗 because the game is symmetric, that is, 𝑏 𝑗𝑖 = 𝑎 𝑖𝑗 for 1 ≤ 𝑖, 𝑗 ≤ 3. All these constraints are obviously tight when 𝑧 𝑖𝑗 = 91 for 1 ≤ 𝑖, 𝑗 ≤ 3. (b) A coarse correlated equilibrium 𝑧 is given by 𝑧 𝑖𝑖 = 31 and 𝑧 𝑖𝑗 = 0 for 𝑖 ≠ 𝑗. The expected payoff to the two players when they let the correlation device play their strategies is zero. A deviating player who does not use the device only knows that all three strategies will be played with probability 13 , against which his expected payoff is − 31 and therefore less, so it is better to participate. This is not a correlated equilibrium (in agreement with the uniqueness of the correlated equilibrium in (a)) because when

63

the device chooses, for example, the strategy pair (𝑅, 𝑅), then each player would know that the other player has been told the same strategy and would deviate to 𝑃 (assuming the other player would follow the recommendation and keep playing 𝑅).

Solution to Exercise 12.3 Consider a correlation device 𝑧 with 𝑎 = 𝑧 32 and 𝑏 = 𝑧 23 = 1 − 𝑎 and all other components 𝑧 𝑖𝑗 = 0 as described. Then the expected payoff as stated on the left-hand side of (12.9) is 2𝑎 + 5𝑏 for player I and 5𝑎 + 2𝑏 for player II. Consider (12.9) for player 𝑖 = I, where player I may deviate and not use the device and then only knows the marginal probabilities that player II plays one of the columns. The marginal probability that player II plays 𝑇, 𝑀, 𝐵 is 0, 𝑎, 𝑏, respectively, so that the constraints (12.9) for 𝑡 𝑖 = 𝑇, 𝑀, 𝐵 state, respectively, 2𝑎 + 5𝑏 ≥ 4𝑎 + 𝑏,

2𝑎 + 5𝑏 ≥ 5𝑏,

2𝑎 + 5𝑏 ≥ 2𝑎.

Similarly, for player 𝑖 = II these constraints state 5𝑎 + 2𝑏 ≥ 4𝑏 + 𝑎,

5𝑎 + 2𝑏 ≥ 5𝑎,

5𝑎 + 2𝑏 ≥ 2𝑏.

In each case, the last two inequalities hold always because 𝑎 ≥ 0 and 𝑏 ≥ 0. With 𝑏 = 1 − 𝑎, the first inequality for player I says 4𝑏 ≥ 2𝑎 or 4 ≥ 6𝑎, and for player II 4𝑎 ≥ 2𝑏 or 6𝑎 ≥ 2, which together defines the two inequalities 23 ≥ 𝑎 ≥ 13 . The extreme case 𝑎 = 23 , 𝑏 = 13 gives payoff 2𝑎 + 5𝑏 = 3 to player I and 5𝑎 + 2𝑏 = 4 to player II, and the other extreme case 𝑎 = 13 , 𝑏 = 23 payoff 4 to player I and 3 to player II. Solution to Exercise 12.4 Consider an 𝑁-player game where every player has two strategies, and a coarse correlated equilibrium 𝑧. According to (12.9), this means that

Õ Õ

Õ

𝑢𝑖 (𝑠 𝑖 , 𝑠−𝑖 ) 𝑧(𝑠 𝑖 , 𝑠−𝑖 ) ≥

𝑠 −𝑖 ∈𝑆−𝑖 𝑠 𝑖 ∈𝑆 𝑖

𝑢𝑖 (𝑡 𝑖 , 𝑠−𝑖 )

𝑠−𝑖 ∈𝑆−𝑖

Õ

𝑧(𝑠 𝑖 , 𝑠−𝑖 )

𝑠 𝑖 ∈𝑆 𝑖

for any strategy 𝑡 𝑖 of any player 𝑖. By taking 𝑠 𝑖 to be the other strategy of player 𝑖, this inequality can written as

Õ

𝑢𝑖 (𝑠 𝑖 , 𝑠−𝑖 ) 𝑧(𝑠 𝑖 , 𝑠−𝑖 ) + 𝑢𝑖 (𝑡 𝑖 , 𝑠−𝑖 ) 𝑧(𝑡 𝑖 , 𝑠−𝑖 )

𝑠 −𝑖 ∈𝑆−𝑖

Õ

≥

𝑢𝑖 (𝑡 𝑖 , 𝑠−𝑖 ) 𝑧(𝑠 𝑖 , 𝑠−𝑖 ) + 𝑧(𝑡 𝑖 , 𝑠−𝑖 ) ,

𝑠 −𝑖 ∈𝑆−𝑖

which simplifies to

Õ 𝑠−𝑖 ∈𝑆−𝑖

Õ

𝑢𝑖 (𝑠 𝑖 , 𝑠−𝑖 ) 𝑧(𝑠 𝑖 , 𝑠−𝑖 ) ≥

𝑢𝑖 (𝑡 𝑖 , 𝑠−𝑖 ) 𝑧(𝑠 𝑖 , 𝑠−𝑖 ) .

𝑠 −𝑖 ∈𝑆−𝑖

These are exactly the two incentive constraints (12.10) for player 𝑖 that define a correlated equilibrium; note that 𝑠 𝑖 and 𝑡 𝑖 can be the strategies of player 𝑖 in either order, and that the incentive constraint for 𝑠 𝑖 = 𝑡 𝑖 holds trivially.

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Solution to Exercise 12.5 If 𝑌 =

𝑎

𝑏

𝑐

𝑑

! and 𝑥> = (𝑥 1 , 𝑥2 ), then the stationarity condition 𝑥>𝑌 = 𝑥> can be written

as (1 − 𝑏)𝑥 1 +

𝑐𝑥2 = 𝑥 1

𝑏𝑥 1 + (1 − 𝑐)𝑥 2 = 𝑥 2 using 𝑎 = 1 − 𝑏 and 𝑑 = 1 − 𝑐. Both equations are equivalent to 𝑐𝑥 2 = 𝑏𝑥1 (note that 𝑌 does not have full rank because 𝑌1 = 1, so the two equations are linearly dependent). If 𝑐 = 𝑏 = 0, then any probability distribution 𝑥 is stationary (which is always the case if 𝑌 is the identity matrix), and 𝑥 is not unique. Otherwise, 𝑐 + 𝑏 > 0 and 𝑥 is unique, because 𝑏 𝑐 𝑏 with 𝑥 1 = 1 − 𝑥2 we can rewrite 𝑐𝑥 2 = 𝑏𝑥 1 as 𝑥2 = 𝑐+𝑏 , which implies 𝑥> = ( 𝑐+𝑏 , 𝑐+𝑏 ), as claimed.

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