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Alex Miller Logan Johnson Midterm Paper

In many cases, mathematical concepts can be demonstrated with playing cards. That is what Martin Gardner demonstrated in an article published in The College Mathematics Journal titled, “Modeling Mathematics with Playing Cards”. As Gardner mentioned, because playing cards have fronts and backs, two colors, four suits, and values ranging from one through thirteen, there are many possibilities that a deck of 52 cards could represent in terms of mathematical concepts. Throughout Gardner’s article, examples of mathematical problems that are easily demonstrated with playing cards are presented. A few of Gardner’s examples will be described further in depth. Sequences and Logical Induction Let a sequence be defined as an ordered list of objects called terms. Let sequence D1 = R1, B2, R3, B4, ..., B26 and let the sequence D2 = B1, R2, B3, R4, ..., R26. From these two sequences, form the new combined sequence C = C1, C2, C3, C4,..., C52 by letting the first -terms be the first . Then, let the

through the

th

terms of C be the first

-terms of D1, where

-terms of D2, where

. Continue in this fashion until all terms from D1 and D2 are used to create C. Prove that from the start of sequence C, the paired terms (C1, C2) consist of an R-term and a B-term, the pair (C3, C4) consist of an R-term and a B-term, ..., and the pair (C51, C52) consist of an R-term and a B-term.


We can prove that the first pair, (C1, C2), will consist of an R and B term by using two cases. Case 1: Both first terms in C will either come from the first two terms in D1 or the first two terms in D2. This will result in an R and B term pair. Case 2: The first pair in C will consist of the first term in D1 and the first term in D2. This will result in an R and B term pair. By using informal, logical induction we can see that after this first pair of terms is created, all pairs of terms created after will all consist of an R and B term. In either case 1 or 2, after the first pair of terms is created, the sequences D1 and D2 look identical to how they were at before the first pair of terms was created. It then follows that either case 1 or case 2 can then be performed again, creating the next pair of terms in C, being (C3, C4). By this creation process of sequence C, it holds true that the paired terms (C1, C2), (C3, C4), ..., (C51, C52) will always consist of an R-term and a B-term.

Set Construction and Manipulation First, let there be two halves of a deck of cards separated so that all the red cards are in one pile and all the black cards are in another pile. Then let there be random n cards removed from the red pile and inserted into the black pile and let the black pile be shuffled. Lastly, remove n amount of cards from the newly shuffled black half and insert them into the red half. Finally, there will be exactly the same amount of red cards in one pile that there are black cards in the other pile. Here is a proof of this theorem. There are equal numbers of black and red cards to begin with. Then there are k red cards in one pile and k black cards in the other pile, where k is a non-negative integer. Since n is the random number of cards removed from the red pile and shuffled into the black pile, n has to be a positive integer less


than or equal to k. When n random cards are removed from the shuffled pile of k black and n red cards, each removed card is either a black or red card. Then the n random cards removed from the shuffled pile consists of x red cards and y black cards, where x and y are non-negative integers x+y=n. After the n random cards that were removed from the shuffled pile are put into the other pile, the number of red cards in the pile that originally consisted of all red cards is k-n+x, and that pile also contains y black cards. The pile that originally consisted of all black cards now has k-y black cards and n-x red cards. Since x=n-y, the number of red cards in the pile that originally consisted of all red cards is k-n+(n-y)=k-y, which is equal to the number of black cards in the pile that originally consisted of all black cards, and the number of red cards in the pile that originally consisted of all black cards is n-(n-y)=y, which is equal to the number of black cards in the pile that originally had all red cards. Since k, n, x, and y were arbitrary, the above theorem holds true for any k, n, x, and y. This can be further be illustrated by set notation and manipulation: Let… R = {R1,R2,R3,…,Rk} = The set of all red cards B = {B1,B2,B3,…,Bk} = The set of all black cards N = {R1,R2,R3,…,Rn | n<k} = The set of n cards removed from the red pile Then… R-N = {Rn+1,Rn+2,Rn+3,…,Rk} = The pile of red cards after n cards removed BUN = { B1,B2,B3,…,Bk,R1,R2,R3,…,Rn} = The Set of black cards combined with the removed red cards Let… M = {B1,B2,…,By,R1,R2,…,Rx | x<n, y<n, x+y=n} = The set of cards removed from BUN


Then… (BUN)-M = {By+1,By+2,By+3,…,Bk,Rx+1,Rx+2,Rx+3,…,Rn} = The set that was BUN with cards in M removed (R-N)UM = { Rn+1,Rn+2,Rn+3,…,Rk,B1,B2,…,By,R1,R2,…,Rx} = The set that was R-N with cards in M added Theorems: (i) (ii)

|((BUN)-M)-B)| = |((R-N)UM)-R)| |((BUN)-M)-R)| = |((R-N)UM)-B)|

Proof: |R| = |B| = |{1,2,3,…,k}| |N| = |{1,2,3,…,n}| |R-N| = |{1,2,3,…,k-n}| |BUN| = |{1,2,3,…,k+n}| |M| = |{1,2,3,…,x+y}| = |{1,2,3,…,n}| = |N| |(BUN)-M| = |{1,2,3,…,k+n-x-y}| = |{1,2,3,…,k+n-n}| = |{1,2,3,…,k}| |((BUN)-M)-B)| = |{1,2,3,…,k+n-x-y-(k-y)| since there are (k-y) black cards removed in (BUN)-M = |{1,2,3,…,k+n-x-y-k+y}| = |{1,2,3,…,n-x}| |(R-N)UM| = |{1,2,3,…,k-n+x+y}| = |{1,2,3,…,k+n-n}| = |{1,2,3,…,k}| |((R-N)UM)-R)| = |{1,2,3,…,k-n+x+y-(k-n+x)}| since there are (k-n+x) red cards removed in (R-N)UM =|{1,2,3,…,k-n+x+y-k+n-x}| = |{1,2,3,…,y}| = |{1,2,3,…,n-x}| since y=n-x. Therefore (i) |((BUN)-M)-B)| = |((R-N)UM)-R)| |((BUN)-M)-R)| = |{1,2,3,…,k+n-x-y-(n-x)}| since there are (n-x) red cards removed in (BUN)-M = |{1,2,3,…,k+n-x-y-n+x}| = |{1,2,3,…,k-y}| |((R-N)UM)-B)| = |{1,2,3,…,k-n+x+y-(y)}| since there are (y) black cards removed in (R-N)UM =|{1,2,3,…,k-n+n-y}| since x+y=n =|{1,2,3,…,k-y}| Therefore (ii) |((BUN)-M)-R)| = |((R-N)UM)-B)|


Probability and Choice This problem deals with the probability of making the correct choice based on odds of what you know and what you don't know. It is a famous problem that has been dubbed "The Monty Hall Problem" after the famous game show host of Let's Make a Deal, Monty Hall. The problem begins with there being three random choices the player can make. Two of the choices result in nothing, while the third choice results in winning. The player does not know which choice will yield a win, but the host of the game knows. After the player makes his choice, the host reveals one of the other two selections, which is always one of the non-winning choices. The player is then allowed to change their selection if they want to the other remaining choice. This is where the confusion, and the proof, comes in on whether it is smarter to always change or stick with your original choice. Here is a proof, part of the proof being Bayes' Theorem. Let the three choices that the player can make be A, B, and C. The probability that choice A is correct is P(A) = 1/3, the probability that choice B is correct is P(B) = 1/3, and the probability that choice C is correct is P(C) = 1/3. Let's say that the player chooses A. The probability that the host reveals choice B if choice A was the winning choice is P(A|B) = 1/2. The probability that the host would reveal choice B if choice B was the correct one would be P(B|B) = 0. The probability that the host would reveal choice B if choice C were the correct one would be (B|C) = 1. Finally, the probability that the host would actually reveal choice B is P(B) = P(A)*(P(B|A)) + P(B)*(P(B|B)) + P(C)*(P(B|C)) = (1/3)(1/2) + (1/3)(0) + (1/3)(1) = (1/6) + (1/3) = 1/2. Now we can use Bayes' Theorem to decide the final outcomes and if you should indeed switch you choice.


Let's examine the odds if you decide to stick with your choice of A. It would be P(A|P(B)) = (P(A)*(P(B|A)))/P(B) = ((1/3)*(1/2))/(1/2) = 1/3. So your chances of winning, if staying with your original selection is 1/3. Which makes sense because since you stayed your odds are the same as when you first chose at the beginning, namely 1/3. Now let's see what would happen, using Bayes' Theorem, if you switched choice and went with C. P(C|P(B)) = (P(C)*(P(B|C)))/P(B) = ((1/3)*(1))/(1/2) = 2/3. As you can see, your chances of winning increase to two thirds if you always switch. The only time you would lose by switching is if you had already chosen the correct one in the first place. However, the chance of this happening is only 1/3, so based on that the chances that you chose the correct one is less than not choosing it. Lastly, here is a diagram that can help explain this age-old problem. This graph has the winning choice being a car and the losing choices being a goat, which was usually the case in the game show Let's Make a Deal.


REFERENCES

Gardner, M. (2000). Modeling mathematics with playing cards. The College Mathematics Journal, 31(3), Retrieved from http://www.jstor.org/stable/2687484 . Pape, S.J., & Tchoshanov, M.A. (2001). The role of representation(s) in developing mathematical understanding. Theory into Practice, 40(2), Retrieved from http://www.jstor.org/stable/1477273?seq=1. Smith, D, Eggen, M, & St. Andre, R. (2011). A transition to advnaced mathematics. Boston, MA: Brooks/Cole.

MAT320 Midterm Paper  

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