Somegamblingproblems
2.1. Inthestandardgambler’sruinproblemwithtotalstake a andgambler’sstake k andthegambler’s probabilityofwinningateachplayis p,calculatetheprobabilityofruininthefollowingcases;
(a) a =100, k =5, p =0.6;
(b) a =80, k =70, p =0.45;
(c) a =50, k =40, p =0 5 Alsofindtheexpecteddurationineachcase.
For p = 1 2 ,theprobabilityofruin uk andtheexpecteddurationofthegame dk aregivenby uk = s k sa 1 sa ,dk = 1 1 2p k a(1 s k) (1 sa)
(a) uk ≈ 0.132, dk ≈ 409.
(b) uk ≈ 0.866, dk ≈ 592.
(c)For p = 1 2 , uk = a k a ,dk = k(a k)
sothat uk =0 2, dk =400.
2.2. Inacasinogamebasedonthestandardgambler’sruin,thegamblerandthedealereachstartwith20 tokensandonetokenisbetonateachplay.Thegamecontinues untiloneplayerhasnofurthertokens.It isdecreedthattheprobabilitythatanygamblerisruinedis 0.52toprotectthecasino’sprofit.Whatshould theprobabilitythatthegamblerwinsateachplaybe?
Theprobabilityofruinis
u = s k s a 1 sa , where k =20, a =40, p istheprobabilitythatthegamblerwinsateachplay,and s =(1 p)/p.Let r = s 20 .Then u = r/(1+ r),sothat r = u/(1 u)and s = u 1 u 1/20
Finally,with u =0.52, p = 1 1+ s = (1 u)1/20 (1 u)1/20 + u1/20 ≈ 0.498999.
2.3. Findgeneralsolutionsofthefollowingdifferenceequations:
(a) uk+1 4uk +3uk 1 =0; (b) 7uk+2 8uk+1 + uk =0; (c) uk+1 3uk + uk 1 + uk 2 =0
(d) puk+2 uk +(1 p)uk 1 =0, (0 <p< 1)
(a)Thecharacteristicequationis m 2 4m +3=0
whichhasthesolutions m1 =1and m2 =3.Thegeneralsolutionis uk = Amk 1 + Bmk 2 = A +3kB, where A and B areanyconstants.
(b)Thecharacteristicequationis 7m 2 8m +1=0, whichhasthesolutions m1 =1and m2 = 1 7 .Thegeneralsolutionis
k = A + 1 7k B.
(c)Thecharacteristicequationisthecubicequation
,
whichhasthesolutions m1 =1, m2 =1+ √2,and m3 =1 √2.Thegeneralsolutionis uk = A + B(1+ √2)k + C(1 √2)k
(d)Thecharacteristicequationisthecubicequation
3 m +(1 p)=(m 1)(
2 + pm (1 p))=0, whichhasthesolutions m1 =1, m2 = 1 2 + 1 2 √[(4 3p)/p]and m3 = 1 2 1 2 √[(4 3p)/p].Thegeneral solutionis uk = A + Bmk 2 + Cmk 3
2.4 Solvethefollowingdifferenceequationssubjecttothegiven boundaryconditions:
(a) uk+1 6uk +5uk 1 =0,u0 =1,u4 =0; (b) uk+1 2uk + uk 1 =0,u0 =1,u20 =0; (c) dk+1 2dk + dk 1 = 2,d0 =0,d10 =0; (d) uk+2 3uk +2uk 1 =0,u0 =1,u10 =0, 3u9 =2u8 .
(a)Thecharactedristicequationis
m 2 6m +5=0, whichhasthesolutions m1 =1and m2 =5.Thereforethegeneralsolutionisgivenby uk = A +5kB.
Theboundaryconditions u0 =1, u4 =0imply A + B =1,A +54 B =0, whichhavethesolutions A =625/624and B = 1/624.Therequiredsolutionis uk = 625 624 5k 624
(b)Thecharacteristicequationis m 2 2m +1=(m 1)2 =0, whichhastherepeatedsolution m =1.Usingtheruleforrepeatedroots, uk = A + Bk.
Theboundaryconditions u0 =1and u20 =0imply A =1and B = 1/20.Therequiredsolutionis uk =(20 k)/20.
(c)Thisisaninhomogeneousequation.Thecharacteristicequationis m 2 2m +1=(m 1)2 =0, whichhastherepeatedsolution m =1.Hencethecomplementaryfunctionis A + Bk.Foraparticular solution,wemusttry dk = Ck2 .Then
if C = 1.Hencethegeneralsolutionis dk = A + Bk k2 .
Theboundaryconditions d0 = d10 =0imply A =0and B =10.Thereforetherequiredsolutionis dk = k(10 k).
(d)Thecharacteristicequationis
whichhastwosolutions m1 =1(repeated)and m2 = 2.Thegeneralsolutionisgivenby uk = A + Bk + C( 2)k
Theboundaryconditionsimply
Thesolutionsoftheselinearequationsare
sothattherequiredsolutionis
2.5. Showthatadifferenceequationoftheform auk+2 + buk+1
, where a,b,c ≥ 0 areprobabilitieswith a + b + c =1,canneverhaveacharacteristicequationwithcomplex roots.
Thecharacteristicequationcanbeexpressedintheform am 3 + bm
+
)m (1 a b)]=0, since a + b + c =1.Onesolutionis m1 =1,andtheotherssatisfythequadraticequation
Thediscriminantisgivenby
since0 ≤ a ≤ 1.
2.6. Inthestandardgambler’sruinproblemwithequalprobabilities p = q = 1 2 ,findtheexpectedduration ofthegamegiventheusualinitialstakesof k unitsforthegamblerand a k unitsfortheopponent.
Theexpectedduration dk satisfies
dk+1 2dk + dk 1 = 2
Thecomplementaryfunctionis A + Bk,andforaparticularsolutiontry dk = Ck2 .Then
if C = 1.Hence
dk = A + Bk k2
Theboundaryconditions d0 = da =0imply A =0and B = a.Therequiredsolutionistherefore dk = k(a k)
2.7. Inagambler’sruinproblemthepossibilityofadrawisincluded.Lettheprobabilitythatthegambler wins,losesordrawsagainstanopponentberespectively, p,p, 1 2p, (0 <p< 1 2 ).Findtheprobabilitythat thegamblerlosesthegamegiventheusualinitialstakesof k unitsforthegamblerand a k unitsforthe opponent.Showthat dk,theexpecteddurationofthegame,satisfies pdk+1 2pdk + pdk 1 = 1.
Solvethedifferenceequationandfindtheexpecteddurationof thegame.
Thedifferenceequationfortheprobabilityofruin uk is uk = puk+1 +(1 2p)uk + puk
Thegeneralsolutionis uk = A + Bk.Theboundaryconditions u0 =1and ua =0imply A =1and B = 1/a,sothattherequiredprobabilityisgivenby uk =(a k)/a Theexpectedduration dk satisfies
Thecomplementaryfunctionis A + Bk.Fortheparticularsolutiontry dk = Ck2.Then
Hence C = 1/(2p).Theboundaryconditions d0 = da =0imply A =0and B = a/(2p),sothatthe requiredsolutionis dk = k(a 2p)/(2p)
2.8. Inthechangingstakesgameinwhichagameisreplayedwitheachplayerhavingtwiceasmanyunits, 2k and 2(a k) respectively,supposethattheprobabilityofawinforthegamblerateachplayis 1 2 .Whilst theprobabilityofruinisunaffectedbyhowmuchistheexpecteddurationofthegameextendedcompared withtheoriginalgame?
Withinitialstakesof k and a k,theexpecteddurationis dk = k(a k).Iftheinitialstakesare doubledto2k and2a 2k,thentheexpecteddurationbecomes,usingthesameformula,
2.9. Aroulettewheelhas37radialslotsofwhich 18 arered, 18 areblackand 1 isgreen.Thegamblerbets oneunitoneitherredorblack.Iftheballfallsintoaslotof thesamecolour,thenthegamblerwinsone unit,andiftheballfallsintotheothercolour(redorblack),thenthecasinowins.Iftheballlandsinthe greenslot,thenthebetremainsforthenextspinofthewheel ormoreifnecessaryuntiltheballlandson aredorblack.Theoriginalbetiseitherreturnedorlostdependingonwhethertheoutcomematchesthe
originalbetornot(thisistheMonteCarlosystem).Showthattheprobability uk ofruinforagambler whostartswith k chipswiththecasinoholdiing a k chipssatisfiesthedifferenceequation
Solvethedifferenceequationfor uk.Ifthehousestartswith ∈1,000,000attheroulettewheelandthe gamblerstartswith ∈10,000,whatistheprobabilitythatthegamblerbreaksthebankif ∈5,000arebetat eachplay.
IntheUSsystemtherulesarelessgeneroustotheplayers.If theballlandsongreenthentheplayer simplyloses.Whatistheprobabilitynowthattheplayerwinsgiventhesameinitialstakes?(seeLuenberger (1979))
Thereisthepossibilityofadraw(seeExample2.1).Ateachplaytheprobabilitythatthegambler winsis p = 18 37 .Thestakeisreturnedwithprobability
orthegamblerlosesafteroneormoregreensalsowithprobability1/74bythesameargument.Hence uk, theprobabilitythatthegamblerlosessatisfies
Thecharactersiticequationis
, whichhasthesolutions m1 =1and m2
Thebetsareequivalentto k =10000/5000=2, a =1010000/5000=202.Theprobabilitythatthe gambler wins is
IntheUSsystem, uk satisfies
inthiscasetheratiois s ′ =19/18.Hencetheprobabilitythethegamblerwinsis
whichislessthanthepreviousvalue.
2.10. Inasingletrialthepossiblescores 1 and 2 canoccureachwithprobability 1 2 .If pn istheprobability ofscoringexactly n pointsatsomestage,thatisscoreafterseveraltrialsisthesumofindividualscores ineachtrial.Showthat
Calculate p1 and p2,andfindaformulafor pm.Howdoes pm behaveas m becomeslarge?Howdoyou interprettheresult?
Let An betheeventthatthescoreis n atsomestage.Let B1 betheeventscore1,and B2 score2. Then
Hence
Thecharacteristicequationis
2.11. Inasingletrialthepossiblescores 1 and 2 canoccurwithprobabilities q and 1 q,where 0 <q< 1 Findtheprobabilityofscoringexactly n pointsatsomestageinanindefinitesuccessionoftrials.Show that
Let pn betheprobability.Then
Thecharacteristicequationis
whichhasthesolutions m1 =1and m2 = (1 q).Hence pn = A + B(q
Theinitialconditionsare
,whichimply
Thesolutionoftheseequationsleadsto A =1/(2 q)and B =(q 1)/(q 2),sothat
2.12. Theprobabilityofsuccessinasingletrialis 1 3 .If un istheprobabilitythattherearenotwoconsecutive successesin n trials,showthat un satisfies
Whatarethevaluesof u1 and u2 ?Henceshowthat
Let An betheeventthattherehave not beentwoconsecutivesuccessesinthefirst n trials.Let B1 be theeventofsuccessand B2 theeventoffailureinasingletrial.Then
Now P(An|B2 )= P(An 1 ):failurewillnotchangetheprobability.Also
Since P(B1 )= 1 3 , P(B2 )= 2 3 ,
, where un = P(An).
Thecharacteristicequationis
whichhasthesolutions
Theinitialconditionsare u1 =1and
Thesolutionsare
2.13. Agamblerwithinitialcapital k unitsplaysagainstanopponentwithinitialcapital a k units.At eachplayofthegamethegamblereitherwinsoneunitorloses oneunitwithprobability 1 2 .Whenever theopponentlosesthegame,thegamblerreturnsoneunitsothatthegamemaycontinue.Showthatthe expecteddurationofthegameis k(2a 1 k) plays.
Theexpectedduration dk satisfies
Theboundaryconditionsare d0 =0and da = da 1,indicatingthereturnofoneunitwhenthegambler loses.Thegeneralsolutionforthedurationis
Theboundaryconditionsimply A =0,A + Ba a 2 = A + B(a 1) (a 1)2 , sothat B =2a 1.Hence dk = k(2a 1 k).
2.14. Intheusualgambler’sruinproblem,theprobabilitythatthegambleriseventuallyruinedis
Inanewgamethestakesarehalved,whilsttheplayersstartwiththesameinitialsums.Howdoesthis affecttheprobabilityoflosingbythegambler?Shouldthegambleragreetothischangeofruleif p< 1 2 ? Byhowmanyplaysistheexpecteddurationofthegameextended?
Thenewprobabilityofruin vk (withthestakeshalved)is,adaptingtheformulafor uk,
=
Given p< 1 2 ,then s =(1 p)/p> 1and s k > 1.Itfollowsthat
Withthischangethegamblerismorelikelytolose. From(2.9),theexpecteddurationofthestandardgameisgivenby
Withthestakeshalvedtheexpectedduration hk is
Theexpecteddurationisextendedby
2.15. Inagambler’sruingame,supposethatthegamblercanwin£2withprobability 1 3 orlose£1with probability 2 3 .Showthat
Compute uk if a =9 for k =1, 2,..., 8.
Theprobabilityofruin uk satisfies
Thecharacteristicequationis
whichhasthesolutions m1 =1(repeated)and m2 = 2.Hence
Theboundaryconditionsare
Thesolutionoftheseequationsis
Thevaluesoftheprobabilities uk for a =9areshowninthetablebelow.
2.16. Findthegeneralsolutionofthedifferenceequation
Areservoirwithtotalcapacityof a volumeunitsofwaterhas,duringeachday,eitheranetinflow oftwounitswithprobability 1 3 oranetoutflowofoneunitwithprobability 2 3 .Ifthereservoirisfullor nearlyfullanyexcessinflowislostinanoverflow.Deriveadifferenceequationforthismodelfor uk,the probabilitythatthereservoirwilleventuallybecomeemptygiventhatitinitiallycontains k units.Explain whytheupperboundaryconditionscanbewritten ua = ua 1 and ua = ua 2.Showthatthereservoiris certaintobeemptyatsometimeinthefuture.
Thecharacteristicequationis
Thegeneralsolutionis(seeProblem2.15)
Theboundaryconditionsforthereservoirare
Thelattertwoconditionsareequivalentto
whichhavethesolutions A =1, B = C =0.Thesolutionis uk =1,whichmeansthatthatthereservoir iscertaintoemptyatsomefuturedate.
2.17. Considerthestandardgambler’sruinprobleminwhichthetotalstakeis a andgambler’sstakeis k, andthegambler’sprobabilityofwinningateachplayis p andlosingis q =1 p.Find uk,theprobability ofthegamblerlosingthegame,bythefollowingalternative method.Listthedifferenceequation(2.2)as
, where s = q/p = 1 2 and k =2, 3,...a.Theboundarycondition u0 =1 hasbeenusedinthefirstequation. Byaddingtheequationsshowthat
Determine u1 fromtheotherboundarycondition ua =0,andhencefind uk.Adaptthesamemethodfor thespecialcase p = q = 1 2
Additionoftheequationsgives
)=(u1 1) s s k 1 s summingthegeometricseries.Thecondition ua =0implies
Hence
sothat
2.18. Acarparkhas100parkingspaces.Carsarriveandleaverandomly.Arrivalsordeparturesofcars areequallylikely,anditisassumedthatsimultaneouseventshavenegligibleprobability.The‘state’ofthe carparkchangeswheneveracararrivesordeparts.Giventhatatsomeinstantthereare k carsinthecar park,let uk betheprobabilitythatthecarparkfirstbecomesfullbefore itbecomesempty.Whatarethe boundaryconditionsfor u0 and u100 ?Howmanycarmovementscanbeexpectedbeforethisoccurs?
Theprobability uk satisfiesthedifferenceequation
Thegeneralsolutionis uk = A + Bk.Theboundaryconditionsare u0 =0and u100 =1.Hence A =0 and B =1/100,and uk = k/100.
Theexpecteddurationofcarmovementsuntilthecarparkbecomesfullis dk = k(100 k).
2.19. Inastandardgambler’sruinproblemwiththeusualparameters,theprobabilitythatthegamblerloses isgivenby
,s = 1 p p .
If p iscloseto 1 2 ,givensayby p = 1 2 + ε where |ε| issmall,show,byusingbinomialexpansions,that uk = a k a
)ε
as ε → 0.(Theorder O terminologyisdefinedasfollows:wesaythatafunction g(ε)= O(ε b)as ε → 0 if g(ε)/εb isboundedinaneighbourhoodwhichcontains ε =0.SeealsotheAppendixinthebook.)
Let p = 1 2 + ε.Then s =(1 2ε)/(1+2ε),and uk = (1 2ε)k(1+2
Applythebinomialtheoremtoeachterm.Theresultis
[Symboliccomputationoftheseriesisausefulcheck.]
2.20. Agamblerplaysagameagainstacasinoaccordingtothefollowingrules.Thegamblerandcasino eachstartwith10chips.Fromadeckof53playingcardswhich includesajoker,cardsarerandomlyand successivelydrawnwithreplacement.Ifthecardisredorthejokerthecasinowins1chipfromthegambler, andifthecardisblackthegamblerwins1chipfromthecasino.Thegamecontinuesuntileitherplayer hasnochips.Whatistheprobabilitythatthegamblerwins?Whatwillbetheexpecteddurationofthe game?
From(2.4)theprobability uk thatthegamblerlosesis
uk = s k s a 1 sa , with k =10, a =20, p =26/53,and s =27/26.Hence u10 = (27/26)10 (27/26)20 1 (27/26)20
Thereforetheprobabilitythatthegamblerwinsisapproximately0.407. From(2.10) dk = 1 1 2p k a(1 s k 1 sa =98 84, forthegivendata.
2.21. Inthestandardgambler’sruinproblemwithtotalstake a andgambler’sstake k,theprobabilitythat thegamblerlosesis
uk = s k s a 1 sa ,
where s =(1 p)/p.Supposethat uk = 1 2 ,thatisfairodds.Express k asafunctionof a.Showthat, k = ln[ 1 2 (1+ s a)] ln s
Given uk = s k s a 1 sa and uk = 1 2 , then1 s a =2(s k s a)or s k = 1 2 (1+ s a).Hence k = ln[ 1 2 (1+ sa)] ln s , butgenerally k willnotbeaninteger.
2.22. Inagambler’sruingametheprobabilitythatthegamblerwinsateachplayis αk andlosesis 1 αk, (0 <αk < 1, 0 ≤ k ≤ a 1),thatis,theprobabilityvarieswiththecurrentstake.The probability uk that thegamblereventuallylosessatisfies uk = αkuk+1 +(1 αk)uk 1,uo =1,ua =0.
Supposethat uk isaspecifiedfunctionsuchthat 0 <uk < 1, (1 ≤ k ≤ a 1), u0 =1 and ua =0.Express αk intermsof uk 1 , uk and uk+1.
Find αk inthefollowingcases:
(a) uk =(a k)/a; (b) uk =(a 2 k2)/a2 ; (c) uk = 1 2 [1+cos(kπ/a)]
Fromthedifferenceequation
(a) uk =(a k)/a.Then
whichistobeanticipatedfromeqn(2.5).
(b) uk =(a 2 k2 )/a2 .Then
(c) uk =1/(a + k).Then
2.23. Inagambler’sruingametheprobabilitythatthegamblerwinsateachplayis αk andlosesis 1 αk, (0 <αk < 1, 1 ≤ k ≤ a 1),thatis,theprobabilityvarieswiththecurrentstake.The probability uk that thegamblereventuallylosessatisfies
Reformulatethedifferenceequationas
, where βk =(1 αk)/αk.Henceshowthat
Usingtheboundaryconditionat k = a,confirmthat
Checkthatthisformulagivestheusualanswerif αk = p = 1 2 ,aconstant.
Thedifferenceequationcanbeexpressedintheequivalentform
where βk =(1 αk)/αk.Nowlisttheequationsasfollows,notingthat u
Addingtheseequations,weobtain
Thecondition ua =0implies
sothat
Finally
Hence
asrequired.
2.24. Supposethatafair n-sideddieisrolled n independenttimes.Amatchissaidtooccurifside i is observedonthe ithtrial,where i =1, 2,...,n
(a)Showthattheprobabilityofatleastonematchis
(b)Whatisthelimitofthisprobabilityas n →∞?
(c)Whatistheprobabilitythatjustonematchoccursin n trials?
(d)Whatvaluedoesthisprobabilityapproachas n →∞?
(e)Whatistheprobabilitythattwoormorematchesoccurin n trials?
(a)Theprobabilityof no matchesis
Theprobabilityofatleastonematchis
Henceforlarge n,theprobabilityofatleastonematchapproaches1 e 1 =(e 1)/e (c)Thereisonlyonematchwithprobability
(e)Probabilityoftwoormorematchesis
2.25. (Kelly’sstrategy)Agamblerplaysarepeatedfavourablegameinwhichthegamblerwinswithprobability p> 1 2 andloseswithprobability q =1 p.Thegamblerstartswithaninitialoutlay K0 (insome currency).Forthefirstgametheplayerbetsaproportion rK0,(0 <r< 1).Henceafterthisplaythestake is K0 (1+ r) afterawinor K0 (1 r) afterlosing.Subsequently,thegamblerbetsthesameproportionof thecurrentstakeateachplay.Hence,after n playsofwhich wn arewinsthestake Sr willbe
Constructthefunction
Whatistheexpectedvalueof Gn(r) forlarge n?Forwhatvaluesof r isthisexpectedvalueamaximum? Thisvalueof r indicatesasafebettingleveltomaximisethegain,althoughataslowrate.Youmight considerwhythelogarithmischosen:thisisknownasa utilityfunction ingamblingandeconomics.It isamatterofchoiceandisabalancebetweenhavingareasonablegainagainsthavingahighriskgain. Calculate r if p =0 55.[Attheextremes r =0 correspondstonobetwhilst r =1 tobetting K0 thewhole stakeinonegowhichcouldbecatastrophic.]
From
itfollowsthat
n(r)= 1 n ln[(1+ r)wn (1 r)n wn = wn n ln(1+ r)+ (n wn) n ln(1 r).
G + n(r)isourchosenutilityfunction.Forlarge n,theratio wn/n isapproximatelytheprobability p,and (n wn)/n theprobability q.Hence Gn(r)approaches
H(r)= p ln(1+ r)+ q ln(1 r)
Thisfunctioniszerowhere r =0andapproaches −∞ as r → 1 .Also
H ′(r)= p 1+ r q 1 r
Hence H ′(0)= p + q =1 > 0,sothattheslopeispositivewhichimpliesthat H(r)hasamaximumforin 0 <r< 1.Thisoccursat H ′(r)=0wherewhere rm =2p 1sothat
H(rm)= p ln(2p)+(1 p)ln(2 2p)
Thefunction Gn(r)isanexampleofautilityfunctionwhichattemptstoavoidpossiblecatastrophic losses,andkeepsgainsmodest.
Asanexamplelet p =0 55.Then rm =2p 1=0 1.Supposethat n =100and wn =55(reflecting theodds).Thenforthisvalueof rm itcanbecalculatedthat K100 (rm) ≈ 1 65K0