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Friday, May 3, 2013

A probability game

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Question: At each turn, if Player A has p coins and Player B has np coins, then the chance of A wins to the chance of B wins is p:np and that whoever wins gain 1 coin from the loser. The game ends if a player gets all the n coins. If currently A has nA coins and B has nnA coins, what is the probability that A is the final winner?



A B
Wealth: p coins Wealth: np coins
Winning chance: p/n Winning chance: 1p/n




Answer: Let Pp denotes the probability that A is the final winner if A has p coins. Then Pp=pnPp+1+npnPp1,P0=0,Pn=1.
Then npn(PpPp1)=pn(Pp+1Pp)
and thus Pp+1Pp=npp(PpPp1)=(np)(n(p1))(n1)p(p1)1(P1P0)=Cn1pP1.
Sum over all Pp from p=0 to r1, we have Pr=r1p=0(Pr+1Pr)=r1p=0Cn1pP1.
Consider the case r=n. 1=n1p=0Cn1pP1=2n1P1
and hence P1=12n1.
The final answer is therefore PnA=nA1p=0Cn1p2n1.

The wealthier or more powerful A is, the higher the chance A wins each battle. A reasonable model, isn't it?

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