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A deterministic probability game


Alice and Bob are going to play a game, with the following rules:
1st Alice picks a probability p, 0 <= p < 0.5
2nd Bob takes any finite number of counters B.
3rd Alice takes any finite number of counters A.
These happen in sequence, so Bob chooses B knowing p, and Alice chooses A knowing p and B.
A series of rounds are then played. Each round, either Bob gives Alice a counter (probability p) or Alice gives Bob a counter (probability 1-p). The game terminates when one player is out of counters, and that player is the loser.
Whom does this game favor?
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2 answers to this question

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Bob has an advantage.

To increase her chances of winning, Alice should pick p as close to 0.5 as possible. Let's say she picks p=0.49, then, on average, for every 49 counters Bob gives her she will give back 51. So, on average, after 100 rounds Bob will have 2 more counters than he started with and Alice will have 2 less. Without doing the math, if Bob picks large enough number of counters, he's almost guaranteed a win. For example, Bob can take p and remove leading "0." (if p=0.4999 then B=4999). It doesn't matter how many counters will Alice pick after that. It will only determine the length of the game, but she will eventually lose.

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After first round:

Bob has given away an expected number of p counters and received expected number of 1-p.

So expected net gain for bob is -p + 1-p = 1-2p.

So expected net loss for Alice is 1 - 2p.

As 2p <1, therefore on an average bob keeps piling up and alice keeps losing. therefore this game is biased towards bob

Edited by m00li

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