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 Fair? (Posted on 2015-09-10)
A and B are playing a game, simultaneously exposing
one or two fingers.
If the total number of fingers is odd, then A pays B that number of dollars.
If itâ€™s even, then B pays A accordingly.

Is it a fair game?

a. Assume random decision by both players.
or
b. Both chose the optimal strategy.

 No Solution Yet Submitted by Ady TZIDON Rating: 4.5000 (2 votes)

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 All's fair in Love and War (spoiler) | Comment 1 of 10
No, it is not fair.  This game favors B.

If B randomly exposes 1 finger with probability 7/12 and 2 with probability 5/12, then on average he expects to win 1/12 of the stake each time.

Using B's strategy,
a) If A always shows one finger, then B's expected win = 7/12*(-2) + 5/12(3) = 1/12
b) If A always shows two fingers, then B's expected win = 7/12*(3) + 5/12(-4) = 1/12
c) So B's expected win is 1/12 no matter what A's random strategy is.

B's mixed strategy of (7/12, 5/12) is calculated by solving for a mix that is indifferent to what A does.

Simple game theory

 Posted by Steve Herman on 2015-09-10 15:55:13

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