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.
Suppose A plays 1 finger with probability x and 2 fingers with probability 1x. Suppose B plays 1 finger with probability y and 2 fingers with probability 1y. Then, A would be expected to win 2xy3x(1y)3(1x)y+4(1x)(1y) dollars.
2xy3x(1y)3(1x)y+4(1x)(1y)=2xy3x+3xy3y+3xy+44x4y+4xy=12xy7x7y+4
If x=7/12, then 12xy7x7y+4=7y49/127y+4=1/12. Therefore, if A plays 1 finger 7/12 of the time and 2 fingers 5/12 of the time, then A will always expect to lose 1/12 dollar. This works regardless of the strategy used by B. Therefore, this is A's best strategy.

Posted by Math Man
on 20150917 16:47:47 