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 it’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 1-x. Suppose B plays 1 finger with probability y and 2 fingers with probability 1-y. Then, A would be expected to win 2xy-3x(1-y)-3(1-x)y+4(1-x)(1-y) dollars.
2xy-3x(1-y)-3(1-x)y+4(1-x)(1-y)=2xy-3x+3xy-3y+3xy+4-4x-4y+4xy=12xy-7x-7y+4
If x=7/12, then 12xy-7x-7y+4=7y-49/12-7y+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.
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Posted by Math Man
on 2015-09-17 16:47:47 |
A's strategy
