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.
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
All's fair in Love and War (spoiler)
