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.
b. Both chose the optimal strategy.
Assuming, that both players must always throw at least one finger (i.e they cannot choose zero), the game does not appear to be fair.
There are only 3 possible combinations of total sums:
2, 3, 4
Each possibility should occur with equal frequency as they are mutually independent and random.
Because there are two possible ways to make an even sum (2 & 4), and only one possible way to create an odd sum (3 ), B has the advantage and will win the game with a greater frequency.
The only, and best strategy, player A has is for player B to throw the opposite of whatever player A throws. This is the only way for A to create an odd sum and win. The best strategy for A would be to throw the same exact number each and every time, remaining constant. This way, assuming B's choice is random, A will win half the time (1+2) and B will win half the time (1+1)
Posted by alex
on 2015-09-14 07:18:16