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Fair? (Posted on 2015-09-10) Difficulty: 3 of 5
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.

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

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 6 of 10 |
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
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