 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Fair? (Posted on 2015-09-10) 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. Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 3 of 10 | a. If A and B play 1 or 2 an equal number of times, the expected gain for A will be (2-3-3+4)/4 = 0 and the game would be fair.

b. If B plays 2 with probability b and 1 with probability (1-b), then, depending on A's strategy (let's call that probability of 2 is a and probability of 1 is (1-a), we get:

2(1-a)(1-b) - 3a(1-b) - 3(1-a)b + 4ab for A's expected gain (B's loss)

The coefficient of a is -2 + 2b -3 + 3b + 3b + 4b = 12b - 5

B can make it irrelevant what A's probability (i.e., a) is, by setting b = 5/12; the expected value of the game remains the same regardless of a.

If b=5/12 and say a=0, we get

2*7/12 - 3*5/12 = -1/12 expected gain for A; that is, a loss for A and gain for B of 1/12 on average.

To verify it's immune to A's changing strategy, let a=1:

-3*7/12 + 4*5/12 = -1/12

The negative expected gain for A is a positive gain for B of 1/12 dollar, no matter what A's strategy, so long as B plays optimally, randomly choosing 1 7/12 of the time and 2 5/12 of the time.

There is no opposing strategy that A can use to defend against B's optimal strategy of playing 2 5/12 of the time and 1, 7/12.

 Posted by Charlie on 2015-09-10 17:16:31 Please log in:

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