In article <36t61m$> in rec.puzzles, David Moews <> wrote:

Taking the payoff to be 1 if A wins and 0 if B wins, and assuming that each player wants to maximize his expected value, this game has value 1-1/e. (This is >1/2, so A does have an edge.)

An optimal strategy for B is to believe all numbers under 1/e and challenge a number x>=1/e with probability 1-1/ex.

An optimal strategy for A is to tell the truth if the random number is >=1/e. Otherwise, A lies, picking a random number from [1/e,1] with probability density function f(x)=e(1/x-1).

The value 1-1/e can easily be verified by showing that it is always attained or bettered for A by our strategy for A, and similarly for B.

David Moews ( )

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