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HMMT 2009 U35

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Von Neumann's Poker: The first step in Von Neumann's game is selecting a random number on [0,1][0,1]. To generate this number, Chebby uses the factorial base: the number 0.A1A2A3A4…0 . A_{1} A_{2} A_{3} A_{4} \ldots stands for ∑n=0∞An(n+1)!\sum_{n=0}^{\infty} \frac{A_{n}}{(n+1)!}, where each AnA_{n} is an integer between 0 and nn, inclusive. Chebby has an infinite number of cards labeled 0,1,2,…\mathbf{0}, \mathbf{1}, \mathbf{2}, \ldots. He begins by putting cards 0\mathbf{0} and 1\mathbf{1} into a hat and drawing randomly to determine A1A_{1}. The card assigned A1A_{1} does not get reused. Chebby then adds in card 2 and draws for A2A_{2}, and continues in this manner to determine the random number. At each step, he only draws one card from two in the hat. Unfortunately, this method does not result in a uniform distribution. What is the expected value of Chebby's final number?

Your score on this problem will be the larger of 0 and ⌊25(1−d)⌋\lfloor 25(1-d)\rfloor, where dd is the positive difference between your answer and the correct answer.

Dataset text NuminaMath olympiads. Quoted with source.

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0.57196

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