12th Annual Harvard-MIT Mathematics Tournament #0iwm

12th Annual Harvard-MIT Mathematics TournamentHard

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Understand the scale

Scale 0.5: Easiest math competition problems, often solvable even for early middle school students without experience. 1: Problems strictly for beginners, often at a moderate middle school level. 1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests. 2: For motivated beginners, harder questions from the previous categories. 2.5: More advanced beginner problems, hardest questions from previous categories. 3: Early intermediate problems that require more creative thinking. 3.5: Tougher early intermediate problems that consistently stretch into higher-level creative thinking or conceptual knowledge. 4: Intermediate-level problems. 4.5: Upper intermediate problems approaching the upper bound for non-invitational sprint math competitions. 5: More difficult AIME problems or simple proof-based Olympiad-style problems. 6: High-level AIME-styled questions or introductory Olympiad-level questions. 7: Tougher Olympiad-level questions, may require more technical knowledge. 8: High-level Olympiad-level questions. 9: Expert Olympiad-level questions. 9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve. 10: Historically notorious problems, generally unsuitable for even very hard competitions because they are exceedingly tedious, long, and difficult.

Problem

Problem:

Von Neumann's Poker: The first step in Von Neumann's game is selecting a random number on $[0,1]$ . To generate this number, Chebby uses the factorial base: the number $0 . A_{1} A_{2} A_{3} A_{4} \ldots$ stands for $\sum_{n=0}^{\infty} \frac{A_{n}}{(n+1)!}$ , where each $A_{n}$ is an integer between $0$ and $n$ , inclusive.

Chebby has an infinite number of cards labeled $\mathbf{0}, \mathbf{1}, \mathbf{2}, \ldots$ . He begins by putting cards $\mathbf{0}$ and $\mathbf{1}$ into a hat and drawing randomly to determine $A_{1}$ . The card assigned $A_{1}$ does not get reused. Chebby then adds in card $2$ and draws for $A_{2}$ , 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 $\lfloor 25(1-d)\rfloor$ , where $d$ is the positive difference between your answer and the correct answer.