Question 77
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.
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.