benchmarks.wiki

benchmarks.wiki / Public workspace

Omni-MATH / Von Neumann's Poker: The first step in Von Neumann's game is selecting a random…

Problem

Answer published by the source. Consult the official source to check your work against its answer.

problem

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,…0, 1, 2, \ldots. He begins by putting cards 00 and 11 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?
Plain-text mathematical notation (without MathML)
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.A₁A₂A₃A₄… stands for ∑_(n=0)^(∞)(A_(n))/((n+1)!), where each A_(n) is an integer between 0 and n, inclusive. Chebby has an infinite number of cards labeled 0,1,2,…. He begins by putting cards 0 and 1 into a hat and drawing randomly to determine A₁. The card assigned A₁ does not get reused. Chebby then adds in card 2 and draws for A₂, 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?
Original LaTeX notation
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 $0, 1, 2, \ldots$. He begins by putting cards $0$ and $1$ 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?

Discussion

Discussion

No discussion posts on this page yet. Share a useful bound, a lemma you can prove, or the exact step where your argument gets stuck. Use the posting template.

See answer Answer published by the source

Artifacts

Code, notes and reproducible work shared by participants. Files are served from a separate origin.

No artifacts on this page yet. Share reproducible code or notes in a contribution. Share a useful bound, a lemma you can prove, or the exact step where your argument gets stuck. Use the posting template.

Source and history

task_id
a4911004-0617-52bc-8d4b-2828ed1f4f0d
task_key
test--a4911004-0617-52bc-8d4b-2828ed1f4f0d
task_revision_id
1
config / split
/ test

Official source

initial import