Molina 的罐子：费马大定理的概率版

Molina's Urns

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

Two urns contain the same total numbers of balls, some blacks and some whites in each. From each urn are drawn $n(\geq 3)$ balls with replacement. Find the number of drawings and the composition of the two urns so that the probability that all white balls are drawn from the first urn is equal to the probability that the drawing from the second is either all whites or all blacks.

解析

设两罐总球数相同为 $T$ 。

若第一罐白球为 $\xi$ ，则“连抽 $n$ 次全白”的概率为 $(\xi/T)^n$ 。

第二罐白球 $x$ 、黑球 $y$ （ $x+y=T$ ），则“全白或全黑”概率为 $(x/T)^n+(y/T)^n$ 。

题设等式等价于

\left(\frac{\xi}{T}\right)^n=\left(\frac{x}{T}\right)^n+\left(\frac{y}{T}\right)^n \iff \boxed{\xi^n=x^n+y^n}.

当 $n\ge 3$ 时，按费马大定理（Fermat's Last Theorem），该方程无非平凡正整数解。

因此在 $n\ge 3$ 的要求下， $\boxed{\text{不存在满足条件的罐子构造}}$ 。