不公平得分游戏

每局你得分概率 $p<1/2$，总局数必须为偶数 $2m$，要赢需得分 $>m$。

由于 $p<1/2$，$\mathrm{Bin}(2m,p)$ 的均值为 $2mp<m$，随着 $m$ 增大，“超过一半”的上尾概率会更小（直观上偏离均值更远；可用 Chernoff 界/大数定律严格化）。

因此应选择最小的偶数局数：

$$\boxed{2\text{ 局}}.$$

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英文解析

Don’t balk just because the game is unfair; after all you are the only one eligible for a prize. Let us call you player $A$ and your opponent player $B$ . Let the total number of plays be $N = 2n$ . On a given play, your chance of winning a point is $p$ , your opponent’s $q = 1 - p$ .

At first blush, most people notice that the game is unfair and therefore that, as $N$ increases, the expected value of the difference ( $A$ ’s points $- B$ ’s points) grows more and more negative. They conclude that $A$ should play as little as he can and still win- that is, two plays.

Had an odd number of plays been allowed, this reasoning based on expected values

would have led to the correct answer, and $A$ should choose only one play. With an even number of plays, two opposing effects are at work: (1) the bias in favor of $B$ , and (2) the redistribution of the probability in the middle term of the binomial distribution (the probability of a tie) as the number of plays increases.

Consider, for a moment, a fair game $\left(p = \frac{1}{2}\right)$ . Then the larger $N$ , the larger $A$ 's chance to win because as $2n$ increases, the probability of a tie tends to zero, and the limiting value of $A$ 's chance to win is $\frac{1}{2}$ . For $N = 2,4,6$ , his probabilities are $\frac{1}{4}, \frac{5}{16}, \frac{22}{64}$ . Continuity suggests that for $p$ slightly less than $\frac{1}{2}$ , $A$ should choose a large but finite number of plays. But if $p$ is small, $N = 2$ should be optimum for $A$ . It turns out that for $p < \frac{1}{3}$ , $N = 2$ is optimum.

Your probability of winning in a game of $2n$ trials is the sum of the probabilities of getting $n + 1, n + 2, \ldots , 2n$ points, a sum given by

$$P_{2n} = \sum_{x = n + 1}^{2n} \binom{2n}{x} p^x q^{2n - x}.$$

In a game of $2n + 2$ plays, the probability of winning at least $n + 2$ points and the game is

$$P_{2n + 2} = \sum_{x = n + 2}^{2n + 2} \binom{2n + 2}{x} p^x q^{2n + 2 - x}.$$

A game composed of $2n + 2$ plays can be regarded as having been created by adding two plays to a game of $2n$ plays. Unless player $A$ has won either $n$ or $n + 1$ times in the $2n$ game, his status as a winner or loser cannot differ in the $2n + 2$ game from that in the $2n$ game.

Except for these two possibilities, $P_{2n + 2}$ would be identical with $P_{2n}$ . These exceptions are: (1) having $n + 1$ successes in the first $2n$ plays, $A$ loses the next two, thus reducing his probability of winning in the $2n + 2$ game by

$$q^{2} \binom{2n}{n + 1} p^{n + 1} q^{n - 1}$$

or (2) having won $n$ plays in the $2n$ game, he wins the next two, increasing his prob-

ability by

$$p^{2}\left( \begin{array}{c}2n \\ n \end{array} \right) p^{n} q^{n}$$

If $N = 2n$ is the optimum, value, then both $P_{N - 2} \leq P_{N}$ and $P_{N} \geq P_{N + 2}$ must hold. The results of the previous paragraph imply that these inequalities are equivalent to the following two inequalities:

$$\begin{array}{l}{{q^{2}\left(\begin{array}{c}{{2n-2}}\\ {{n}}\end{array}\right)p^{n}q^{n-2}\leq p^{2}\left(\begin{array}{c}{{2n-2}}\\ {{n-1}}\end{array}\right)p^{n-1}q^{n-1},}}\\ {{q^{2}\left(\begin{array}{c}{{2n}}\\ {{n+1}}\end{array}\right)p^{n+1}q^{n-1}\geq p^{2}\left(\begin{array}{c}{{2n}}\\ {{n}}\end{array}\right)p^{n}q^{n}.}}\end{array}$$

After some simplifications, which you may wish to verify (we exclude the trivial case $p = 0$ ), we reduce inequalities (1) to

$$(n - 1)q \leq np; \quad nq \geq (n + 1)p$$

These inequalities yield, after a little algebra, the condition

$$\frac{1}{1 - 2p} - 1 \leq 2n \leq \frac{1}{1 - 2p} + 1.$$

Thus unless $1 / (1 - 2p)$ is an odd integer, $N$ is uniquely determined as the nearest even integer to $1 / (1 - 2p)$ . When $1 / (1 - 2p)$ is an odd integer, both adjacent even integers give the same optimum probability. And we can incidentally prove that when $1 / (1 - 2p) = 2n + 1$ , $P_{2n} = P_{2n + 2}$ .

Consequently for $p = 0.45$ , we have $1 / (1 - 0.9) = 10$ as the optimum number of plays to choose.

This material is abbreviated from "Optimal length of play for a binomial game," Mathematics Teacher, Vol. 54, 1961, pp. 411- 412.

P.
G. Fox originally alluded to a result which gives rise to this game in "A primer for chumps," which appeared in the Saturday Evening Post, November 21, 1959, and discussed the idea further in private correspondence arising from that article in a note entitled "A curiosity in the binomial expansion-and a lesson in logic." I am indebted to Clayton Rawson and John Scarne for alerting me to Fox's paper and to Fox for

helpful correspondence.