你抛 5 枚、我抛 4 枚

toss four coins

专题: Probability / 概率
难度: L4
来源: QuantQuestion

个人笔记

题目详情

概率题：你抛 5 枚、我抛 4 枚：赢的概率。

英文原题

I am going to toss four coins. You are going to toss five coins. You win if you get strictly more heads than I do. What is the probability that you win?

解析

答案为 $\boxed{1/2}$ 。

把游戏改写为等价形式：双方都先抛 4 枚，若你严格多于我则你赢、若少于则你输；若恰好相等，则你再抛第 5 枚硬币作为“加赛”，正面则你赢，反面则你输。

在前 4 枚阶段，双方完全对称，胜负各占一半；在平局时加赛也对称，因此总体你赢的概率为 $1/2$ 。

英文解析

I give an elegant answer first, and then a hammer- and- tongs solution that could be useful for variations of the game.

FIRST SOLUTION

The original game is stated as "You toss five coins and I toss four coins. You win if you get strictly more heads." This game is isomorphic to an analytically simpler game.15 "You toss four coins and I toss four coins. Whomever gets the most heads wins immediately. If we are tied, however, then you toss one more coin to decide the outcome." By observation, both stages of this second game are completely symmetric. So, we each have a probability one half of winning.

SECOND SOLUTION

A simple sketch of the joint density of our outcomes, and an appeal to symmetry, gives the answer as $1 / 2$ without any calculation. I derive the answer graphically, with more details than you need.

The key to this more complex solution is to recognize that when $p = 1 / 2$ , the binomial density $B(N, p)$ that counts the number of heads of either player is symmetric, and that two people tossing coins independently of each other means that the joint density is just the product of the marginals.16

To make the intuition clear, I will consider first the simpler case where you toss only two coins and I toss only one. I give all the fine details to make it clear, which may be useful in variations of the game, but in fact, no calculation is needed. Each toss is a Bernoulli trial with probability of success (i.e., a head) equal to $1 / 2$ . The tosses are independent, so the count of heads $Y$ that you get is distributed binomial: $f_{Y}(y) = P(Y = y) = \left( \begin{array}{c}N \\ y \end{array} \right)p^{y}q^{N - y}$ , where $N = 2, p = \frac{1}{2}, q \equiv 1 - p = \frac{1}{2}$ . So, $f_{Y}(y) = \left( \begin{array}{c}2 \\ y \end{array} \right)\left(\frac{1}{2}\right)^{y}\left(\frac{1}{2}\right)^{2 - y} = \left( \begin{array}{c}2 \\ y \end{array} \right)\left(\frac{1}{2}\right)^{2}$ . Similarly, with $X$ as $Y$ is the number of heads in two tosses of a fair coin. $X$ is the number of heads in one toss of a fair coin. The binomial marginal densities $f_{X}(x)$ and $f_{Y}(y)$ are shown in square brackets. The tosses are all independent. So, the binomial joint density $f_{XY}(x, y)$ is the product of the marginals. The sum of the shaded cells is $P(Y > X) = 1 / 2$ .

Table 4.2: Binomial Joint Density: Two Tosses versus One Toss

The number of heads in one toss of a fair coin, $P(X = x) = f_{X}(x) = \left( \begin{array}{c}1 \\ x \end{array} \right)\left(\frac{1}{2}\right)^{1}$ . The marginals are thus given by

f_{Y}(y) = \left\{ \begin{array}{ll}\frac{1}{4}, & y = 0, \\ \frac{1}{2}, & y = 1, \\ \frac{1}{4}, & y = 2, \end{array} \right. \quad \text{and} \quad f_{X}(x) = \left\{ \begin{array}{ll}\frac{1}{2}, & x = 0, \\ \frac{1}{2}, & x = 1 \end{array} \right.

The random variables $Y$ and $X$ are statistically independent of each other, so the joint density is just the product of the marginals $P(X = x, Y = y) = f_{XY}(x, y) = f_{X}(x) \cdot f_{Y}(y)$ as shown in Table D.1. Simple visual inspection of Table D.1 shows that in the simple case of two and one tosses, we get $P(Y > X) = 1 / 2$ (i.e., the sum of the shaded cells).

Although I gave the formal calculation of the marginal and joint densities in this simplified case, this calculation is not needed. The two marginal Binomial distributions shown in square brackets in Table D.1 are symmetric because $p = 1 / 2$ .

With both marginals symmetric, it must be that the sum of the joint probabilities in the shaded cells and the sum of the joint probabilities in the un- shaded cells are the same. The two sums must add to one, so the sum of the shaded cells must be $P(Y > X) = 1 / 2$ .

Table D.2 shows an analogous table for the case of you tossing five coins and me tossing four coins. Again, I have given all the details of the marginal and joint densities, but in fact, no numbers are required to do the calculation. I have given them only to fill in the fine details to back up the answer.

Table 4.3: Binomial Joint Density: Two Tosses versus One Toss

$Y$ is the number of heads in five tosses of a fair coin. $X$ is the number of heads in four tosses of a fair coin. The binomial marginal densities $f_{X}(x) = \left( \begin{array}{c}4 \\ x \end{array} \right)\left(\frac{1}{2}\right)^{4}$ and $f_{Y}(y) = \left( \begin{array}{c}5 \\ y \end{array} \right)\left(\frac{1}{2}\right)^{5}$ are shown in square brackets. The tosses are all independent. So, the binomial joint density $f_{XY}(x,y)$ is the product of the marginals. The symmetrical marginals mean that the joint density is symmetric in all respects. If you take a pair of scissors and cut out the shaded cells as a single inverted staircase, then rotate that through 180 degrees, the contents perfectly match the contents of the un- shaded cells. So, by symmetry, $P(Y > X) = 1 / 2$ .