Problem 13.1 Easy Free 1 variation

Expected Maximum of Two Dice

Two fair six-sided dice are rolled, and you take the larger of the two results (if they tie, that common value). Find the expected value of this maximum.

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $M$ be the maximum of the two dice. For $k\in\{1,\dots,6\}$ , $M\le k$ exactly when both dice show at most $k$ . By independence,

\mathbb{P}(M\le k)=\left(\frac{k}{6}\right)^2 .

Differencing this distribution function gives

\mathbb{P}(M=k)=\frac{k^2-(k-1)^2}{36}=\frac{2k-1}{36}.

So the probabilities are $\frac{1}{36},\frac{3}{36},\dots,\frac{11}{36}$ . Hence

\mathbb{E}[M]=\frac{1}{36}\sum_{k=1}^{6}k(2k-1)=\frac{2\cdot 91-21}{36}=\frac{161}{36}.

Alternative. For a variable taking values in $\{1,\dots,6\}$ , the expectation is the sum of the tail probabilities:

\mathbb{E}[M]=\sum_{k=1}^{6}\mathbb{P}(M\ge k)=\sum_{j=0}^{5}\left(1-\frac{j^2}{36}\right).

This equals $6-\frac{55}{36}=\frac{161}{36}$ .

Generalisation. For two fair $n$ -sided dice the same tail sum gives

\mathbb{E}[M]=n-\frac{(n-1)(2n-1)}{6n}.

\boxed{\mathbb{E}[\max]=\frac{161}{36}\approx 4.47}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 13.1a

Let $X$ and $Y$ be the results of two independent rolls of a fair $30$ -sided die labelled $1$ through $30$ . Find $\mathbb{E}[|X-Y|]$ .

Work it out, then compare with the answer.

Solution

Write $M=\max(X,Y)$ and $m=\min(X,Y)$ . Then $|X-Y|=M-m$ .

Since $M+m=X+Y$ , we have $\mathbb{E}[m]=(n+1)-\mathbb{E}[M]$ , with $n=30$ . Therefore

\mathbb{E}|X-Y|=2\,\mathbb{E}[M]-(n+1).

The tail sum from the main solution gives $\mathbb{E}[M]=n-\frac{(n-1)(2n-1)}{6n}$ . Substituting,

\mathbb{E}|X-Y|=(n-1)-\frac{(n-1)(2n-1)}{3n}=\frac{(n-1)(n+1)}{3n}=\frac{n^2-1}{3n}.

With $n=30$ this is $\frac{899}{90}$ .

\boxed{\mathbb{E}|X-Y|=\frac{899}{90}\approx 9.99}

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