What is the average of rolling two dice and only taking the value of the higher dice roll?

Question

What is the average result of rolling two dice, and only taking the value of the higher dice roll?

To make sure the situation I am asking about is clear, here is an example: I roll two dice and one comes up as a four and the other a six, the result would just be six.

Would the average dice roll be the same or higher than just rolling one dice?

Answer 1

The number of ways to roll a number $x$ under your definition would be $2(x-1) + 1$.

Therefore the expected value would be $$E[X] = \sum_{x=1}^6\frac{2(x-1)+1}{36}x = \frac{1}{36}\sum_{x=1}^6(2x^2 - x) = \frac{161}{36} \approx 4.47$$ So the average is considerably higher than the average of a single die, being $3.5$.

Answer 2

For $k=1,\dots,6$ there are $k^2$ ways to get two numbers less than or equal to $k$. To get two numbers whose maximum is $k$ I must get two numbers that are less than or equal to $k$, but not two numbers that are less than or equal to $k-1$, so there are $k^2-(k-1)^2=k^2-(k^2-2k+1)=2k-1$ ways to get two numbers whose maximum is $k$. Thus, the probability of getting a maximum of $k$ is

$$\frac{2k-1}{36}\;,$$

and the expected value of the maximum is

$$\begin{align*} \sum_{k=1}^6k\cdot\frac{2k-1}{36}&=\frac1{36}\sum_{k=1}^6\left(2k^2-k\right)\\ &=\frac1{18}\sum_{k=1}^6k^2-\frac1{36}\sum_{k=1}^6k\\ &=\frac{6\cdot7\cdot13}{18\cdot6}-\frac{6\cdot7}{36\cdot2}\\ &=\frac{91}{18}-\frac{21}{36}\\ &=\frac{161}{36}\\ &=4.47\overline{2}\;. \end{align*}$$

Of course this is larger than the expected value of $\frac72=3.5$ for a single roll of a die: picking the maximum of the two numbers can be expected to bias the result upwards.

Answer 3

I know this is an old question, but wanted to provide another perspective. It still involves some computation.

Let $A$ and $B$ be the maximum and minimum of two die rolls $X_1, X_2$ respectively, we have $$ \mathbb{E}[A+B] = \mathbb{E}[X_1+X_2] = 7, \quad \mathbb{E}[A-B]=\mathbb{E}[|X_1-X_2|]. $$ To compute the expected range $\mathbb{E}[A-B]$ note that among $36$ outcomes there are $2\times k$ outcomes for the range to be $6-k$. For example, for the range to be $2$, two rolls must be $(1,3), (2,4), (3,5), (4,6)$ and the factor of $2$ accounts for symmetry. Hence $$ \mathbb{E}[A-B] = \frac{2}{36}\left(1\times 5+2\times 4+3\times 3+\cdots+5\times1\right) = \frac{35}{18}, $$ and consequently $\mathbb{E}[A] = (7+\frac{35}{18})/2=\frac{161}{36}$.