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

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?

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With regard to your final question: what does your intuition tell you? –  Benjamin Dickman Oct 29 '12 at 3:51

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 $3.5$.

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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.

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@Matthew: Thanks for catching the typo. –  Brian M. Scott Oct 29 '12 at 4:38

This is very much delayed, but consider the case with an $n$-sided die. As has already been observed, the expected value of the maximum of two $n$-sided die is

$${1 \over n^2} \sum_{k=1}^n (2k^2-k)$$

and we can write out this sum explicitly. In particular, we can expand to get

$${1 \over n^2} \left( \left( 2 \sum_{k=1}^n k^2 \right) - \sum_{k=1}^n k \right)$$ and recalling the formulas for those sums, this is

$${1 \over n^2} \left( {2n(n+1)(2n+1) \over 6} - {n(n+1) \over 2} \right)$$

or after some rearrangement

$${(n+1)(4n-1) \over 6n}.$$

In particular this is approximately $2n/3$. This could have been guessed if you know that the expectation of the maximum of two uniform random variables on $[0, 1]$ has the beta distribution $B(2,1)$, which has mean $2/3$.

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