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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
up vote 8 down vote accepted

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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Where did $2(x - 1) + 1$ come from? (Answers below give explanations, but I'm wondering where you got it, since it seems to have come out of thin air.) – Andrey M. Mishchenko Feb 26 at 18:25
@AndreyM.Mishchenko To roll an $x$ with this scheme, you either (i) roll an $x$ with die 1, and a number less than $x$ with die 2, or (ii) roll an $x$ with die 2, and a number less than $x$ with die 1, or (iii) roll an $x$ with both dies. The possibilities for (i) and (ii) are $x-1$, so the total is $2(x-1)+1$. – EuYu Feb 26 at 19:27
Cool, makes sense. I asked because I think that's fundamentally the hard part of the problem, not the summation that follows it. – Andrey M. Mishchenko Feb 27 at 0:20

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


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

I'll have a go and answer this the maths-lite way (though there are a number of answers with more mathematic rigor and .. dare I say it vigor posted here already).

Dice Results...

Note that there is:

  • 1 result with a face value 1
  • 3 results with a face value 2,
  • 5 results with a face value 3,
  • 7 results with a face value 4,
  • 9 results with a face value 5, and
  • 11 results with a face value 6

The Average is defined to be: $$\text{Average} = \frac{\text{Sum of the Results}}{\text{Total number of Results}}$$

The Sum of the Results is: $$\begin{eqnarray} \text{Sum} &=& (1 \times 1) + (3 \times 2) + (5 \times 3) + (7 \times 4) + (9 \times 5) + (11 \times 6) \nonumber \\ &=& 1 + 6 + 15 + 28 + 45 + 66 \nonumber \\ &=& 161 \nonumber \end{eqnarray}$$

The Total number of Results is: $ 6 \times 6 = 36$

So the Average is: $$\text{Average} = \frac{161}{36} \approx 4.472$$

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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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