24
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ With regard to your final question: what does your intuition tell you? $\endgroup$ Commented Oct 29, 2012 at 3:51

5 Answers 5

20
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ @Matthew: Thanks for catching the typo. $\endgroup$ Commented Oct 29, 2012 at 4:38
20
$\begingroup$

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). The black dice represent the dice rolled, the white dice represent the max of the two dice in the respective row, column.

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

$\endgroup$
1
  • 1
    $\begingroup$ This is awesome. $\endgroup$ Commented Dec 15, 2020 at 17:03
17
$\begingroup$

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 $2/3$.

$\endgroup$
4
  • 1
    $\begingroup$ Just posting to say I've visited this exact same question a couple times over the last two years and even though this isn't the accepted answer I find it far more useful. $\endgroup$ Commented Feb 7, 2017 at 20:46
  • 2
    $\begingroup$ 2 1/2 years later I'm still coming back to this answer $\endgroup$ Commented Jul 24, 2019 at 15:24
  • 2
    $\begingroup$ End of 2022 and still coming back to this answer <3 $\endgroup$ Commented Nov 18, 2022 at 18:04
  • $\begingroup$ I agree, this is the best asnwer. Getting rid of the sum-operators is so good! Now we can quickly go through all dice D4, D6, ..., D20 (or impossible dice like a D3) and find that, astonishingly: The difference to the "regular" expected value scales ALMOST linearly with the number of dice-faces! Something beautiful happens if we take this formula, subtract the regular exp.value, multiply out everything and simplify: It simplifies to just n/6 - (1/6n) . Now normalizing to n yields: 1/6 - (1/6n^2) - so it's ALMOST a proportionality factor of 1/6, but smaller by 1/6n^2. $\endgroup$ Commented Nov 17, 2023 at 12:39
16
$\begingroup$

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 $3.5$.

$\endgroup$
3
  • 5
    $\begingroup$ 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.) $\endgroup$ Commented Feb 26, 2016 at 18:25
  • 3
    $\begingroup$ @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 $2(x-1)+1$. $\endgroup$
    – EuYu
    Commented Feb 26, 2016 at 19:27
  • 1
    $\begingroup$ Cool, makes sense. I asked because I think that's fundamentally the hard part of the problem, not the summation that follows it. $\endgroup$ Commented Feb 27, 2016 at 0:20
2
$\begingroup$

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 $\mathbb{E}[A] = (7+\frac{35}{18})/2=\frac{161}{36}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .