The mean value of a selected die from an infinite series of rolls

If I roll a pair of dice an infinite number of times, and always select the higher value of the two, will the expected mean of the highest values exceed 3.5?

It would seem that it must be since if I rolled a million dice, and selected the highest value each time, the odds are overwhelming that sixes would be available in each roll. Thus, the expected mean would have to be something like 5.999999999999...

However, I can't seem to figure out what the expected value would be with my example using just 2 dice. Can someone help me arrive at a number? Would it barely exceed 3.5? Is this even something that can be calculated?

• Can you enumerate the sample space? List out the possibilities for the 2-dice example. – soakley Oct 27 '15 at 3:09

The experiment can also be simulated. This approach is useful when enumeration is difficult (like rolling 3 dice).

# fix the seed for reproducibility
set.seed(123)

# simulate pair of dice
rolls = matrix(sample(1:6, 2000000, replace=T), ncol=2)

# compute expected value
mean(apply(rolls, 1, max))
[1] 4.471531


There is no need to use simulation for this, the general case is quite easy to analyse. Let $n$ be the number of dice and $X$ be the maximum roll made when rolling the $n$ dice.

It follows that $$P(X \leq 1) = \left(\frac{1}{6}\right)^n$$ and in general $$P(X \leq k) = \left(\frac{k}{6}\right)^n$$ for $k$ between 1 and 6. Therefore we can obtain $$P(X = k) = P(x \leq k) - P(x \leq k-1)=\left(\frac{k}{6}\right)^n-\left(\frac{k-1}{6}\right)^n.$$

So we can write down the probability distribution in closed form. Doing this for $n = 2$ you obtain the expected value 4.472222.

• Notice that, in the limit, $P(X = 6) = 1^n - (\frac{5}{6})^n \rightarrow 1$ as $n \rightarrow \infty$, so this formula also confirms your intuition from your question. – Matthew Drury Oct 27 '15 at 14:21

I suggest just working through the trivial case to see the answer.

Possible results from rolling two dice generate a 6x6 matrix: $$\begin{bmatrix} (1,1) & (1,2) & ... \\ (2,1) & (2,2) & ... \\ (3,1) & (3,2) & ... \\ ... \end{bmatrix}$$

The expected value of the sum is 7. This is the case because the rolls are identical independent drawings, so they may be summed. The expectation of rolling a fair cubical die is 3.5.

But you are asking about maximization. Now let us enumerate the maximization from rolling two dice. Again, it is a 6x6 matrix: $$\begin{bmatrix} 1 & 2 & ... \\ 2 & 2 & ... \\ 3 & 3 & ... \\ ... \end{bmatrix}$$

Calculate the expected value, like so: $$E[x] = \Sigma(xP(x)) = 1/36(1) + 1/36(2) + ... + 1/36(6) \approx 4.47$$.

Notice that rolling $n$ dice is (in a probabilistic sense) equivalent to rolling one die $n$ times. So for rolling $n$ dice you can see how the matrix changes and how the resulting expectation changes, too.

Assuming each of the 36 combinations has an equal probability, we just have to add the values of each of the 36 combinations and divide by 36 to get the average:

1. 1 possibility: 11
2. 3 possibilities: 12, 21, 22
3. 5 possibilities: 13, 23, 31, 32, 33
4. 7 possibilities: 14, 24, 34, 41, 42, 43, 44
5. 9 possibilities: 15, 25, 35, 45, 51, 52, 53, 54, 55
6. 11 possibilities: 16, 26, 36, 46, 56, 61, 62, 63, 64, 65, 66

(1*1 + 2*3 + 3*5 + 4*7 + 5*9 + 6*11) / 36 = 4.47222..

Troll Dice Roller is the tool for finding dice probabilities. He has a paper explaining the implementation, but it's pretty academic.

max(2d6) yields

1 - 2.8%
2 - 8.3%
3 - 13.9%
4 - 19.4%
5 - 25%
6 - 30.6%
Average value =    4.47222222222