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Let's say I roll N twenty sided dice. What is the expected value of the highest and lowest rolls?

Bonus: Is there a more general formula if I wanted to find the expected value of N dX dice, where X is the number of sides on each dice? Or what if each if it were a mixture of dice of different numbers of sides?

I wanted to figure this out for a D&D session, but its been a long time since my class in probability theory, lol

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  • $\begingroup$ The short exact answer is: 21 $\endgroup$ – wolfies Jun 3 at 4:52
  • $\begingroup$ And for a T-sided dice, the exact general solution is: $T+1$. $\endgroup$ – wolfies Jun 3 at 14:07
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You might find this answer enlightening: https://math.stackexchange.com/a/150633/26091. In short, the expected maximum of $n$ random numbers on the interval $[0,1]$ is $\frac{n}{n+1}$. Your dice are discrete rather than continuous, but scaling this by $20$ (or whatever number of sides you have) for a rough approximation would be pretty good. In other words, the answer is approximately $$20\frac{n}{n+1}.$$

You can check that for one die, this gives $10$, which is close to the actual average of $10.5$, and as $n$ increases, it approaches $20$ (as it clearly should).

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To refresh your memory, this falls under order statistics in probability theory.

Let $X_i$ be a random variable that represents a roll. We can write the maximum as $X_{\max} = \max\{X_1, X_2, X_3, ... X_N\}$

Let's use the CDF of $X_{\max}$ to find the PDF of $X_{\max}$

$$F_{X_{\max}} = P(X_{\max} \leq x)$$

This is the same thing as asking, what is the probability that every die rolled is less than or equal to $X_{\max}$. Since these are iid, we have:

$$F_{X_{\max}}(x) = P(X_{\max} \leq x) = P(X_1\leq x)P(X_2\leq x)...P(X_N\le x)$$

If the dice aren't the same, you'd have to compute each of the CDFs independently, and continue to the next step. Since they are the same, we have that $F_{X_\max}(x) = P_1(X_1 \leq x)^N=x^N/20^N.$

The PMF is now just the difference between $F(x) - F(x-1)$. In this case that's:

$$p(X_\max=x) = {x^N-(x-1)^N \over 20^N}$$

To find the expected value, we just take a weighted sum over the possible rolls

$$ E[X_\max] = \sum_{x=1}^{20}x{x^N-(x-1)^N \over 20^N}$$

You can use similar reasoning to find the expectation of the minimum.

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I'm only answering the part about rolling a bunch of identical dice. Let $X$ be the largest outcome and $Y$ the smallest outcome among $N$ independent rolls of an $s$-sided die with sides numbered from $1$ to $s.$


For $E(X)$ I copy my answer to this old question:

Define $X_i$ to be the random variable whose value is $1$ if $X\ge i$ and $0$ otherwise; then $X=\sum_{i=1}^sX_i$ and $$E[X]=E\left[\sum_{i=1}^sX_i\right]=\sum_{i=1}^sE[X_i]=\sum_{i=1}^sP(X_i=1)=\sum_{i=1}^sP(X\ge i)=\sum_{i=1}^s[1-P(X\lt i)]=\sum_{i=1}^s\left[1-\left(\frac{i-1}s\right)^N\right]=s-s^{-N}\sum_{i=1}^s(i-1)^N=\boxed{s-s^{-N}\sum_{i=1}^{s-1}i^N}.$$


Now for $E(Y)$. Since the outcome of a single roll is distributed symmetrically about its mean $\frac{1+s}2$, we have $E(X)+E(Y)=1+s$, so that $$E(Y)=1+s-E(X)=\boxed{1+s^{-N}\sum_{i=1}^{s-1}i^N}.$$

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