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I'm trying to create a little tool for comparing dice roll probabilities (for table top gaming). I've got a way of calculating "Roll $n$D$s$ pick highest" and "Roll $n$D$s$ and sum" (where $n$ is the number of dice and $s$ is the number of sides on the dice), but I haven't been able to work out how to do "Roll $n$D$s$ and add the highest and lowest".

This answer shows the probability formula for the Nth ordered discrete random variable having a given value.

I assumed that I could get from there to "Roll $n$D$s$ and add the highest and lowest = T" by summing the probabilities of each way to achieve that score (e.g. 5 from $T_1=1,T_n=4$ and $T_1=2,T_n=3$†), where the probability for a pair is the probability of both the case $k=1$ being the lower score and the case $k=n$ being the higher score (i.e. $Pr(X_{(1)} = T_1) \cap Pr(X_{(n)} = T_2)$).

But that isn't working. My best guess is because results are actually related rather than independent, and so $Pr(X_{(n)} = 4)$ includes times when $Pr(X_{(1)} > 1)$ and $Pr(X_{(1)} = 1)$ includes cases where $Pr(X_{(n)} \neq 4)$, and so it's not as simple as ANDing the probabilities of the individual cases.

So, how can I calculate "Probability of rolling $X$ on $n$D$s$ when adding the highest and lowest", or at least $Pr(X_{(1)} = T_1 \cap X_{(n)} = T_2)$ (i.e. both conditions hold on the same roll)?

(A continual WIP JavaScript implementation is currently at https://ibboard.co.uk/dice-roll-compare/temp.html but could be in any state when you read this!)

† or a special case for when $T_1 = T_2$, because there's only one way to roll that on any number of dice! Although the special case might not be necessary if I fix the calculations.

IBBoard
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As you write, the case $T_1=T_2$ is easy, since there's exactly one roll that achieves it. So assume $T_2\gt T_1$. Then the probability that all rolls lie in $[T_1,T_2]$ is

$$ \left(\frac{T_2-T_1+1}s\right)^n\;. $$

But this also includes events where either $T_1$ or $T_2$ isn't actually rolled, so we have to subtract both of those cases:

$$ \left(\frac{T_2-T_1+1}s\right)^n-2\left(\frac{T_2-T_1}s\right)^n\;. $$

But now the cases where neither $T_1$, nor $T_2$ was actually rolled have been subtracted twice, so we have to add them back in once:

$$ \mathsf P(X_{(1)}=T_1\land X_{(n)}=T_2)=\left(\frac{T_2-T_1+1}s\right)^n-2\left(\frac{T_2-T_1}s\right)^n+\left(\frac{T_2-T_1-1}s\right)^n\;. $$

joriki
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    Well, that was easier than I'd expected. And much easier than the modification of the formula from the Q that I was just going to try! I'd tried thinking through ways to calculate all possibilities in the range and then remove some, but had never got it to work generically across all scores/dice. For anyone else who reads this, my special case for $T_2 = T_1$ is then just $\left(\frac{1}s\right)^n$. – IBBoard Jul 29 '18 at 19:31