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A few days ago while playing a tabletop RPG I rolled 10d20s (10 20 sided dice) and got 156 and I'd like to calculate the odds of that roll.
After I determine the odds of the sum I'd also like to calculate the odds of getting certain values on some of the individual dice.
Essentially what I'm asking for is how to calculate the probability of rolling 10d20 where at least 4 dice are 20s, at least 1 die is greater than or equal to 19, and at least 1 die is greater than or equal to 18.

Here's what I understand (maybe):
The average roll for a d20 is 10.5. So, if you rolled 10d20 you're most likely to get a sum around the value 105. The odds of getting a sum exponentially decreases the further away from 105 the value is and thus we have a bell curve (something something central limit theorem?).

Here's what I need:
I'd prefer to calculate it myself so what I'm looking for is possibly an equation that already exists for this sort of thing or at least be pointed in the right direction.

J. Doe
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  • Have a look at this question. – Paul Aljabar Jan 05 '18 at 12:29
  • For ten such dice, the total should be very nearly normal. Some years ago (when computers were clunkier) $Z = \sum_{i=1}^{12}U_i - 6,$ for $X_i$ iid UNIF(0,1), was routinely used to simulate $Z \sim$ NORM(0,1). – BruceET Jan 05 '18 at 14:47
  • If you’re more interested in getting the answer than how to get there, let someone else do the tedious part of the work: have a look at anydice.com. – amd Jan 06 '18 at 06:50

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An $n$-sided die has mean $\frac{n+1}{2}$ and standard deviation $\sqrt{\frac {n^2-1}{12}}$. Thus the 20-sided die has mean $10.5$ and standard deviation $\sqrt {\frac {133}{4}}$, and 10d20 has mean $105$ and standard deviation $\sqrt{\frac {665}{2}}≈18.23$.

Using these values we can calculate z-scores for $155.5$ and $156.5$; these come out to $2.769$ and $2.824$ respectively. Passing these through the normal distribution table, we get approximately $0.000438$ probability of getting 156.

Doing it the exact way, tabulating probabilities for each count of dice up to 10, we get $0.000468$ probability of getting 156. So the fake way was pretty close.

Dan Uznanski
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  • Great! Thanks. That's the first part of the problem done. Any insight on how to calculate the probability of rolling 10d20 where at least 4 dice are 20s, at least 1 die is greater than or equal to 19, and at least 1 die is greater than or equal to 18? – J. Doe Jan 05 '18 at 13:09
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    Thing to look up: multicombination. That one you'll need to do explicitly, though some shortening is possible: you can disregard order in generation and just figure out how many permutations a particular layout admits, and collapse everything from 1 to 17 into a single highly probable element. – Dan Uznanski Jan 05 '18 at 13:38