Question

Assume a game in which one rolls 20, 8-sided die, for a total number of 8^20 possible outcomes. To calculate the probability of a particular event occurring, we divide the number of ways that event can occur by 8^20.

One can calculate the number of ways to get exactly 5 dice of the value 3. (20 choose 5) gives us the number of orders of 3. 7^15 gives us the number of ways we can not get the value 3 for 15 rolls.

number of ways to get exactly 5, 3's = (20 choose 5)*7^15.

The answer can also be viewed as how many ways can I rearrange the string 3,3,3,3,3,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 (20 choose 5) times the total number of values we the zero’s (assuming 7 legal values) 7^15 (is this correct).

• Question 1: How can I calculate the number of ways to get exactly 5 dice of the same value(That is, for all die values).
  Note: if I just naively use my first answer above and multiply bt 8, I get an enormous amount of double counting?

  I understand that I could solve for each of the cases (5 1’s), (5, 2’s), (5, 3’s), … (5’s, 8) sum them (more simply 8*(5 1’s) ). Then subtract the sum of number of overlaps (5 1’s) and (5 2’s), (5 1’s) and (5 3’s)… (5 1’s) and (5, 2’s) and … and (5, 8’s) but this seems exceedingly messy. I would a generalization of this in a way that scales up to large numbers of samples and large numbers of classes.

• How can I calculate the number of ways to get at least 5 dice of the same value?

  So 111110000000000000000 or 11110100000000000002 or 11111100000001110000 or 11011211222222223333, but not 00001111222233334444 or 000511512252363347744.

I’m looking for answers which either explain the math or point to a library which supports this (esp python modules). Extra points for detail and examples.

Answer 1

Double counting can be solved by use of the Inclusion/Exclusion Principle

I suspect it comes out to:

Choose(8,1)*P(one set of 5 Xs) 
- Choose(8,2)*P(a set of 5 Xs and a set of 5 Ys) 
+ Choose(8,3)*P(5 Xs, 5 Ys, 5 Zs) 
- Choose(8,4)*P(5 Xs, 5 Ys, 5 Zs, 5 As)

P(set of 5 Xs) = 20 Choose 5 * 7^15 / 8^20
P(5 Xs, 5 Ys) = 20 Choose 5,5 * 6^10 / 8^20

And so on. This doesn’t solve the problem directly of ‘more then 5 of the same’, as if you simply summed the results of this applied to 5,6,7..20; you would over count the cases where you have, say, 10 1’s and 5 8’s.

You could probably apply inclusion exclusion again to come up with that second answer; so, P(of at least 5)=P(one set of 20)+ … + (P(one set of 15) – 7*P(set of 5 from 5 dice)) + ((P(one set of 14) – 7*P(one set of 5 from 6) – 7*P(one set of 6 from 6)). Coming up with the source code for that is proving itself more difficult.