Probabilies of rolling n dice to add up to a specific sum

Question

I try to visualize a bernoulli chain. With variables p q and s.

• p probability for success
• q probability for failure
• s length of the chain

For fixed p q and s the chain is easiliy calcultated. However s in my case is determined by a roll of n dice which each have d sides:

• n = number of dice rolled
• d = number of sided of the dice (3, 4, 6, 8, 10, 12, 20)

So lets take this example:

• p = 0.25
• q = 0.75
• n = 10
• d = 3

So s can be anywhere from 10 to 30. I want to cacluclate a probability vecotor with

[No success, exactly 1 success, exactly 2 Successes..., exactly 30 successes]

I think I have to calculate the probability of each possible s, multiplie it with the actual bernoulli formula. And add the probabilities for i.e. 5 successes from each possible outcome of s. Is this correct?

I really struggle to get the probabilities of each sum. I get it for 2 or 3 dice I could just go through all possible results and well add it but for 20 dice it breaks my computer ;)

Answer 1

There is a stars and bars approach...

Lets start with an example:

Suppose you roll 5 six-sided dice and you want to now the probability that they sum to 20.

You can visualize this as 20 pips getting divided into 5 bins, with no more than 6 pips going into any one bin, and every bin receiving at least one pip.

20 pips over 5 bins is with every bin getting at least 1 is ${20-1 \choose 5-1}$

Now we need to excluding the possibilities that there are more than 6 allocated to any 1 bin.

$-{5\choose 1}{13 \choose 4}$

We have over-corrected, and must count the cases with more than 6 going to 2 bins.

$+{5\choose 2}{7 \choose 4}$

${19 \choose 4} -{5\choose 1}{13 \choose 4} + {5\choose 2}{7 \choose 4} = 651$

and there are $6^5$ ways to roll $5$ six sided dice.

$\frac {651}{6^5}$

How to generalize this...

How many ways to roll $m, n$-sided dice, with sum equal to $y.$

Let $k = \lfloor \frac {y-m}n \rfloor$

$\sum_\limits{i=0}^{k} (-1)^i{m\choose i}{y-1 - in\choose m-1}$