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Four fair 6-sided dice are rolled. The sum of the numbers shown on the dice is 8. What is the probability that 2's were rolled on all four dice?
The answer should be 1/(# of ways 4 numbers sum to 8) However, I can't find a way, other than listing all possibilities, to find the denominator.
Possibly brute force is nearly the best available option here. \begin{align} 2+2+2+2 & & & 1 \text{ way} \\[6pt] 1+2+2+3 & & & 12\text{ ways} \\[6pt] 1+1+2+4 & & & 12\text{ ways} \\ 1+1+3+3 & & & 6\text{ ways} \\[6pt] 1+1+1+5 & & & 4\text{ ways} \end{align}
So you get $1/35$.
I proved a relevant theorem here ( for some reason this was closed )
Theorem 1: The number of distinct $n$-tuples of whole numbers whose components sum to a whole number $m$ is given by
$$ N^{(m)}_n \equiv \binom{m+n-1}{n-1}.$$
to apply this to your problem , you need to take into account that the theorem allows a minimum value of zero, but the dice show a minimum value of one. So you can just add one to each die so we are looking for the number of 4-tuples whose components sum to 4. I guess we are lucky that no element could possibly be greater than 6, so it will work for 6 sided dice.
The denominator is thus given by $$\binom 73 = 35 $$ which I have verified by enumeration.
