Conditional Probability

Question

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?

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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.

Answer 1

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 $1/35$.

Answer 2

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.