What is the probability that the sum of $2$ dice rolls is a multiple of $3$? What about for $3$ dice rolls? For $n$ dice rolls?

So I have the first part of this solution worked out by writing out all the combinations of $2$ dice rolls (I won't write them here). There are $6^2=36$ total and the following multiples of $3$ are rolled: $$3\rightarrow2 ways$$$$6\rightarrow5ways$$$$9\rightarrow4ways$$$$12\rightarrow1way$$So adding up all the ways over the total rolls is the probability:$$\frac{2+5+4+1}{6^2}=\frac{12}{36}=\frac{1}{3}=0.33$$The second part and the general solution is what trips me up. I know there are $6^3$ possible rolls for $3$ dice, and we're now including the multiples $15$ and $18$, but how do I figure out how many ways there are to roll each multiple without counting all $216$ possibilities? How would I apply this to a general solution with $n$ dice?


It is always $\frac{1}{3}$. To see this, roll the first $n-1$ dice. Now add the last die, which will make the sum divisible by 3 exactly $\frac{1}{3}$ of the time.

  • $\begingroup$ Oh wow, I never thought of it that way. That sort of negated my work, but it's definitely true. $\endgroup$
    – Jodo1992
    May 6 '16 at 2:41

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