A fair die is rolled nine times. What is the probability that 1 appears three times, 2 and 3 each appear twice, 4 and 5 once and 6 not at all?
My approach is fairly simple. The dice is fair, so we have a total of $6^9$ possible strings. Consider the set $\{1,1,1,2,2,3,3,4,5\}$. From this set is a total possible number of combinations of
$$\frac{9!}{3!2!2!}=30240$$
Thus, the probability of the above occuring is $30240/6^9$
My question is whether or not I have properly accounted for the probability of $4$ and $5$, and whether or not I should account for their probabilty of being rolled by considering the probability of never rolling a $6$.
Could someone explain why I should or should not have these concerns?
Best Answer
This is correct, other than getting your computation wrong. Check wythatoras' comment for the correct value.
As there is only a single instance of 4 and 5 in the permuted set, 9! does not double count on the possible positions of 4 or 5.
The probability of never rolling 6 is similarly accounted for- if you roll three 1s, two 2s, two 3s, a 4 and a 5, no dice remain unrolled. If all the other conditions are satisfied, not rolling a 6 comes automatically.