Rolling a fair die 18 times, what is the probability of rolling 1,2,3,4,5,6 (each) three times?

Question

The problem statement and solution is below. I understand that each die toss is independent but it doesn't make sense to me why the probability is multiplied by the number of sequences, as in the solution.

When you roll a die 18 times, it is possible to get a countable number of sequences such that a 1,2,3,4,5,6 each appear 3 times. Wouldn't the probability that each number appears exactly 3 times be equal to the ratio of the number of sequences in which 1,2,3,4,5,6 appear 3 times to the total number of sequences you can have (6^18)?

Answer 1

Wouldn't the probability that each number appears exactly 3 times be equal to the ratio of the number of sequences in which 1,2,3,4,5,6 appear 3 times to the total number of sequences you can have (6^18)?

Yes, and the number of sequences is exactly what's calculated with the multinomial coefficient. You choose the positions in the sequence for the $1$'s, then for the $2$'s, and so forth. The collection of these sequences is counted by the multinomial coefficient.

Answer 2

This questions ask you for the number of possible sequences that meet your conditions.

So, consider this: The sequence 1,2,3,4,5,6 is not the same sequence as 1,3,2,6,5,4. However, each number has occurred exactly once. Since order matters, you must treat this as a permutation.

You could roll something like: {(1,2,3,4,5,6) (1,2,3,4,5,6) (1,2,3,4,5,6)} or something like {(1,1,1,2,2,2) (3,3,3,4,4,4) ...} and as you can see both meet your criteria and both are different.

So, $$(\frac{1}{6})^{18}$$

is the probability of one of these possible sequences occurring.

$$\frac{18!}{(3!)^6}$$

gives the possible number of these sequences.

$$(\frac{1}{6})^{18}\frac{18!}{(3!)^6}$$

yields the probability of any such sequence occurring.