# rolling a single die ten times

I have the following problem on a homework assignment for my Probability theory course:

You roll a single six sided die ten times. What is the probability that you roll four 1's, three 2's, and three 3's?

I was hoping somebody could check my idea since I haven't done this sort of problem in a while.

So my idea is to count the total number of ways you could get this outcome. My way of counting it is to first count the number of ways the four 1's could show up on the ten rolls. So we get ${10 \choose 4}$ ways this could happen. Now of the remaining 6 rolls I want to count the ways you could get three 2's, which is ${6 \choose 3}$. Now only three rolls remain so they have to be the three 3's.

So what I'm getting is that there are ${10 \choose 4}{6 \choose 3}$ ways to get such an outcome.

What I'm not sure of is if the probability would then be: $$\frac{{10 \choose 4}{6 \choose 3}}{6^{10}},$$ or if the probability should be, $$\frac{{10 \choose 4}{6 \choose 3}3^4 2^3 1^3}{6^{10}}.$$

It's also possible that I could be completely wrong.

Any help is greatly appreciated.

• You're on the right track. What are your arguments for either of the last two expressions? Why should the factor $3^42^31^3$ be there (or not)? – Servaes Aug 27 '15 at 2:15
• I don't see what you mean by "four of them have 3 options, three of them have only 2 options, and then the remaining three have only one option.". Four of them should be $1$'s, three of them $2$'s and then three of them $3$'s, right? Then again, I'm very sleepy... – Servaes Aug 27 '15 at 2:21
• Yeah, I think you are right. I think it should be the first option listed in the posting. In my reasoning for the second option I don't ensure the outcome I desire. I did a couple examples with a smaller number of die to confirm it for myself. Thanks for the help! – User112358 Aug 27 '15 at 2:26
• This is simply an example of a multinomial distribution: en.wikipedia.org/wiki/Multinomial_distribution – Brent Aug 27 '15 at 5:01
• Your first method is correct. – N. F. Taussig Aug 27 '15 at 8:56

(1) Your answer has to be less than $(1/2)^{10} = 0.0009765625$. That would be the probability all 10 rolls showed 3 or fewer spots.
(2) Your first answer computes to 0.00006946032, which $is$ smaller than the answer in 1.
(4) Finally, the multinomial formula in Wikipedia, as per @Brent's Comment is $$\frac{10!}{4!\times 3! \times 3!}\left(\frac{1}{6}\right)^{10},$$ which is easily seen to be the same as (2).