Rolling a balanced die 10 times

Question

A balanced die is rolled 10 times by 5 different people. What is the probability that at least two people rolled exactly 4 twos, 3 fours and 3 sixes?

So I know I'm looking for $P(X\ge 2)= 1-P(X=0)+P(X=1)$.

But I'm having trouble defining the event X.

I've tried letting X be the even that someone rolls 4 twos, 3 fours and 3 sixes but I can't tell how it's distributed. Any help is appreciated.

Answer 1

You need 2222444666, but not necessarily in that order. This can be done in $$\frac{10!}{4!3!3!} = 4200$$

different ways. The number of total ways the dice can be rolled are $6^{10}$, so the odds of a "success" are $$\frac{4200}{6^{10}} \approx 6.946 * 10^{-5}$$

Like you said, you need to take the odds of nobody getting it and the odds of exactly one person getting it, and subtracting that from 1. So now you need the odds of nobody rolling that. That'd be $$\left(1-\frac{4200}{6^{10}}\right)^5 \approx 0.99965$$

Then you need the odds of exactly one person getting it. The $\binom{5}{1}$ comes from choosing one out of the five to be the one who succeeds, as it can be any of them.

$$\binom{5}{1}\left(\frac{4200}{6^{10}}\right)\left(1-\frac{4200}{6^{10}}\right)^4 \approx 3.472 * 10^{-4}$$

So then you have $$1-\left(1-\frac{4200}{6^{10}}\right)^5-\binom{5}{1}\left(\frac{4200}{6^{10}}\right)\left(1-\frac{4200}{6^{10}}\right)^4 \approx 4.824 * 10^{-8}$$

Answer 2

What you are looking for is the Multinomial distribution. The pmf for the Multinomial is given by: $$ P(x_1, x_2, \ldots, x_k;n;p_1,p_2,\ldots,p_k)=\dfrac{n!}{x_1!x_2!\ldots x_k!}p_1^{x_1} p_2^{x_2}\ldots p_k^{x_k} $$ In this case, you'd like to compute first the probability that one person gets the arrangement you want: $$ P(T=4,F=3,S=3;10;1/6) = \dfrac{10!}{4!3!3!}\left(\frac{1}{6}\right)^{10} = 6.946 \times10^{-5} $$ Now we feed this into a Binomial distribution with the probability of success as above, and with $5$ trials: $$ \begin{aligned} P(X=0) &= (1-6.946 \times10^{-5})^{5} \\ P(X=1) &= 5 (6.946 \times10^{-5})(1-6.946 \times10^{-5})^4 \\ P(X\ge 2) &= 1- P(X=0)-P(X=1) = 4.82407 \times 10^{-8} \end{aligned} $$