# A probability question that involves $5$ dice

For five dice that are thrown, I am struggling to find the probability of one number showing exactly three times and a second number showing twice.

For the one number showing exactly three times, the probability is: $${5 \choose 3} \times \left ( \frac{1}{6} \right )^{3} \times \left ( \frac{5}{6}\right )^{2}$$

However, I understand I cannot just multiply this by $${5 \choose 2} \times \left ( \frac{1}{6} \right )^{2} \times \left ( \frac{5}{6}\right )^{3}$$ as this includes the probability of picking the original number twice which allows the possibility of the same number being shown $5$ times. I am unsure of what to do next, I tried to write down all the combinations manually and got $10$ possible outcomes so for example if a was the value found $3$ times and $b$ was the value obtained $2$ times one arrangement would be '$aaabb$'. However I still am unsure of what to do after I get $10$ different possibilities and I am not sure how I could even get the $10$ different combinations mathematically. Any hints or advice on what to do next would be much appreciated.

-

HINT: I think that this one is at least as easy to do by counting outcomes. There are $\binom53$ ways to choose which $3$ dice show one number and which $2$ show another. For each of these arrangements there are $6$ choices for the number on the $3$ dice and then $5$ choices for the number on the other $2$ dice. Thus, there are $\binom53\cdot6\cdot5$ successful outcomes. How many equally likely outcomes are possible?

-
So I just divide the successful outcomes over the total number of possible outcomes? I think that answers it – Johnmgee Mar 9 '13 at 18:22
@Johnmgee: That’s right. – Brian M. Scott Mar 9 '13 at 18:27

First, I assume they wil all come out in neat order, first three in a row, then two in a row of a different number. The probability of that happening is $$\frac{1}{6^2}\cdot \frac{5}{6}\cdot\frac{1}{6} = \frac{5}{6^4}$$ The first die can be anything, but the next two have to be equal to that, so the $\frac{1}{6^2}$ comes from there. Then the fourth die has to be different, and the odds of that happening is the $\frac{5}{6}$ above, and lastly, the last die has to be the same as the fourth.

Now, we assumed that the three equal dice would come out first. There are other orders, a total of $\binom{5}{3}$. Multiply them, and you get the final answer $$\binom{5}{3}\frac{5}{6^4}$$

You might reason another way. Let the event $A$ be "there are exactly 3 of one number" and $B$ be "There are exactly 2 of one number". Then we have that $$P(A\cap B) = P(A) \cdot P(B|A) = {5 \choose 3} \cdot \frac{1}{6^3} \cdot\frac{5^2}{6^2} \cdot \frac{1}{5} = \binom{5}{3}\frac{5}{6^4}$$ Where $P(A)$ you allready calculated on your own, and $P(B|A)$ is the probability that there is an exact pair given that there is an exact triple, which again is the probability that the two last dice are equal, and that is $\frac{1}{5}$.

-

We can work directly with probabilities, or else we can count. Label the dice A, B, C, D, E. Record the result of tossing as an ordered $5$-tuple of numbers. There are $6^5$ such $5$-tuples, all equally likely.

We now count the number of $5$-tuples that have $3$ of one kind and $2$ of another, like a Full House in poker.

The number we have $3$ of can be chosen in $\binom{6}{1}$ ways. (Of course, this is simply $6$.) For every way to choose the number we have $3$ of, there are $\binom{5}{1}$ ways to choose the number we have $2$ of. And for every choice that we have made, the locations of the number we have $3$ of can be chosen in $\binom{5}{3}$ ways. If we wish, we can say that then the locations of the number we have $2$ of can be chosen in $\binom{2}{2}$ ways. However, this number is $1$, so it can be left out.

Thus the total number of "favourable" configurations is $\binom{6}{1}\binom{5}{1}\binom{5}{3}$.

-