# A fair die is rolled five times. Find the probability that one shows twice, three shows twice, and six shows once.

I am working on the following problem:

A fair die is rolled five times. Find the probability that one shows twice, three shows twice, and six shows once.

The answer in the solution is given : $$\left(\frac{5!}{1! 2! 2!}\right) \left(\frac{1}{6}\right)^6$$

But I think it should be $$(1/6)^5$$.

(Ref: Probability, Random variables and Stochastic processes - Papulis(Ch-4/Exercise-31)

• The answer in the solution is correct except that exponent $6$ must be $5$. Probably a typo. Oct 12, 2019 at 9:19
• Just to confirm it should be (1/6)^5, correct ? Oct 12, 2019 at 9:21
• Not just on its own. If you mean that exponent $6$ must be replaced by $5$ then: yes. Oct 12, 2019 at 9:22
• Yea, i mean with the combinatorics part Oct 12, 2019 at 9:24

Since the die is fair, there are $$6$$ equally likely outcomes for each of the five rolls, so there are $$6^5$$ possible outcomes for the five rolls.
We have a sequence of five outcomes. For the favorable cases, choose two of the five positions in the sequence for the ones and two of the remaining three positions in the sequence for the threes. The final open position in the sequence must be filled with a six. Therefore, there are $$\binom{5}{2}\binom{3}{2} = \frac{5!}{3!2!} \cdot \frac{3!}{2!1!} = \frac{5!}{2!2!1!}$$ favorable sequences.
Hence, the probability that two ones, two threes, and a six are obtained when a fair die is rolled five times is $$\frac{\dbinom{5}{2}\dbinom{3}{2}}{6^5} = \left(\frac{5!}{2!2!1!}\right)\left(\frac{1}{6^5}\right)$$
• That is correct. There are $$\binom{5}{2, 2, 1} = \frac{5!}{2!2!1!}$$ possible sequences of two ones, two threes, and a six. Oct 12, 2019 at 9:29
The probability that the dice show up in the specific order 1,1,2,2,6 is certainly $$\frac1{6^5}$$. But you can arrange those five rolls in multiple ways so that the same results come up. The multinomial factor in front is what counts that.