A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements?
I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! 🙂
Best Answer
Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the same number is rolled six times. Thus, the probability of exactly two numbers appearing in six rolls of the die is $$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$
Exactly three numbers appear:
There are $\binom{6}{3}$ ways for three of the six numbers to appear. On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers. However, we have counted the sequences in which not all three of the numbers appear. There are $\binom{3}{2}$ ways for two of the numbers to appear. For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times. Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear. The number of sequences in which exactly one of the three numbers appears is $3$. Hence, the probability that exactly three of the numbers appear is $$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$
My thanks to @bof for clarifying my thinking about the first problem. Any errors that remain are entirely my responsibility.