Say I roll a 6-sided die six times, what's the probability that I roll:
- 3 different no.s – twice each?
- some number exactly 4 times?
My approach:
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Which no.s come up in the rolls can be selected in $\binom63$ ways and those no.s can be arranged in $\frac{6!}{2!2!2!}$ ways. So the probability is $\frac{\binom63\frac{6!}{2!2!2!}}{6^6}$ ways.
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The number that comes up $4$ times can be selected in $6$ ways, the rolls (e.g. first roll? second roll? etc.) in which that number comes up can be selected in $\binom64$ ways. For the other two rolls, we can select any of the remaining $5$ numbers, which makes for $25$ ways to specify the remaining two rolls. This makes for a total of $6\binom64\times5\times5$ possibilities.
Another way is consider 2 scenarios: either only $2$ numbers come up, or $3$. If $2$, the no.s can be selected in $\binom62$ ways and they can be arranged in $\frac{6!}{4!2!}$ ways. If $3$, the no.s can be selected in $\binom63$ ways and they can be arranged in $\frac{6!}{4!}$ ways. This makes for a total of $\binom62\frac{6!}{4!2!}+\binom63\frac{6!}{4!}$ possibilities.
Is the first part correct? And I'm confused about the second part. Both seem reasonable enough to me but I'm undercounting / overcounting in one of the two approaches.
Best Answer
I think there's a typo in the first part. Three different numbers thrice each is $9$ outcomes. Perhaps you meant to write twice each? That seems to be what you computed, correctly.
The first computation in the second part looks correct to me. The second computation is certainly wrong, because you say that if two numbers show up they can be chosen in ${6\choose2}$ ways. That would count $1$ coming up twice and $2$ four times as the same as $2$ coming up twice and $1$ four times, which isn't so.
Of course, you make a similar error in counting the cases where $3$ numbers show up.