<s>[INST] Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die I'm trying to calculate the probability based on the size of the event space divided by the size of the sample space $P=\frac{|E|}{|S|}$
I know that $|S|=6^6$, but am not sure what exactly the event space consists. Currently my thoughts are that we have 6 choices for our favorable event(the triples) and for the remaining 3 numbers we have $5\times5\times4=100$, $4$ because we do not want to include the possibility of having 3 same numbers two times, and to consider all possible arrangements, there are then $\frac{6!}{3!1!1!1!}$ possibilities.
This leads to our final equation of : $P=\frac{|E|}{|S|}=\frac{6!}{3!1!1!1!} \times \frac{6\times100}{6^6}$
But the problem is that this exceeds 1, which is clearly wrong but I couldn't really figure out what is the fix for my equation.
Thanks:)
 [/INST] Since there are six possible values for each of the six rolls, there are indeed $6^6$ elements in the sample space.
For the favorable cases, since we want to calculate the number of cases in which exactly one number appears three times, there are two possibilities:

*

*One number appears three times and three other numbers each appear once.

*One number appears three times, a second number appears twice, and a third number appears once.

One number appears three times and three other numbers each appear once:  There are six ways to select the number which appears three times, $\binom{6}{3}$ ways to select which three of the six positions in the sequence of rolls that number occupies, $\binom{5}{3}$ ways to select which three of the other five numbers each appear once, and $3!$ ways to arrange those three distinct numbers in the remaining positions of the sequence.  Hence, there are $$\binom{6}{1}\binom{6}{3}\binom{5}{3}3!$$
such cases.
One number appears three times, a second number appears twice, and a third number appears once:  There are six ways to select the number which appears three times, $\binom{6}{3}$ ways to select which three of the six positions in the sequence of rolls that number occupies, five ways to select which of the remaining numbers appears twice, $\binom{3}{2}$ ways to select which two of the three positions in the sequence of rolls that number occupies, and four ways to select which of the remaining numbers fills the remaining position in the sequence.  Hence, there are
$$\binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
such cases.
Since the above cases are mutually exclusive and exhaustive, the number of favorable cases is
$$\binom{6}{1}\binom{6}{3}\binom{5}{3}3! + \binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
Hence, the probability that one number appears exactly three times in six rolls of a fair six-sided die is
$$\frac{\dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{3}3! + \dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{1}\dbinom{3}{2}\dbinom{4}{1}}{6^6}$$
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