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:)
 A: First, select the number that appears exactly $3$ times: $6$ ways. We have $\binom{6}{3}$ ways to place them.
Now, in the rest of the $3$ positions, you have $5$ options. So, count = $$6 \cdot \binom{6}{3} \cdot 5^3$$
But, in this we are also counting the number arrangements of $3 + 3$ (so two numbers appear $3$ times), so subtract $$\binom{6}{2} \cdot 2 \cdot \binom{6}{3}$$ This is subtracting the number of ways to pick $2$ numbers and place them in $3 + 3$.
EDIT: My explanation was wrong and thanks to @N.F.Taussig for pointing it out. I will just put their comment that explains where I am wrong:

The factor of $2\binom{6}{2}\binom{6}{3}$ is twice the number of arrangements in which two numbers each appear three times. That is what we want to subtract since those patterns are counted twice among the $\binom{6}{1}\binom{6}{3}5^3$ arrangements in which a number appears three times, and we don't want to count such arrangements at all. However, your wording suggests that the term you are subtracting is the number of arrangements in which two numbers each appear three times, which is not the case.

So, answer = $$\frac{120 \cdot 125 - 40 \cdot 15}{6^6} = 0.31$$

Currently my thoughts are that we have $6$ choices for our favorable event (the triples)

That is correct.

and for the remaining $3$ numbers we have $5×5×4=100$, $4$ because we do not want to include the possibility of having $3$ same numbers two times

You are ordering the combinations later on, but when multiplying $5 × 5 × 4$ like that, you are not just picking the elements but also ordering them, something which you are doing later on. So, this over-counts. Also, it's not $5 × 5 × 4$, What if the first element (when you multiply by $5$) is the count-$3$-element itself?

and to consider all possible arrangements, there are then $\frac{6!}{3!1!1!1!}$ possibilities.

This is true only if the counts look like $6 = 3 + 1 + 1 + 1$ but what if $6 = 3 + 2 + 1$?
Let me know if you have any questions)))
A: 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
$$\frac{\dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{3}3! + \dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{1}\dbinom{3}{2}\dbinom{4}{1}}{6^6}$$
A: Number of cases when there are exactly three 1's is $N_1={6\choose 3}5^3$. Number of cases when there are exactly three 1's and exactly three 2's is $N_2={6\choose 3}$. Number of cases when there are exactly three 1's and no more threes is $N_3=N_1-5N_2={6\choose 3}(5^3-5)$. Number of cases when exactly 1 number is rolled exactly 3 times is $N_4=6N_3={6\choose 3}(5^3-5)6=14400$. Probability is $\frac{N_4}{6^6}=\frac{25}{81}\approx 0.31$
A: For patterns matching stipulations,  we shall use the format
[Choose numbers for pattern]$\times$[Permute]
$3-2-1\; pattern:$
$\left[\binom61\binom51\binom41\right]\times\left[\frac{6!}{3!2!}\right] = 7200$
$3-1-1-1\; pattern:$
$\left[\binom61\binom53\right]\times\left[\frac{6!}{3!}\right]= 7200$
Thus $Pr = \dfrac{7200+7200}{6^6} \approx 0.31$
