Probability of 2 dice getting same number, when done $p$ times (number for each time different)

Question

The probability of throwing two sets of dice with same number is $1/n$ where $n$ is the number of distinct values on dice, and the number can be any of those values. The probability of the same event recurring twice, the same sets of dice having the same number (still any number, not the same number as the first example) should then be $1/n * 1/n$. The probability of this occurring yet another time, $1/n * 1/n * 1/n$. The probability of this occurring $p$ times, $(1/n)^p$. Is this correct?

Answer 1

Yes, that is correct. However, this means that ALL of the numbers are the same (e.g. 6, 6, 6, 6, 6).

If you mean that only $2$ can be the same, then out of it occurring $p$ times you need to choose $2$. Do this in $\binom{p}{2}$ ways, and then the other $p-2$ die have to have to consist the rest of the $n-1$ numbers. The way to find the number of ways is $$(n-1)(n-2)\dots(n-p+2).$$

This is because there are $p-2$ die left, and we normally would do $(n-1)!$, except it will cut off when we multiply a total of $p-2$ times.

In this case, the answer is $\binom{p}{2}\cdot (n-1)(n-2)\dots(n-p+2).$