Average count of overlap between random samples If I randomly sample n numbers from 1..m, x times, what would be the average count of numbers that overlap in every sample?
Example:

*

*Variables: n = 4, m = 12, x = 3

*Samples: 4 8 1 11, 5 4 12 1, 1 3 7 4

*Numbers that overlap in every sample: 2 (1 and 4)

I think I have a general intuition on how you might do this for x = 2. My thought is that you want to find the probabilities for all possible counts, 0..n, and then you could use those probabilities to do a weighted average of the counts? And to do those probabilities, you'd do something like 1 / choose(count, m), except for count = 0 which is 1-(1/choose(n,m))? It's been a while since I took stats so I wouldn't be surprised if that intuition is wrong though.  But even if it is right, I'm still not sure how you would go about doing it for x > 2/generalizing for all x.
From comments:

What is the probability a particular number, e.g. $1$, appears in the first sample?

It would be $n/m$, so for my example, $30\%$ probability that $1$ appears in the first sample

What is the probability it appears in all $x$ samples?

I think it would be $(n/m)^x$

What is the expected number of numbers that appear in all $x$ samples?

I can understand that I should now have the information needed to answer this, but I think this is beyond my current capabilities.

You have found the probability a particular number appears in all $x$ samples to be $(n/m)^x$
Linearity of expectation means that the expected number of numbers that appear in all $x$ samples is simply that probability multiplied by the number of possible numbers $(m)$, giving an expectation of $$m\left(\frac n m\right)^x = \frac{n^x}{m^{x-1}}$$
For example, when $x=1$ it gives the obviously correct $n$.  For $x=2$ it gives $\frac{n^2}{m}$ which you might want to check.  For your example of $n = 4$, $m = 12$, $x = 3$ it gives $\frac 49 \approx 0.444$ as the expected number
