how many overlaps if we choose items randomly Suppose that I have $n$ items $X_1, X_2, \ldots, X_n$. I am going to choose $p$ items randomly from them. Alice also chooses $p$ items out of the $n$ items.
What is the probability of having $k$ overlaps between my choice and Alice's choice?

my trial: The probability of each being chosen is $p/n$. So the prob of each being chosen twice is $(p/n)^2$. So this is a binomial process $B(n, (p/n)^2)$.
Therefore the probability of having $k$ overlaps is $\binom{n}{k} (p/n)^{2k} (1-(p/n)^2)^{n-k}$.

Is this correct? The point that I am confusing is that if we think each item will be chosen wp $p/n$, then it may violate our construction "choosing $p$ items in total". Because nothing will be chosen wp $(1-p/n)^n$?
 A: You can assume, without loss of generality, that one of the people specifically chooses items $\{X_1, \cdots, X_p\}.$
This is because the number of matches between the two people is unaffected by (i.e. independent of) which items the first person selected.
Therefore, the probability of having exactly $k$ matches can be expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where $~\displaystyle D = \binom{n}{p}.$
The number of ways that the second person can select $k$ items from the first $p$ items and $(p-k)$ items from the last $(n-p)$ items is
$$N = \binom{p}{k} \times \binom{n-p}{p-k}.$$
Putting this all together, I get
$$\frac{N}{D} = \frac{\binom{p}{k} \times \binom{n-p}{p-k}}{\binom{n}{p}}$$
$$=~ \frac{p!}{k! \times (p-k)!} \times \frac{(n-p)!}{(p-k)! \times  (n+k-2p)!} \times \frac{p! \times (n-p)!}{n!}$$
$$=~ \frac{(p!)^2 \times [(n-p)!]^2}{[(p-k)!]^2 \times (k!) \times [(n+k-2p)!] \times n!} \tag1 .$$

However:
The above analysis begs the question: 
Does the OP's (i.e. original poster's) computation agree with mine?
If not, what mistake did the OP make?


my trial: The probability of each being chosen is $p/n$. So the prob of each being chosen twice is $(p/n)^2$. So this is a binomial process $B(n, (p/n)^2)$.


Therefore the probability of having $k$ overlaps is $\binom{n}{k} (p/n)^{2k} (1-(p/n)^2)^{n-k}$.

Basically, the OP's computation is wrong for only one reason: the individual events are not independent events.
For example, suppose that you have $n$ independent Bernoulli trials, each with a probability of success of $p$, where $q = (1-p)$.
Then, the probability of having exactly $k$ successes is in fact 
$~\displaystyle \binom{n}{k} \times p^k \times q^{n-k}.$
However, suppose (for example) that item $X_1$ is selected twice.  This implies that the probability of item $X_2$ being selected twice is slightly smaller than normal.  So, if the OP's method is to be used, you have to deal with the complications of dependent events.  I strongly suspect that this makes the algebra a mess.

Finally, there is a separate issue:
Either the analysis at the start of my answer is valid, or it isn't.  If it is valid, then any alternative method of attack, that is also valid, must result in the same computation.
