Expected number of targets hit with paired shooters

Question

Consider $n$ targets, with 2 shooters aiming at each target, as shown in the picture below. If we randomly choose $k$ out of the $2n$ shooters to fire, what is the expected number $E(k)$ of targets that will be hit?

I can of course solve this for simple cases and make some general observations, e.g.:

• Trivially $E(1)=1$, for any $n$.

• For $k=2$, fix one shooter, and the probability that the other randomly-chosen one is his 'partner' is $\frac{1}{2n-1}$, in which case one target is hit; otherwise, two targets are hit. From these probabilities the expectation easily follows.

• If $k>n$ then at least one target has both its shooters firing at it. Remove it and consider the problem with $n-1$ targets. Do this recursively until $k\leq n$. So the only uniquely interesting cases are $3 \leq k \leq n$.

I am wondering if this can be solved in the general case. The expectation will suffice for me, but of course the exact probability distribution would be even nicer.

Answer 1

The probability that some target is hit, is $$1-\frac{\binom{2n-2}{k}}{\binom{2n}{k}}=1-\frac{(k-2n)(k-2n+1)}{2n(2n-1)}=\frac{k^2-4kn+k}{2n-4n^2}$$

The expected number of hit targest is $n$ times this probability.

So, $$E(X)=\frac{k^2-4kn+k}{2-4n}$$

It should be much more difficult to calculate the probabilities that $m$ targets are hit.

Answer 2

Concentrate on the first target. The probability it is hit is $$1- \dfrac{{2n-2 \choose k}}{{2n \choose k}}= 1- \dfrac{(2n-k)(2n-k-1)}{(2n)(2n-1)}= \dfrac{k(4n-k-1)}{(2n)(2n-1)}.$$

By linearity of expectation, the expected number of targets hit is $n$ times this, i.e. $$\dfrac{ k(4n-k-1)}{4n-2}$$ giving $E[1]=1$ and $E[2n]=n$ as you might expect.