Expected number of targets hit by 10 shooters, 2 bullets each, 20 targets

Question

Consider a competition with 10 clay shooters. Each shooter has equal ability and they all use identical shotguns. Each shotgun is loaded with two bullets, therefore each shooter can shoot twice. One throws 20 clay targets simultaneously, each shooter picks a target at random and shoots at it once, all the shooters shooting at the same time. Then each shooter picks one of the remaining targets at random, again independently of the other shooters, and shoots at it. This ends the shooting.

What is the expected number of targets to get hit?

I don't know how to answer question like this because the question doesn't give what is the probability of a shooter can hit the target.

Let's say the first shooter hit a target, then the second shooter hit $\frac{19}{20}$ of a target because I think he has a $\frac{1}{20}$ chance of shooting at the same target. The third hunter, well, if there's one target is hit so far, that's a $\frac{1}{20}$ chance, and then the third shooter hit $\frac{19}{20}$ of a target. If there are two targets get hit, then the third shooter hit $\frac{18}{20}$ of a target, but this looks pretty messy and I'm not even sure this is the correct approach. Is there any way to solve it?

Answer 1

Assume that there are $t$ targets and $s$ shooters. The first target is aimed at by the first shooter for her first shot with probability $1/t$ and hit with probability $p$ hence the first target is not hit by the first shot of the first shooter with probability $$q(t)=1-p/t.$$ By independence, the first target is not hit by the first shot of any shooter with probability $q(t)^s$. Thus, the number $T$ of targets not hit by any first shot is such that $$E(T)=t\cdot q(t)^s.$$ Assume now that $t\gt s$, hence $T\geqslant1$ almost surely. Applying once more the reasoning above, one sees that the mean number of targets not hit after two rounds of shots is $r=E(T\cdot q(T)^s)$, that is, $$r=E\left(T\cdot\left(1-\frac{p}T\right)^s\right).$$ Unfortunately, $r$ seems to depend on the full distribution of $T$, not only on its mean $E(T)$, and the hypothesis that $t=2s$ does not seem to help.

Exact computations when $t=4$ and $s=2$ yield $$r=4-4p+\tfrac12p^2+\tfrac16p^3+\tfrac1{24}p^4,$$ a formula which does not point to the possibility of huge simplification when $t=2s$ for some general $s$.

Answer 2

This problem is much more nuanced than I'd originally anticipated. The problem is that neither shot can be viewed as a set of independent trials, and thus not as a binomial process.

Letting $p\left(T_{i} = m\right)$ be the probability that the $i$th target is selected by $m$ shooters, and $p\left( H|m\right)$ be the probability that a target is hit if selected by $m$ shooters, then we can construct the tree $$\sum\limits_{{m_1}} {p\left( {{T_1} = {m_1}} \right)\left[ {\left. {\begin{smallmatrix} {p\left( {H|{m_1}} \right)}&{{n_1} = 1} \\ {1 - p\left( {H|{m_1}} \right)}&{{n_1} = 0} \end{smallmatrix}} \right|\,\,\,\sum\limits_{{m_2}} {p\left( {{T_2} = {m_2}|T_1 = m_1} \right)} \left[ {\left. {\begin{smallmatrix} {p\left( {H|{m_2}} \right)}&{{n_2} = 1} \\ {1 - p\left( {H|{m_2}} \right)}&{{n_2} = 0} \end{smallmatrix}} \right|\,\, \ldots } \right]} \right]}$$ by tracing back through the tree over all sums, which yields the probability of each tuple $\left( {{n_1},{n_2},{n_3}, \ldots ,{n_{20}}} \right)$ with $n_i = 0,1$ representing the number of hits on target $i$.

The problem is that $n_1,n_2,\ldots$ aren't IID and hence their sum isn't binomially distributed. I don't know what the conditional distribution looks like.

Let $p$ be the probability of shot success.

The expected value as a function of $p$ has the following form for some low numbers of shooters, with ${\rm number\ of\ targets = 2 \cdot number\ of\ shooters}$: $$\begin{array}{*{20}{l}} {}&{\rm after\ first\ shot}&{\rm after\ second\ shot} \\ {{n_{{\text{tgt}}}} = 2,\,\,{n_{{\text{s}}}} = 1}&p&{2p} \\ {{n_{{\text{tgt}}}} = 4,\,\,{n_{{\text{s}}}} = 2}&{2p - \tfrac{1}{4}{p^2}}&{4p - \tfrac{1}{2}{p^2} - \tfrac{1}{{2 \cdot 3}}{p^3} - \tfrac{1}{{{2^3} \cdot 3}}{p^4}} \\ {{n_{{\text{tgt}}}} = 6,\,\,{n_{{\text{s}}}} = 3}&{3p - \tfrac{1}{2}{p^2} + \tfrac{1}{{{2^2} \cdot {3^2}}}{p^3}}&\begin{gathered} 6p - {p^2} - \tfrac{{11}}{{{3^2} \cdot 5}}{p^3} - \tfrac{{23}}{{{2^3} \cdot 3 \cdot {5^2}}}{p^4} + \\ \,\,\,\,\,\,\,\,\,\tfrac{{71}}{{{2^5} \cdot {3^2} \cdot {5^2}}}{p^5} + \tfrac{{17 \cdot 37}}{{{2^6} \cdot {3^4} \cdot {5^2}}}{p^6} \\ \end{gathered} \\ {{n_{{\text{tgt}}}} = 8,\,\,{n_{{\text{s}}}} = 4}&{4p - \tfrac{3}{4}{p^2} + \tfrac{1}{{{2^4}}}{p^3} - \tfrac{1}{{{2^9}}}{p^4}}&\begin{gathered} 8p - \tfrac{3}{2}{p^2} - \tfrac{{17}}{{{2^3} \cdot 7}}{p^3} - \tfrac{{433}}{{{2^8} \cdot {7^2}}}{p^4} + \tfrac{{43 \cdot 307}}{{{2^7} \cdot 3 \cdot 5 \cdot {7^3}}}{p^5} + \\ \,\,\,\,\,\,\,\,\,\tfrac{{66533}}{{{2^7} \cdot {3^2} \cdot {5^2} \cdot {7^3}}}{p^6} + \tfrac{{13 \cdot 61 \cdot 113}}{{{2^{10}} \cdot {3^2} \cdot {5^3} \cdot {7^3}}}{p^7} - \tfrac{{11 \cdot 109 \cdot 5059}}{{{2^{14}} \cdot {3^3} \cdot {5^3} \cdot {7^3}}}{p^8} \\ \end{gathered} \end{array}$$ $$\begin{array}{*{20}{c}} {{n_{{\text{tgt}}}} = 10,\,\,{n_{{\text{s}}}} = 5}&{5p - {p^2} + \tfrac{1}{{2 \cdot 5}}{p^3} - \tfrac{1}{{{2^3} \cdot {5^2}}}{p^4} + \tfrac{1}{{{2^4} \cdot {5^4}}}{p^5}} & {\hspace{140pt}} \end{array}$$

Hopefully somebody else will take a crack at it.