Heads or Tails

The players are not going to accept the possibility that one of their opponents will have a better chance of winning than them, so they are bound to play for a situation where everyone has the same chance of winning. If $N(x)$ people choose $x$ then the probability that any one of those people wins is $\frac{1}{N(x)}{10 \choose x} 2^{-10} \hspace{2cm} 0 \le x \le 10$ and so these probabilities must all be equal. Thus we must have (if there are $M$ players in total) $N(x) \; = \; \frac{M}{2^{10}}{10 \choose x} \hspace{2cm} 0 \le x \le 10$ and so the expected number of survivors of a round is $\mathbb{E}[N] \; = \; \sum_{n=0}^{10} \frac{M}{2^{20}}{10 \choose x}^2 \; = \; \frac{M}{2^{20}}{20 \choose 10}$ and so the expected proportion of survivors is ${20 \choose 10}2^{-20}$ which is $\boxed{18}$ % to the nearest percentage point.

The probability of getting $k$ heads out of $n$ coins follows the binomial distribution and it is given by $Pr(X=k) = \displaystyle {n \choose k} p^k (1-p)^{n-k}$ , where $p$ is the probability of getting a head in a toss.

For 10 fair coins, we have $n=10$ and $p = \frac 12$ and $Pr(X=k)$ $\displaystyle = {10 \choose k} \left(\frac 12\right)^k \left(\frac 12\right)^{10-k}$ $= \dfrac {10 \choose k}{2^{10}}$ .

Let the large number of people before a round be $N$ . As the players are extremely smart the distribution of number of players $N_k$ selecting $k$ heads in the next toss follows the $Pr(X=k)$ . The average number of players who make it through a round is the expected value of number of winners in a round as given below.

$\begin{aligned} E[X] & = \sum_{k=0}^{10} Pr(X=k)N_k = \sum_{k=0}^{10} \frac {{10 \choose k}^2}{2^{20}} N = \frac {20 \choose 10}{2^{20}}N \end{aligned}$

$\begin{aligned} \implies \frac {E[X]}N & = \frac {20 \choose 10}{2^{20}} = \frac {184756}{1024^2} \approx 0.18 = 18\% \end{aligned}$

$\implies P = \boxed{18}$