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On maximizing the number of heads when you need to set as...
Wouter van Doorn · 2024-06-21 · via math.PR updates on arXiv.org

You play the following game: you start out with $n$ coins that all have probability $p$ to land heads. You toss all of them and you then need to set aside at least one of them, which will not be tossed again. Now you repeat the process with the remaining coins. This continues (for at most $n$ rounds) until all coins have been set aside. Your goal is to maximize the total number of heads you end up with. In this paper we will prove that there exists a constant $p_0 \approx 0.5495021777642$ such that, if $p_0 < p \le 1$ and the number of remaining coins is large enough, then it is optimal to set aside exactly one coin every round, unless all coins landed heads. When $\frac{1}{2} \le p \le p_0$, it is optimal to set aside exactly one coin every round, unless at most one coin came up tails. Let $v_{n,p}$ be the expected number of heads you obtain when using the optimal strategy. We will show that $v_{n,p}$ is larger than or equal to $v_{n-1,p} + 1$ for all $n \ge 3$ and all $p \ge \frac{1}{2}$. When $p < \frac{1}{2}$ there are infinitely many $n$ for which this inequality does not hold. Finally, we will see that for every $p$ with $0 < p \le 1$ the sequence $n - v_{n,p}$ is convergent.