A coin is flipped $n$ times and you win if it comes up heads at least $k$ times. The coin is unusual in that you're allowed to pick the probability $p_i$ that it comes up heads on the $i$th flip, subject only to the constraint that $\sum_i p_i \le b$, where $b$ is some predetermined "budget" that you have. Moreover, you are allowed to wait until you've seen the results of the first $i-1$ flips before choosing the value of $p_i$. Given $n$, $k$, and $b$, what is your optimal strategy, and what is your probability of winning?

One colorful way to state the problem is that if you're a sports team tasked with winning a best-of-$n$ series and you have limited resources (e.g., a limited bullpen for the World Series of major league baseball), how should you budget them?

Naturally, if $b\ge k$, you can simply pick $p_i=1$ for $k$ of the $n$ flips, and win with probability 1. So the question is interesting only if $b\lt k$.

I've circulated this problem informally among colleagues, who have obtained miscellaneous partial results but not a full solution. It would take too much space to summarize all the partial results, but let me mention some of the highlights.

Even the "non-adaptive case," where you're

*not*allowed to see the results of your flips before choosing $p_i$, is not trivial. The best strategy is to divide the budget evenly over $r$ flips for some $r$, but the exact value of $r$ is more complicated than you might think. For a given $r$, the probability of $k$ successes is $$\sum_{m=k}^r {r \choose m} \left({b\over r}\right)^m\left(1-{b\over r}\right)^{r-m}.$$ From this it appears that if $b\lt k-1$ then we should choose $r=n$, and if $k-1 \le b \lt k$ then $r\approx (k-1)/3(b-k+1)$, but we have a proof only in special cases.In the actual stated problem, let's let $d=k-b$, the

*deficit*. Then, at least in the small-deficit case, the best general strategy we have so far is to make an initial coin flip with probability $1-\lbrace d\rbrace$ (where $\lbrace d\rbrace$ denotes the fractional part of $d$), and then take $p_i=1/2$ until we find ourselves in a situation where we can "clinch" the win by taking the remaining $p_i=1$. (It's possible to analyze this strategy quantitatively but I'll omit the details here.) In particular, one can show that adaptive strategies significantly outperform non-adaptive strategies.If $b$ is small then one can show that the best non-adaptive strategy is within a constant factor of optimum. For example if $b\le 1$, then one can show that the overall winning probability $p$ satisfies $${1\over 4}{n\choose k}\left({b\over n}\right)^k \le p \le {n\choose k}\left({b\over n}\right)^k.$$ The upper bound is actually true for all $b$ and the lower bound can be derived from the best non-adaptive strategy.

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