Let $n,\ell\in\mathbb N$ for some $n\gg \ell\gg 1$.

The goal is to pick two sequences of numbers, $x_1,\ldots,x_\ell$ and $y_1,\ldots,y_\ell$ such that $$\Sigma_{i=1}^\ell x_i = n\quad{}\mbox{and}\quad{} \Sigma_{i=1}^\ell y_i = \ell\cdot n.$$

Next, there are $\ell$ bins such that bin $i$ contains $x_i$ blue balls. Given a random permutation $\sigma:[\ell]\to[\ell]$, $y_{i}$ is the number of red balls in bin $\sigma(i)$.

From every non-empty bin, a random ball is chosen uniformly, and we win the game if all sampled balls are red.

How can we select $x,y$ to maximize our winning odds?

If we pick $x_1,\ldots,x_\ell=n/\ell$ and $y_1,\ldots,y_\ell=n$, we have that in each bin a red ball will be sampled with probability $1-1/(\ell+1)$ and the chance of getting only red balls would be $(1-1/(\ell+1))^{\ell}\approx 1/e$.

As GMB answered, it's possible to improve this by setting $x_1=\ldots=x_{\ell-1}=0$ and $x_\ell=y_1=\ldots=y_\ell=n$, and getting a success probablity of 1/2.

Is this optimal?

Formulated as a math expression:

What is $\max_{x,y\mid \Sigma_{i=1}^\ell x_i = n, \Sigma_{i=1}^\ell y_i = \ell\cdot n} \quad{} \mathbb E_\sigma\; \left[\prod_{i=1}^\ell \frac{y_{\sigma^{-1}(i)}}{y_{\sigma^{-1}(i)}+x_i}\right]$?


1 Answer 1


I think $x_1 = \dots = x_{\ell-1} = 0, x_{\ell} = y_1 = \dots = y_{\ell} = n$ is better. You are guaranteed to choose red in all bins except for bin $\ell$, and bin $\ell$ has $n$ red balls and $n$ blue balls, so your overall odds of success are $1/2 > 1/e$.

  • $\begingroup$ You're right, missed that. Any thoughts on whether this is optimal? $\endgroup$
    – R B
    Commented May 24, 2019 at 22:39
  • $\begingroup$ @RB I would guess that it is. I don't have a formal proof, but some very sketchy calculations suggest that you can always improve a solution by moving a ball from one bin to another if it increases the variance in the x's or decreases the variance in the y's. So this solution would be the only unimprovable one. $\endgroup$
    – GMB
    Commented May 25, 2019 at 20:58

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