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We have $n$ boxes. And initially there are $x_1, x_2, x_3, \ldots, x_n$ marbles in each box. We randomly (with equal probabilities) select one of the boxes. We take one marble from it and we put it into another (different from the origin) box chosen randomly (with equal probabilities). We continue this process until one of the boxes become empty. How many operations we do on average?

It is not a homework. I don't know whether a closed form solution exists. My current results are:

\begin{align}{} x_1 x_2 & \text{ for } n=2\\\ \frac{3x_1 x_2 x_3}{x_1 + x_2 + x_3} & \text{ for } n=3 \end{align}

I have crossposted in artofproblemsolving. This problem is related and maybe (or not) useful.

Update2: As i learned: this problem has been studied before. As usual :) It seems very hard even for $n=4$. No explicit solution is known, only asymptotics for the case $f(x,x,x,x)$. Nevertheless the solution is much much more easier if we change slightly the problem. For example.

Big thanks to Viktor for pointing the reference!

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Roah
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  • Looking at what you have for $n=2$ and $n=3$, my guess would be $\frac{\binom{n}{2}}{\displaystyle \sum_{i,j,i \neq j}\frac{1}{x_i x_j}}$ –  Mar 13 '11 at 20:33
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    It's beautiful but it seems that it does not pass the following test for $n=4$: $f(x,x,x,x)=f(x-1,x+1,x,x)+1$. – Roah Mar 13 '11 at 21:21
  • Are there any bounds/constrains for $x_i$ or for $\sum x_i$? – Eelvex Mar 13 '11 at 23:40
  • $x_{i}$ are natural numbers – Roah Mar 14 '11 at 06:28
  • Could You provide a proof for the case when $n=2$. It is a bit suprising for me that when we have $x_{1}=1$ and $x_{2} \to \infty$ then expected number of operations goes to $\infty$. Intuitively there shouldn't be big difference between case $x_{1}=1 \text{ } x_{2}=1000$ and $x_{1}=1 \text{ } x_{2}=10^{6}$. I'm not saying that this result is wrong, it is just suprising for me. – Tomek Tarczynski Mar 14 '11 at 20:06
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    @Tomek The case $n=2$ is the usual one dimensional gambler's ruin problem. It is true, but not intuitively obvious, that for a symmetric random walk the expected number of steps needed to hit 0 is infinite, even starting at 1. Unless we hit 0 very soon, the random walk wanders far into the positives and then takes a long (but finite!) number of steps to reach 0. –  Mar 15 '11 at 00:22
  • @Tomek Tarczynski: See http://math.stackexchange.com/questions/23808/coin-toss-question/23845. – joriki Mar 15 '11 at 02:38
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    That is N-player ruin problem. See: Y.Swan - A Matrix-Analytic Approach to the N-Player Ruin Problem – Viktor Mar 15 '11 at 23:23
  • @solomoan: Wow -- you're setting almost your entire reputation on this? :-) – joriki Mar 16 '11 at 16:17
  • @Roah: I think it would be good to explicate in the question what the "slight change" is -- if I understand correctly, the difference to the Ross paper is that he waits until one player has all the money, and you stop when one player has no money? I'm surprised that that change should make the problem easier; I would have expected it to get harder. Also, you write "For example" -- do you have any other "slight changes" in mind that lead to solved (or solvable) variations of the problem? – joriki Mar 16 '11 at 16:21
  • @joriki: Yes, that is the only difference to the Ross paper; that surprises me too. There is another variation which is also solved for n=3 but not for n>3 (as far as i know now). We randomly choose a box and we add there (n-1) marbles, one from each of the remaining boxes. – Roah Mar 16 '11 at 17:11
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    There is a paper arguing for why this is unlikely to have any closed form solution for N>3. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aap/1046366109 – TROLLHUNTER Mar 18 '11 at 07:42
  • I don't have access to that paper, but I bet that the essence is that there is no Riemann mapping theorem in higher dimensions. – Douglas Zare Apr 11 '11 at 16:30

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Here are some results for very small numbers, when there are $n$ variables: $$ \begin{align*} f(1,1,1,\ldots,1) &= 1, \\ f(2,1,1,\ldots,1) &= \frac{n}{n-1}, \\ f(3,1,1,\ldots,1) &= \frac{n^3-2n^2+3n}{n^3-3n^2+4n-2} = \frac{n}{n-1} \cdot \frac{n^2-2n+3}{n^2-2n+2}, \\ f(2,2,1,\ldots,1) &= \frac{n^3-n^2+2n}{n^3-3n^2+4n-2} = \frac{n}{n-1} \cdot \frac{n^2-n+2}{n^2-2n+2}. \end{align*} $$

Yuval Filmus
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