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Given $N$ urns, what is the expected number of balls that need to be thrown until one of the urns has $\beta$ more balls than all the other urns?

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closed as off-topic by heropup, metamorphy, José Carlos Santos, user91500, Paul Frost Nov 10 at 12:24

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    $\begingroup$ This is a Markov chain with infinitely many states. I doubt there's a closed-form solution. For specific values of $N$ and $\beta$, you could try simulation. (I'm assuming the question means that some urn has at least $\beta$ more balls than any other urn.) $\endgroup$ – saulspatz Nov 8 at 18:21
  • $\begingroup$ @saulspatz The states are not infinite, because we can always remove $x_m$ balls from all urns (where $x_m$ is the amount of balls of the least populated urn) and get an equivalent state. They are too many states, still. $\endgroup$ – leonbloy Nov 8 at 23:45
  • $\begingroup$ I assume the statement is equivalent to "... until $x_1 = x_2 + \beta$ where $x_1,x_2$ are the count of balls in the most populated and second most populated urns". $\endgroup$ – leonbloy Nov 8 at 23:46
  • $\begingroup$ @leonbloy I thought that too, at first, but why can't we have one urn with $0$ balls, and $N-1$ urns with $\alpha$ balls, $\alpha =0,1,2,\dots$? $\endgroup$ – saulspatz Nov 9 at 1:51
  • $\begingroup$ Wouldn't this Markov chain be terminating? We may define a single absorption state. $\endgroup$ – Dranithix Nov 9 at 10:31