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We have n urns. Repeat next process k times: choose 2 distinct urns, throw ball into each. What is the probability of choosing 2 urns with at least 1 ball in it? (e.g. we have 8 urns. Then choose 3 times: (1, 2), (7, 4), (8, 2). So we have 2 balls in 2-nd urn and 1 ball in each of urns 1, 4, 7, 8.)

Probability of urn being with at least 1 ball after k trials: $P=1-(\frac{n-2}{n})^k$. If one chooses 2 urns independently: $P=(1-(\frac{n-2}{n})^k)^2$. What will be the probability if one chooses 2 distinct urns at the same time?

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The probability that one iteration leaves both urns empty is

$$ \frac{\binom{n-2}2}{\binom n2}=\frac{(n-2)(n-3)}{n(n-1)}\;, $$

so the probability of both urns being empty is

$$ \left(\frac{(n-2)(n-3)}{n(n-1)}\right)^k\;. $$

The probability of a particular one of the two urns being empty is

$$ \left(\frac{n-2}n\right)^k\;. $$

Thus the probability of neither urn being empty is

$$ 1-2\,\left(\frac{n-2}n\right)^k+\left(\frac{(n-2)(n-3)}{n(n-1)}\right)^k\;. $$

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    $\begingroup$ Thank you very much! This helps a lot for understanding. $\endgroup$ Commented May 5, 2016 at 6:05

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