Question: Let $\omega_1,\omega_2,\cdots,\omega_k$ be $k$ distinct events. When we perform a random test, the outcome $\omega_1,\omega_2,\cdots,\omega_k$ occurs with equal possibility $\frac{1}{k}$. Now we perform the random test independently for $m$ times, then what is the possibility that each event occurs at least once?

Example: For $k=6$, the question is, after we throw a dice for $m$ times, what is the possibility that each of the numbers $1,2,3,4,5,6$ occurs at least once?

My Solution: Consider the inclusion-exclusion principle. The possibility is $1$ minus the possibility that one event doesn't occur, plus the possibility that two events don't occur, minus the possibility that three events don't occur, ... Then the result is $$ P=\sum\limits_{i=0}^{k}{(-1)}^i\binom{k}{i}{\left(\frac{k-i}{k}\right)}^m $$ My Question: Is there a more elegant way to solve this problem (which yields a simpler expression rather than a summation)?

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    $\begingroup$ This is related to Coupon collector's problem (Wikipedia link). I'm not familiar with this topic, but answers in Math.SE (see this or this) seem to indicate that there are no simpler way to compute the value. $\endgroup$
    – VTand
    Jan 8, 2023 at 9:23
  • $\begingroup$ @VTand thans for your comment. I found that my formula is like the Stirling number of the second kind, except that the latter is divided my $k!$ while mine is not. $\endgroup$
    – Soha
    Jan 8, 2023 at 9:58

1 Answer 1


Many problems can be fitted into a balls-in-bins type. Look at it as the probability of putting distinct balls into distincr bins with no bin empty. Firstly counting the # of ways m balls can be put in k identical bins, the answer (by the very definition of Stirling numbers of the second kind, is $\large{m\brace k}$

But since our bins are distinct, they can be permuted in $k!$ ways, so the favourable count is $k!\large{m\brace k}$ against a total of $k^m$ ways

thus $Pr = \huge\frac{k!\Large{m\brace k}}{k^m}$


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