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Consider a finite multiset:

$S = \{ \{a, a\}, \{\}, \{ a, c, d \}, \ldots, \{ a, t, u \}\}$

where $\forall s \in S$, $0 \leqslant \|s\| \leqslant n$, and where each multiset $s$ contains only elements from a finite set of terms: $s \subseteq \Gamma$.

For any random $s \in S$ and term $\gamma \in \Gamma$, what is the pmf which describes the probability of $\gamma$ occurring in $s$ a total of (each of) $[0,\|s\|]$ times?


Update:

Because $S$ is arbitrary (may contain multisets $s$ of any size from $[0,n]$), @Raskolnikov has suggested a counting approach using the following:

$\frac{\# \text{ of multisets containing } n \text{ ocurrences of } \gamma}{\# \text{ of multisets of cardinality at least } n} \;$

This calculates the probability of some $\gamma \in \Gamma$ occurring $n$ times in any $s \in S$ where $\|s\| \geqslant n$.

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  • $\begingroup$ You won't have a standard distribution for your example if your set $S$ is arbitrary. If on the other hand $S$ contained all possible multisets between size $0$ and $n$, then it is possible to find a standard formula. Anyway, whatever the case is, all you have to do is compute:$$\frac{\# \text{ of multisets containing } \gamma}{\# \text{ of multisets}} \; .$$ $\endgroup$ – Raskolnikov Jul 18 '13 at 5:30
  • $\begingroup$ $S$ is definitely arbitrary (doesn't contain all possible multisets). @Raskolnikov I don't quite understand how computing what you've suggested helps... for example, how do I compute the probability that $\gamma=a$ occurs in some random $s \in S$ a total of 2 times, for all $s$ where $\|s\| \geqslant 2$? $\endgroup$ – sharky Jul 18 '13 at 5:40
  • $\begingroup$ By counting. I can't answer the question any more precisely if you don't specify precisely what the multiset $S$ looks like. $\endgroup$ – Raskolnikov Jul 18 '13 at 5:41
  • $\begingroup$ I think I understand, thanks - so perhaps it should be calculated as: $\frac{\# \text{ of multisets containing } n \text{ ocurrences of } \gamma}{\# \text{ of multisets of cardinality at least } n} \;$ ...? $\endgroup$ – sharky Jul 18 '13 at 5:49
  • $\begingroup$ Sure, that would be the way to go. $\endgroup$ – Raskolnikov Jul 18 '13 at 6:05

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