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There are $2$ ways to unravel $\Pr(A\ \cap\ B)$ using conditional probability, either $\Pr(A\ |\ B)\Pr(B)$ or $\Pr(B\ |\ A)\Pr(A)$.

Unraveling $\Pr(C\ \cap\ D\ \cap\ E)$ gets a bit trickier. In the first stage of applying conditional probability in the three-event case, there is freedom in choosing which two of the letters to group together, as well as the same freedom as in the two-event case of which letter or group to condition on...

  • Grouping $C$ and $D$ yields either $\Pr((C\ \cap\ D)\ |\ E)\Pr(E)$ or $\Pr(E\ |\ (C\ \cap\ D))\Pr(C\ \cap\ D)$.
  • Grouping $C$ and $E$ yields either $\Pr((C\ \cap\ E)\ |\ D)\Pr(D)$ or $\Pr(D\ |\ (C\ \cap\ E))\Pr(C\ \cap\ E)$.
  • Grouping $D$ and $E$ yields either $\Pr((D\ \cap\ E)\ |\ C)\Pr(C)$ or $\Pr(C\ |\ (D\ \cap\ E))\Pr(D\ \cap\ E)$.

In the second stage of applying conditional probability in the three-event case, there is another instance of conditioning freedom, at least for each of the above expressions on the right. For example, $\Pr(E\ |\ (C\ \cap\ D))\Pr(C\ \cap\ D)$ yields either $\Pr(E\ |\ (C\ \cap\ D))\Pr(C\ |\ D)\Pr(D)$ or $\Pr(E\ |\ (C\ \cap\ D))\Pr(D\ |\ C)\Pr(C)$. Either of these looks like a textbook three-event probability-of-intersection formula, so I don't think I'm supposed to further expand. As for the above expressions on the left, I'm less sure if I need to further expand or how. For example, does $\Pr((C\ \cap\ D)\ |\ E)\Pr(E)$ yield $\Pr((C\ |\ D)\ |\ E)\Pr(D)\Pr(E)$?

Without this, I appear to have $9$ ways to unravel a probability of a three-event intersection. Intuitively, I am expecting the actual number to be an integer multiple of six, perhaps $12$, as there are six ways to permute the names of three events. For an $n$-event intersection, I expect the number to be a not-necessarily-constant integer multiple of $n!$. How many different maximally-unraveled expressions for a probability of intersections are there, and what steps beyond what I've already explained are necessary to find them?

user10478
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    $\Pr((C\mid D)\mid E)$ is nonsensical. The pipe, $\mid$ , is not a set connective. Nor is something like $C\mid D$ an event. The conditional probability function has only one pipe between the event and the condition over which it is being evaluated. – Graham Kemp May 19 '23 at 03:56
  • $\Pr(C \cap D \cap E)= \Pr(C \cap D \mid E),\Pr(E)=\Pr(C\mid D \cap E),\Pr(D \mid E),\Pr(E)$ and similar expressions work, where $\cap$ has priority over $\mid$ – Henry May 28 '23 at 00:13

1 Answers1

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We can compute the numbers in a recursive way. Let $N_k$ denote the number of ways to unravel the probability of intersections of $k$ events. We have $N_1 = 1$ and $N_2 = 2$. For $k \geq 2$, we have $N_k = \Pr[\bigcap_{i \in [k]} A_i] = \Pr[\bigcap_{i \in S} A_i \cap \bigcap_{j \in [k] \setminus S} A_j] = \Pr[\bigcap_{i \in S} A_i \mid \bigcap_{j \in [k] \setminus S} A_j] \Pr[\bigcap_{j \in [k] \setminus S} A_j], \forall \emptyset \neq S \subsetneq [k]$, where $\Pr[\bigcap_{i \in S} A_i \mid \bigcap_{j \in [k] \setminus S} A_j]$ has $N_{|S|}$ ways and $\Pr[\bigcap_{j \in [k] \setminus S} A_j]$ has $k - |S|$ ways. Therefore, we have $$N_k = \sum_{s = 1}^{k-1} \binom{k}{s} N_s N_{k-s},$$ which gives $N_k = (2k - 2)! / (k - 1)!$.

See https://oeis.org/A001813

See also Double conditional probability

Vezen BU
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