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The example in question is from Rick Durrett's "Elementrary Probability for Applications", and the setup is something like this:

Let $A$ be the event "Alice and Betty have the same birthday", $B$ be the event "Betty and Carol have the same birthday", and $C$ be the event "Carol and Alice have the same birthday".

Durrett goes on to demonstrate that each pair is independent, since for example,

$P(A \cap B) = P(A)P(B)$.

However, he concludes that $A, B$, and $C$ are not independent, since

$P(A \cap B \cap C) = \frac{1}{365^2} \neq \frac{1}{365^3} = P(A)P(B)P(C)$.

I understand the reasoning here, and that one can generally show that arbitrary events $X$ and $Y$ are not independent by showing that $P(X\cap Y) \neq P(X)P(Y)$.

I am a little new to probability, though, and don't understand why exactly $P(A \cap B \cap C) = \frac{1}{365^2}$.

My progress so far:

I do see why $P(A) = P(B) =P(C) = \frac{1}{365}$, and thus why $P(A)P(B)P(C) = \frac{1}{365^3}$.

It seems like the sample space $\Omega = \{ (a, b, c) \mid a,b,c \in [365] \}$ -- i.e., all of the possible triples of numbers from 1 to 365, where 1 denotes January 1st, 2 denotes January 2nd, etc. From that, I can conclude $|\Omega| = 365^3$, but I'm not sure where to go from here.

It seems like once a single birthday is chosen, the rest are completely determined if they're all equal to each other - is this a good direction to go in?

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up vote 5 down vote accepted

There are $365$ tuples of the form $(a,a,a)$ out of $365^3$ tuples in the sample space, hence the probability.

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Ah! Yep, that's exactly what I was missing. That clears it up perfectly, thanks! – dzackgarza Jan 10 at 0:55

The event $A\cap B\cap C$ happens if and only if the event $A\cap B$ happens. And you know how to find $\Pr(A\cap B)$.

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