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I'm trying to calculate the probability that at least two people out of three have the same birthday (simplified Birthday Problem). I know that calculating it using the complementary event is easiest, but I want to calculate it directly. So here's an approach that works:

$P(A) = 3\cdot\frac{1}{356}\cdot\frac{364}{356} + \frac{1}{365^2}$ where I account for the three ways 2 people can have the same birthday and the other a different birthday and plus for the event that all three have the same birthday which has a probability of $\frac{1}{365^2}$.

But here's a calculation that doesn't work: Let's first define the events:

$E_{1}:$ person 1 and person 2 have the same birthday.

$E_{2}:$ person 1 and person 3 have the same birthday.

$E_{3}:$ person 2 and person 3 have the same birthday.

Now we can calculate the probability of at least two people having the same birthday as: $$P(E_{1}\cup E_{2}\cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}\cap E_{2}) - P(E_{1}\cap E_{3}) - P(E_{2}\cap E_{3}) - P(E_{1}\cap E_{2}\cap E_{3}) $$

with $P(E_{i}) = \frac{1}{365}$ , $P(E_{i}\cap E_{j}) = \frac{1}{365^2}$ and $P(E_{1}\cap E_{2} \cap E_{3}) = \frac{1}{365^2}$

but this produces the wrong answer 0.008189153 instead of the correct one: 0.008204165.

Why does the second calculation produce a wrong answer? What am I not taking into consideration?

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    $\begingroup$ Why did you subtract $P(E_1 \cap E_2 \cap E_3)$? You should have added it. $\endgroup$ – fleablood Dec 21 '18 at 17:30
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You have done your inclusion-exclusion principle incorrectly. Instead of:

$$P(E_{1}\cup E_{2}\cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}\cap E_{2}) - P(E_{1}\cap E_{3}) - P(E_{2}\cap E_{3}) - P(E_{1}\cap E_{2}\cap E_{3}) $$

It should have been:

$$P(E_{1}\cup E_{2}\cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}\cap E_{2}) - P(E_{1}\cap E_{3}) - P(E_{2}\cap E_{3}) \color{red}+ P(E_{1}\cap E_{2}\cap E_{3}) $$

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