I am trying to come up with an explanation of the probability of birthday collision.
$P$(no collision among t people) = $(1− \frac{1}{365}) · (1-\frac{2}{365}) ··· (1-\frac{t-1}{365})$
For one person, the probability of no collision is 1, which is trivial since a single birthday cannot collide with anyone else’s. For the second person, the probability of no collision is 364 over 365, since there is only one day, the birthday of the first person, to collide with:
$P$(no collision among 2 people) = $(1− \frac{1}{365})$
If a third person joins the party, he or she can collide with both of the people already there, hence:
$P$(no collision among 2 people) = $(1− \frac{1}{365})·(1−\frac{2}{365})$
While it is clear how we get the probability of collision for 2 people, it is not intuitive to me that how we get the probability of collision between 3 people. I'd expect the probability would be $(1−\frac{2}{365})$. For example, when you roll the dice, the probability of $6$ is $\frac{1}{6}$, and the probability for 5 or 6 is $\frac{2}{6}$. It is not $\frac{1}{6}$·$\frac{2}{6}$ which seems to be the case in birthday collisions.
I'd appreciate an answer with an intuitive explanation.