The first shared birthday arrives on person ten

Question

People enter a room one at a time, birthdays independent and uniform over $365$ days.

What is the probability that the first shared birthday appears exactly when the tenth person walks in? (That is, the first nine are all distinct, and the tenth matches someone.)

Accepted Answer

The event has two parts that must both happen: the first nine birthdays are all different, and then the tenth person collides with one of them. Part 1, the first nine all distinct. Seat them one at a time, each avoiding the days already used: P(	ext{first 9 distinct}) = \frac{365}{365}\cdot\frac{364}{365}\cdots\frac{357}{365} \approx 0.9054. Part 2, given nine distinct birthdays are on the board, the tenth person matches one of those nine days: P(	ext{10th matches} \mid 	ext{9 distinct}) = \frac{9}{365} \approx 0.02466. Multiply the two parts: P(	ext{first collision at person 10}) \approx…

The first shared birthday arrives on person ten

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