# Probability of someone being born the same day as you in a group of N people.

Provided that I went to a class with $N$ people, where everyone was born the same year in the same city where there are $K$ hospitals, what's the probability that someone was born in the same hospital and in the same day as me?

My approach would be as follows:

$$\left( 1- \left(\frac{364}{365} \right)^N \right) \cdot \left(1-\left(\frac{K-1}{K} \right)^N \right)$$

But I'm not really sure that this is the correct answer :)

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How do you take into account leap (bissextile) years ? –  wap26 Jul 4 '12 at 8:31
I don't, just assume we were all born on a normal year :) –  Hallucynogenyc Jul 4 '12 at 8:34
Your expression is the probability that someone was born on the same day as you, and that someone (not necessarily the same person) was born in the same hospital as you. If you want to know the probability that at least one person was born on the same day and in the same hospital as you, see Andre Nicolas's answer. –  ShreevatsaR Jul 4 '12 at 8:44
I was born on August 1, 1988 at 2:00 pm and my name is Brian Wayne Miller. There was another baby born at the exact same hospital on August 1, 1988 at 3:00 pm and his name was Brian Wayne Miller. What are the odds of that??? –  user37310 Aug 6 '12 at 3:50
Since it actually happened, the odds are pretty good. But if you want to ask a question, you should post it as a question, not as an answer to some other question. –  Gerry Myerson Aug 7 '12 at 3:27

Under appropriate independence assumptions, and assumptions about equality of hospitals, the probability that someone chosen at random was born on the same day as you and in the same hospital is $$\frac{1}{365}\cdot\frac{1}{K}=\frac{1}{365K}.$$ So the probability the person was born on a different day or a different hospital than you is $$1-\frac{1}{365K}.$$ The probability that all $N$ were born on a different day or different hospital than you is $$\left(1-\frac{1}{365K}\right)^N.$$ So the probability that at least one person "double matches" with you is $$1-\left(1-\frac{1}{365K}\right)^N.$$

Remark: This is exactly the same reasoning as the pure days problem, except that we now have $365K$ abstract "days."

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+1 Your answer, as always, is crystal clear. –  Galois Group Aug 6 '12 at 4:37
@FortuonPaendrag: Thank you. I (usually) try for clarity. With mathematics, it is in principle possible! –  André Nicolas Aug 6 '12 at 4:45