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There are 400 people in a room. I pick two people at random. What is the probability that they have the same birthday?

I know that there must be two people in the room who share the same birthday through pigeonhole principle.

But if I pick two people at random I am not sure how to calculate the probability.

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    $\begingroup$ Welcome to Math SE. FYI, a related problem is the Birthday problem. $\endgroup$ – John Omielan Feb 28 at 19:04
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    $\begingroup$ This is easier than the famous birthday problem linked in the other comment. If we assume (for simplicity, and in the absence of any other data) that each of $365$ possible birthdays is equally likely, then the first person's birthday is irrelevant, and the question is just the probability that the second person's birthday is that one. So $1/365$. It's a little trickier if you take Leap Days (like tomorrow) into account. $\endgroup$ – Toby Bartels Feb 28 at 19:06
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    $\begingroup$ @amWhy Did you read the previous comment? $\endgroup$ – NCh Feb 29 at 0:57
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The way the problem is stated, the number of people in the room is irrelevant. The probability that any two people have the same birthday is approximately $\frac{1}{365.25}$ assuming a uniform distribution of birthdays over time.

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    $\begingroup$ If you are going to say $365.25$ rather than $365$ and talk only about days of the year while ignoring years themselves., then why not use the more accurate $365.2425$? If you were to insist on leap days having some influence, then having a guess as to the age and current year will influence our expectation that there is the chance that one of our selected people were actually born on a leap day $\endgroup$ – JMoravitz Feb 28 at 19:20
  • $\begingroup$ @JMoravitz The only recent year that messes up 365.25 is 1900. I don't think there are too many people around born that year. The original question did not have year of birth. $\endgroup$ – herb steinberg Feb 28 at 21:49

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