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How many people do you need in the room?

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The question you opened

How many people share a birthday

You want a better than even chance that two people in a room share a birthday.

How many people do you need in the room?

23 people

Twenty three people make 253 pairs, and the chance that all 253 pairs differ falls just below half. A match arrives 50.7 percent of the time.

The answer is 23, and the reason is that the question is about pairs. With n people there are n times n minus 1, over 2, pairs. At 23 people that is 253 pairs, each one a separate chance for a match. The clean way to compute it is to ask for the opposite.

The chance that all birthdays differ is 365 over 365, times 364 over 365, times 363 over 365, and so on for 23 terms. That product is about 0.493, so the chance of at least one match is about 0.507. At 22 people the match probability is 47.6 percent, so 23 is the first room where a shared birthday is more likely than not.

The intuition fails because people count themselves against everyone else and stop there. One person against 22 others is 22 chances, which really would need a much larger room. The other 231 pairs never enter the mental picture, and they are the majority of the evidence. A different question has a different answer, and mixing the two causes most of the confusion.

The chance that someone shares your birthday specifically passes half only at 253 people, which is where the number 183 people expect belongs, roughly. Any match is easy. A match with you is hard. The same shape governs hash collisions in computing, where it sets how long an identifier has to be before duplicates become unlikely, and it explains why coincidences feel supernatural.

A large group generates an enormous number of pairs, and any one of them producing something striking is close to certain. Assumptions worth stating: births are treated as uniform across the year and leap days are ignored. Real birthdays cluster slightly by season, which makes matches marginally more likely, so 23 is if anything a safe answer.

Technique: Count the pairs

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