The Birthday Problem and the 23 Person Answer

With 23 people in a room the probability that two share a birthday passes 50 percent, driven by the number of pairs rather than the number of days.

With 23 people in a room, the probability that at least two share a birthday exceeds 50 percent, reaching approximately 50.7 percent, a number that surprises almost everyone the first time they compute it because 23 is such a small fraction of the 365 days available.

Take n people, and assume each person's birthday is drawn independently and uniformly from 365 possible days, ignoring leap years and any seasonal skew in real birth rates. The question asks for the smallest n at which the probability that at least two people in the room share a birthday exceeds one half, and for the exact probability at n equals 23.

Computing the probability of a match directly is awkward, since a match could occur between any pair among the n people and those events overlap heavily. The standard move is to compute the probability of no match at all and subtract from 1, since no-match is a single clean event: every person's birthday differs from every other person's birthday.

Line the n people up one at a time. The first person can have any birthday, a certainty. The second person must avoid the first person's day, which happens with probability 364 divided by 365. The third person must avoid both of the first two people's days, an event of probability 363 divided by 365, given the first two already differ. Continuing this pattern, the k-th person must avoid k minus 1 already-used days, giving a factor of 365 minus k plus 1, all divided by 365.

The probability that all n birthdays are distinct is therefore the product of 365 divided by 365, times 364 divided by 365, times 363 divided by 365, and so on down to 365 minus n plus 1, divided by 365. For n equal to 23, this product works out to approximately 0.4927. Subtracting from 1 gives approximately 0.5073, just over one half.