Difficulty: Medium · Puzzle 5 of 6 · Topic: Probability Puzzles
How many people are needed so that probability of at least one shared birthday exceeds 50%? Ignore leap years and assume uniform birthdays.
Compute complement first: all birthdays distinct.
1) For n people, P(no match) = product_{k=0 to n-1} (365-k)/365.
2) P(at least one match) = 1 - P(no match).
3) Evaluate around n=23: P(no match) approx 0.4927.
4) So P(match) approx 1 - 0.4927 = 0.5073.
5) At n=22 this value is still below 0.5, so threshold is 23.Answer: 23 people are needed for probability > 50% of a shared birthday.
Checked by running a model of the rule in a Python model. A computational check: the exact product formula gives 0.4757 at 22 people and 0.5073 at 23, so 23 is the threshold; 57 people reach 99% (P(56) = 0.988).
Linear intuition ('365/2 = 183 people'): the number of pairs grows as n(n-1)/2, so by n=23 there are 253 pairs, each a 1/365 chance.
Approximate n for 99% collision probability. Model answer: n = 57 (P(57) = 0.990, P(56) = 0.988). The rule of thumb n ~ sqrt(2*365*ln(1/(1-P))) gives about 58, a slight overestimate of the exact 57.
It's easier to compute the chance of no shared birthday, which is the product of 365 minus k over 365 for k from zero to n minus one. At 23 people that product is about 0.4927, so the chance of a match is about 0.507, just over half. At 22 people it is still below half, so the threshold is 23. The trap is linear intuition like 183 people; what grows is the number of pairs, and 23 people already have 253 pairs.