The Duplicate Birthday

The teacher wins. With 23 students, P(shared birthday) ≈ 50.7%. The counterintuition: 23 people create 253 pairs, not 23 comparisons. Pair count grows quadratically, driving the probability past 50% at just 23 people.

The paradox arises because we intuitively measure 'my birthday vs everyone else's' — a 1-vs-22 comparison. But the actual question counts all pairwise comparisons: 23 people create C(23,2) = 253 distinct pairs, each with a small but non-trivial chance of matching. Step 1: Calculate P(no shared birthdays) — the complement. Person 1: any birthday = 365/365. Person 2: must differ from person 1 = 364/365. Person 3: must differ from persons 1 and 2 = 363/365. And so on. Step 2: P(all different) = (365 x 364 x 363 x ... x 343) / 365^23. Step 3: Computing: 365/365 x 364/365 x 363/365 x ... Multiply 23 terms. Result ≈ 0.4927. Step 4: P(at least one match) = 1 − 0.4927 ≈ 0.5073, or roughly 50.7%. Step 5: The probability crosses 50% at exactly 23 people. At 30 people it rises to 70%, and at 57 people it exceeds 99%. Step 6: The key: 23 people form 253 pairs. Each pair has a 1/365 ≈ 0.27% chance of sharing a birthday. 253 small chances accumulate to just over 50% — like 253 lottery tickets each with 0.27% odds. Key Insight: People compare themselves against the group (23 vs 365 ≈ 6%). The question actually measures pair collisions — and 23 people create 253 pairs. Pair count grows as n(n-1)/2, which shoots up far faster than the number of people, pushing the probability past 50% sooner than intuition expects.