Birthday Matches in a Small Room

Each card represents a person whose birthday is assigned to one of the possible days. Run adds people gradually; Step lets you inspect each arrival. When a day label repeats, every revealed card with that label is marked as a match. The room starts empty so the first comparisons remain visible.

The usual version uses 365 dates, with no leap day, and makes every date equally likely. The seven-day setting is a deliberately smaller model, not a week-long year. The seed makes the simulated assignments repeatable, so a shared link reproduces the same group.

The second person creates one pair. The third creates two more, because they can match either earlier person. By the time there are 23 people, there are 23 × 22 / 2 = 253 possible pairs. The question concerns a match in any one of them, not just a match to a person selected in advance.

Pair counting explains why the chance grows quickly, but dividing 253 by 365 does not give the probability of at least one match. That quotient is the expected number of matching pairs. A single room can contribute several pairs, so expected pair count and match probability differ.

Instead of counting all the ways a match could occur, count the ways everyone could have a different day. The first person has all 365 dates available. If they remain distinct, the second must avoid one date, the third must avoid two, and each later arrival has fewer unused dates.

These are conditional probabilities. We multiply them along the path where no previous match has occurred. We are not asserting that all the pair comparisons are mutually independent. Subtracting the resulting no-match probability from one gives the event we wanted.

For $n$ independent assignments to $d$ equally likely dates, with $n\le d$,

$P(\text{at least one match})=1-{\prod }_{i=0}^{n-1}\frac{d-i}{d}.$

For $n>d$, the probability is one by the pigeonhole principle. For zero or one person it is zero. If the question instead asks whether anyone else matches the first person in the group, then

$P(\text{a match to person one})=1-{\left(1-\frac{1}{d}\right)}^{n-1}(n\ge 1).$

At $n=23$ and $d=365$, these are approximately 50.73% and 5.86%. The two curves describe different events under the same model, so there is no contradiction between them.

There are $d^n$ ordered assignments of dates to $n$ distinguishable people. Under the stated model they all have equal probability. If $n\le d$, precisely $d(d-1)\cdots (d-n+1)$ assignments have distinct dates: choose any date for the first person, any unused one for the second, and continue. Dividing by $d^n$ gives the no-match product.

For a match to the first person, condition on their assigned day. Each of the other $n-1$ people avoids that particular date with probability $(d-1)/d$, independently, giving the second formula. The test suite also enumerates small date spaces and compares their counts with the first formula.

A 23-person room can have no match even though the model chance exceeds one half. That outcome does not refute the formula, and one room with a match does not verify it. The formula is proved by counting, not inferred from the visible room.

Real birthday dates are not perfectly uniform, and some groups contain dependence, for example twins. Applying the formula to those populations requires considering those assumptions. The page displays at most 80 people and shows rounded probabilities; it does not estimate a real population distribution or claim the seed is a source of truly random birthdays.