# Is there a way to *directly* calculate the probability that two or more people (of four chosen) have the same birthday?

You randomly choose four people. What is the probability that two or more of them have the same birthday? It's easy to calculate the probability that none of them have the same birthday (($$365*364*363*362)/365^{4})$$, and then subtract that number from $$1$$ to get the answer.

But is there a way to directly calculate the probability, instead of calculating the opposite event and subtracting from one? The choosing of the people is presumably an independent event so I thought calculating the probability directly wouldn't be difficult, but whatever I've tried has not given the right answer...

• Sure, it's just a bit messy. Pick a pattern, like $AABC$...meaning that exactly two have a common birthday (on day $A$) and that the other two have distinct birthdays. Compute the probability of that. Then do $AAAB$, $AAAA$, and $AABB$. (to first order, you can probably ignore those last three cases).
– lulu
Oct 18, 2019 at 14:55
• Nov 3, 2020 at 10:15

$$={4 \choose 2}(\frac{1*364*363}{365^3})+{4 \choose 2}(\frac{1*1*364}{365^3})(\frac{1}{2})+{4 \choose 3}(\frac{1*1*364}{365^3})+{4 \choose 4}(\frac{1*1*1}{365^3}) = .01635591246$$
• Are you certain that's correct? $.016355912$ seems correct, given OP's calculation, but I get a slightly smaller result ($.01633$) after computing your proposed solution. Also, shouldn't the last combination be ${4\choose 4}$? although it still results in a smaller number. Oct 18, 2019 at 16:18
• While you result is now correct, you need to explain the fourth term, which should be $\binom{4}{2}(\frac{1*364*1}{365^3})(\frac{1}{2})$ Oct 18, 2019 at 20:30