# Birthday Problem Proof?

I was looking at the Birthday Problem (the probability that at least 2 people in a group of n people will share a birthday) and I came up with a different solution and was wondering if it was valid as well. Could the probability be calculated with this formula: $$1-(364/365)^{n(n+1)/2}$$

The numbers don't seem to perfectly match up with the normal proof, but I don't see the flaw in my logic, so if someone could clear it up, that would be much appreciated.

To find the formula, I found the probability that one person didn't share a birthday first, which is: $$(364/365)^{n-1}$$ for the first person, $$(364/365)^{n-2}$$ for the next, and so on. The probability that none of them do would be the product, and considering exponent laws, would be $$(364/365)^{n(n+1)/2}$$. We subtract that from $$1$$ to find the converse of our statement.

• The flaw when $n=367$ is the same flaw for $n=3$ Jun 23, 2020 at 13:01

Here, what you are doing is calculating the probability of event $$A_{ij}$$, where person $$i$$ does not share a birthday with person $$j$$, for every pair of $$i$$ and $$j$$. But the problem is that the $$A_{ij}$$ are not mutually independent. For example, given $$A_{ij}$$ and $$A_{jk}$$ both to be false, $$A_{ik}$$ is most certainly false as well. In plain English, if I told you that $$i$$ and $$j$$ in fact did share the same birthday, and so did $$j$$ and $$k$$, then the probability of $$i$$ and $$k$$ sharing the same birthday is non longer $$1/365$$, but in fact $$1$$. This suffices to show that the events are not mutually independent, so you cannot justify the step where you claim the probability of none of the pairs sharing birthdays is a product of probabilities. And for good reason as well, because unfortunately the formula you gave is not the correct one.