1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The flaw when $n=367$ is the same flaw for $n=3$ $\endgroup$ – Henry Jun 23 at 13:01
3
$\begingroup$

The problem is that the probability of the conjunction of two events is the product of individual probabilities only if the events are independent. And more generally, for your line of logic to be correct, all events you are considering must be mutually independent from one another. That is, the outcome of any event is independent, regardless of the outcomes of any of the other events.

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.