What is the probability that 3 or more people share a common birthday, in a group of 160 people?
Approach:
We have:
$P(X\geq 3)= 1-[P(X=0)+P(X=2)]$.
(where $X\geq i$ means at least $i$ people share a common birthday, whereas $X=i$ means exactly $i$ people share a common birthday, while considering the possibility of multiple groups of i people sharing a common birthday, for eg the case involving birthdates A A B B C C D E F G is counted in $X=2$, A A B B B C C D E F is not.)
The cases for $X=0$ and $X=2$ can be dealt together by using the following argument:
Consider $x$ groups of $2$ people each, i.e. $2x$ people put into $x$ groups of $2$. There are ${160 \choose 2x}(2x)!/(2!)^x$ ways to create such groups. Now, we just have to select $x$ dates out of $365$ and assign it to these groups, so we have ${365\choose x}x!$ ways of doing this. We then assign dates to the remaining $160-2x$ people, and there are $365-x$ dates left. So we have ${365-x \choose 160-2x}(160-2x)!$ ways of doing it.
The sample space is $365^{160}$,and $x$ can run from $0$ to $80$, ($x=0$ corresponds to the case where everyone has a different birthday), so it appears to me that:
$$P(X=0)+P(X=2)=\sum_{x=0}^{80} \dfrac{{160 \choose 2x}(2x)!/(2!)^x*{365\choose x}x!*{365-x \choose 160-2x}(160-2x)!}{365^{160}}$$
However, it seems that something has gone seriously wrong with my reasoning, as Wolfram estimates this sum to be about $10^{61}$…
This expression does give the correct values for the trivial cases $x=0$ and $x=1$…and there doesn't seem to be any occurrence of double counting with this kind of approach..
What is possibly going wrong then?
Edit: I have seen multiple variations of question, and most of them use the poisson's formula to arrive at a numerical estimate….However more than finding the correct numerical answer I wish to know the flaw in this particular approach of mine.
Best Answer
There are ${160 \choose 2x}(2x)!/(2!)^x$ ways to choose an ordered sequence of $x$ groups of $2$. But you don't care about the order: for you, the events
are the same. So you should instead have $\binom{160}{2x} \cdot \frac{(2x)!}{x! (2!)^x}$ here, dividing by $x!$.