# What is the probability of 4 person in group of 18 can have same birth month?

This is not a class assignment.

It so happened that 4 team members in my group of 18 happened to share same birth month. Lets say June. . What are the chances that this could happen. I’m trying to present this as a probability problem in our team meeting.

Here is my attempt:

• All possible outcome $$121812^{18}$$
• 4 people chosen among 18: 18$$C4C_4$$
• Common month can be chosen in 1 way: 12$$C1C_1$$

So the probability of 4 people out of 18 sharing the same birth month is $$18C4∗12C11218\frac{18C_4 * 12C_1}{12^{18}}$$ = very very small number.

Questions:

1. Is this right way to solve this problem?
2. What the probability that there is exactly 4 people sharing a birth month?
3. What the probability that there is at least 4 people (4 or more people) sharing a birth month?

Please note: I know that all months are not equal, but for simplicity lets assume all months have equal chance.

You can see your argument is not correct by applying it to the standard birthday problem, where we know the probability is 50% at 23 people. Your argument would give $$\frac{{23\choose 2}{365\choose 1}}{365^{23}}\frac{{23\choose 2}{365\choose 1}}{365^{23}}$$, which is very small. The usual argument is to say that if we are going to avoid a coincidence we have $$365-(k-1)365-(k-1)$$ choices for the $$kk$$th person’s birthday, so the probability of no coincidence in $$KK$$ people is $$\prod_{k=1}^K \frac{365-k+1}{365}\prod_{k=1}^K \frac{365-k+1}{365}$$
Unfortunately, there is no such simple argument for more than two coincident birthdays. There is only one way (up to symmetry) for $$kk$$ people to have no two-way coincidence, but there are many, many ways to have no four-way coincidence, so the computation as you add people is not straightforward. That’s why R provides pbirthday() and why it is still only an approximation. I’d certainly hope this wasn’t a class assignment.
The reason your argument is not correct is that it undercounts the number of ways you can get 4 matching months. For example, it’s not just that you can choose any month of the 12 as the matching one. You can also relabel the other 11 months arbitrarily (giving you a factor of 11! ). And your denominator of $$12^{18}12^{18}$$ implies that the ordering of the people matters, so there are more than $$18\choose 418\choose 4$$ orderings that have 4 matches.