So I'm working on a probability problem:

In Exercise 19 assume it is equally likely that a person is born in any given month of the year.

  1. b) What is the probability that in a group of $n$ people chosen at random, there are at least two born in the same month of the year?

I know that the probability of two people being born in the same month is $1/12$ (from part a of this problem) and that the pigeonhole principle shows that some $n≥12$ will guarantee that two people within the group were born within the same month.

I don't know what to do with this information though, especially when looking at the book answer: $1 − 11/12 · 10/12 ··· (13−n)/12$. It doesn't make sense to me that a group ($n$) containing only 1 person with this formula would have a 100% chance of being born in the same month as someone else in the group. You need two people to compare birth months, so how can this be right? I thought that after 12 people, you'd be guaranteed to find two people with matching birth months. If I were to make a formula, it would look more like: $(n-1)·(1/12)$

  • $\begingroup$ Look up "the birthday problem" or "the birthday paradox" and replace $365$ by $12$. $\endgroup$ – David G. Stork Apr 18 at 18:59

Your confusion may be related to the empty product

Your expression could have started $1 - \frac{12}{12}\frac{11}{12}\cdots\frac{13-n}{12}$.

Then with $n=1$ it would give a probability of $1 - \frac{12}{12}=0$ that at least two born in the same month of the year, which is the obvious result with only one person

For other values of $n$ you would get the following probabilities that at least two born in the same month of the year, which is decidedly non-linear. As with the classic birthday or collision problem, the probability exceeds $\frac12$ well before you have half as many individuals as there possible birthdays

 n      Prob
 1      0
 2      0.08333
 3      0.23611
 4      0.42708
 5      0.61806
 6      0.77720
 7      0.88860
 8      0.95358
 9      0.98453
10      0.99613
11      0.99936
12      0.99995
13+     1

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