#math trivia: What are the chances that a given pair of students are born in the same month? Assume all 12 months equally likely

— Burt Kaliski Jr. (@modulomathy) October 8, 2012

Assuming all 12 months are equally likely, the chances that a given pair of students are born in the same month is 1/12. This also assumes, of course, that the two students’ birth months are independent of one another. If the students were twins, they’d almost certainly be born in the same month.

#math trivia, cont’d: What are the chances that at least two students in a class of 20 are born in the same month?

— Burt Kaliski Jr. (@modulomathy) October 12, 2012

This is a trick question. There’s a 100% chance that at least two students in a class of 20 are born in the same month. At least two students must be born in *some* month, because if at most one student were born in every month, there could be at most 12 students in the class. This is an example of the Pigeonhole Principle, which states that if there are *n* items (“pigeons”), each associated with one of *k* properties (“holes”), and *n* > *k*, then there must be a property that has at least two items associated with it (“two pigeons in the same hole”).

Beyond the trivia: The classic non-trick question about shared birthdays in classes comes up when considering both birth month and day. Here, the question is “What are the chance that at least two students in a class of 20 have the same birthday?” With *n* = 20 and *k* = 365 (or 366), the Pigeonhole Principle doesn’t apply. Instead, the chances are governed by the Birthday Paradox, which states that the probability that there is at least one property has at least two items associated with it is approximately *k*^{2}/2*n* when *n* << *k*. (Read “<<” as “much smaller than.”) The rationale is as follows: The probability that a given pair of items are associated with the same property, assuming all properties are equally likely, is 1/*n*. There are approximately *k ^{2}*/2 pairs of items. Assuming the probability that one pair matches is independent of the property that any other pair does (which is a simplification), the probability for one pair can be multiplied by the number of pairs, given

*k*

^{2}/2

*n*.

The reason this is called the Birthday Paradox is that it a surprisingly small class is likely to have two students with the same birthday. In particular, a class of just 23 students has a more than 50% chance of having a repeated birthday. (Observe that (23*23)/(2*365) = .724, so our approximation is pretty close.) That’s quite a small number of students considering that it’s fewer than a month’s worth of birthdays! Although the chances that any given birthday occurs is less than 10%, but the chances that a pair of birthdays matches is over 50%, hence the “paradox”.