Posted by: dschadewald | October 12, 2008

The Birthday Problem

I am taking Stats 110 (Probability Theory) this semester and in almost every class the professor presents a problem that has a very surprising answer.  One of the first of these problems, and so far the most interesting one, is called the birthday problem.  

I discovered that this is a very famous problem once I googled it as the problem tricks almost every person when it is first presented to them.  The problem can be stated in many different ways, but the one I find most compelling is, “How many people do there need to be in a room such that there is a 50% chance that two of the people in the room have the same birthday?”  When I first heard this problem I thought that the number would be somewhere below 180 but above 100.  The answer, however, is that only 23 people need to be in the room for there to be a one half chance that two have the same birthday.  Sounds crazy doesn’t it?  Well here is how the problem works:

Instead of trying to figure out the probability of one match and two matches and three matches, etc. (since we have to find the probability of AT LEAST one match), we can use a trick.  We will just find the probability that there are no matches and subtract that probability from one.  Given that there are 365 days in a year (we will forget about leap years for simplicity), that all 365 days are equally likely, and that there are n number of people in the room, then the probability of no match is simply:

(365*364*…(365-n+1)) / (365^n)

The reason this problem is so confusing is because we think of the problem as just having, for example, 25 people in the room.  Instead you should think of the problem as having 25 choose 2 (300) pairs of people in the room and you need to compare the pairs and look for a match.  

A cool java simulator is can be found here:  You simply type in the number of “people” you want to have in the room and then the simulator generates random birthdays and sees if there is a match.

A few probabilities are listed below:

if n = 23, 50.7%

if n = 57, 99%

if n = 100, 99.9999%



  1. This is one of the oldest tricks we mathematicians have when we’re teaching probability. 🙂 My dad uses it in his psychology class too. I would probably have used it on you guys too, if you hadn’t stolen my thunder! Nice post.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s


%d bloggers like this: