# Happy Birthday!

OU math professor and friend of the blog, Ravi Shankar, had his birthday on Monday (which, by the way, was the summer solstice!).  You should email him a belated Happy Birthday!

In his honor, and in the spirit of our previous (and future) posts on probability, we thought we should mention the famous Birthday Problem:

If you are in a math class of 50 people and the professor offers to bet you $100 that two people have the same day as their birthday, should you take the bet? Of course, even without doing the math you should be suspicious. To paraphrase a quote: Ha ha! You fool! You fell victim to one of the classic blunders – The most famous of which is “never get involved in a land war in Asia” – but only slightly less well-known is this: “Never go against a math professor when money is on the line”! Ha ha ha ha ha ha ha! Ha ha ha ha ha ha ha! Ha ha ha… The key of course is that the professor is not betting that you and someone else will have the same birthday, but just that two people in the class will have the same birthday. For the first scenario, the odds are 49/365 or approximately 13.5% (to see this, you see that each of the 49 other people in the class have a 1/365 chance that they’ll have the same birthday as you*). In the second scenario, it is easier to count the odds that no two people will have the same birthday; that is, that all 50 birthdays are different. Well, the odds that two people have different birthdays is (365/365)*(364/365) (the first person can have any birthday, the second person can have any birthday except the day of the first person’s birthday). If there are three people, then the odds are (365/365)(364/365)(363/365) (the first person can have any birthday, the second any birthday except the first person’s birthday, the third person any day except the two birthdays of the first two), etc. In general, if you have n people, then the odds that no two of them have the same birthday is given by $\bar{p}(n)=\frac{365*364*363* \cdots * *(365-n+1)}{365^{n}}.$ So the odds that two people out of a group of n people have the same birthday is given by $1- \bar{p}(n).$ Note that this gives the odds as a fraction of 1. If you want it as a percentage, multiply by 100. So back to our question, what is the odds that two people in a class of 100 have the same birthday? It’s a shocking 97%! Your suspicions were right! Your professor is trying to con you out of your$100!

In a group of 100 people, the odds of two having the same birthday are 99.99996%!

The probability that a pair of people in a group will have the same birthday (from Wikipedia)

We’ll end with a pigeonhole principle question:  How many people do you need to have in the room to ensure that there is a 100% chance that two people will have the same birthday?

* = not counting leap years