The Birthday Problem
The subject does say statistics, I couldn’t find probability so I found the closest relative.
This infamous problem confuses the heck out of most people when first come across. The question at hand is:
suppose you are at a birthday party and you make a bet with a friend. You say you can find at least two people at the party that share the same birthday. What would be the probability of at least two people sharing the same birthday, given a certain amount of people are at the party?
Before we solve this, we must make some assumptions.
1) We assume all birthdays are independent and equally likely. That is, any day of the year is equally likely to have a birthday as any other day of the year and
2) We will assume, for sake of argument, that we are in a 365 day year.
Since we are dealing with a probability question, we must find two things:
1) All possible events of our sample space and
2) All possible events of interest (at least two people sharing a birthday)
Let’s find 1)
Our sample space consists of all possible birthday combinations of the people at the party. We have 23 people and 365 available days of the year. If I select the first person, he has 365 possible birthdays. I choose the second person, he too has 365 possible birthdays, and so on. Thus by the m,n rule we have 365*365*365…*365 = 365^23 which is the size of our sample space. So this will go in our denominator of our probability statement.
Let’s find 2)
Two count the number of ways at least two people share the same birthday we could simply tackle the brute force method. Count all possible ways exactly two people share a birthday + all possible ways exactly three people share a birthday + so on until 23 people share a same birthday. But that would take a while. Here is where we need to incorporate a key component to this problem: complementary events. If we can find the probability of the complementary event, all we need to do is subtract that from 1 and we get our answer. So the complementary event for AT LEAST two people sharing the same birthday is exactly NO ONE sharing the same birthday. We do this by similar counting principles above:
We choose our first person, he has 365 possible birthdays. Person two has 364 possible birthdays, and person three has 363 possible birthdays and so on down to the last person. We do this since they must not share a birthday. So this is, by the m,n rule, 365*364*363*362*…341 = 365P23 = 365 permute 23 = 365!/(365-23)! = 365!/342! where ! is the factorial symbol.
Now we have our complementary event: 365!/342! and our sample space: 365^23
Now we divide and subtract 1 from it:
1-((365!/342!)/(365^23)) = 50%
So the probability of at least two people sharing the same birthday out of a group of 23 people is about 50%, assuming independence of birthdays and each birthday being equally likely to occur. One of my favorite math problems. It also may help to draw out some lines where you can write on the lines. The lines act like place holders to help you visualize what’s going on, it helps me. Hope this helps not just with this problem but with counting principles and setting up probability questions.
About The Author
|I am a recently graduated student from Northern Arizona University, (Flagstaff, AZ) with a degree in mathematics. Over the years I have helped many of my friends in math. I recently tutored a friend in preparation for the ASVAB test. I tutored a college friend in a psychology statistics course. She ...|