Why Study Math - Probability and the Birthday Paradox

Joe Pagano

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When I decided to become a mathematics major in college, I knew that in order to complete this degree, two of the required courses—besides advanced calculus—were Probability Theory and Math 52, which was statistics. Although probability was a course I was looking forward to, given my penchant for numbers and games of chance, I quickly learned that this theoretical math course was no walk in the park. This notwithstanding, it was in this course that I learned about the birthday paradox and the mathematics behind it. Yes, in a room of about twenty-five people the odds that at least two share a common birthday are better than 50-50. Read on and see why.

The birthday paradox has to be one of the most famous and well known problems in probability. In a nutshell, this problem asks the question, “In a room of about twenty-five people, what is the probability that at least a pair will have a common birthday?" Some of you may have intuitively experienced the birthday paradox in your everyday lives when talking and associating with people. For example, do you ever remember talking casually with someone you just met at a party and finding out that their brother had the same birthday as your sister? In fact, after reading this article, if you form a mind-set for this phenomenon, you will start noticing that the birthday paradox is more common than you think.

Because there are 365 possible days on which birthdays can fall, it seems improbable that in a room of twenty-five people the odds of two people having a common birthday should be better than even. And yet this is entirely the case. Remember. The key is that we are not saying which two people will have a common birthday, just that some two will have a common date in hand. The way I will show this to be true is by examining the mathematics behind the scenes. The beauty of this explanation will be that you will not require more than a basic understanding of arithmetic (see Arithmetic Magic ) to grasp the import of this paradox. That’s right. You will not have to be versed in combinatorial analysis, permutation theory, complementary probability spaces—no not any of these! All you will need to do is put your thinking cap on and come take this quick ride with me. Let’s go.

To understand the birthday paradox, we will first look at a simplified version of the problem. Let’s look at the example with three different people and ask what the probability is that they will have a common birthday. Many times a problem in probability is solved by looking at the complementary problem. What we mean by this is quite simple. In this example, the given problem is the probability that two of them have a common birthday. The complementary problem is the probability that none have a common birthday. Either they have a common birthday or not; these are the only two possibilities and thus this is the approach we will take to solve our problem. This is completely analogous to having the situation in which a person has two choices A or B. If they choose A then they did not choose B and vice versa.

In the birthday problem with the three people, let A be the choice or probability that two have a common birthday. Then B is the choice or probability that no two have a common birthday. In probability problems, the outcomes which make up an experiment are called the probability sample space. To make this crystal clear, take a bag with 10 balls numbered 1-10. The probability space consists of the 10 numbered balls. The probability of the entire space is always equal to one, and the probability of any event that forms part of the space will always be some fraction less than or equal to one. For example, in the numbered ball scenario, the probability of choosing any ball if you reach in the bag and pull one out is 10/10 or 1; however, the probability of choosing a specific numbered ball is 1/10. Notice the distinction carefully.

Now if I want to know the probability of choosing ball numbered 1, I can calculate 1/10, since there is only one ball numbered 1; or I can say the probability is one minus the probability of not choosing the ball numbered 1. Not choosing ball 1 is 9/10, since there are nine other balls, and 1 - 9/10 = 1/10. In either case, I get the same answer. This is the same approach—albeit with slightly different mathematics—that we will take to demonstrate the validity of the birthday paradox.

In the case with three people, observe that each one can be born on any of the 365 days of the year (for the birthday problem, we ignore leap years to simplify the problem). In order to get the denominator of the fraction, the probability space, to calculate the final answer, we observe that the first person can be born on any of 365 days, the second person likewise, and so on for the third person. Therefore the number of possibilities will be the product of 365 three times, or 365x365x365. Now as we mentioned earlier, to calculate the probability that at least two have a common birthday, we will calculate the probability that no two have a common birthday and then subtract this from 1. Remember either A or B and A = 1-B, where A and B represent the two events in question: in this case A is the probability that at least two have a common birthday and B represents the probability that no two have a common birthday.

Now in order for no two to have a common birthday, we must figure the number of ways this can be done. Well the first person can be born on any of the 365 days of the year. In order for the second person not to match the first person’s birthday then this person must be born on any of the 364 remaining days. Likewise, in order for the third person not to share a birthday with the first two, then this person must be born on any of the remaining 363 days (that is after we subtract the two days for persons 1 and 2). Thus the probability of no two people out of three having a common birthday will be (365x364x363)/(365x365x365) = 0.992. Thus it is almost certain that nobody in the group of three will share a common birthday with the others. The probability that two or more will have a common birthday is 1 - 0.992 or 0.008. In other words there is less than a 1 in 100 shot that two or more will have a common birthday.

Now things change quite drastically when the size of the people we consider gets up to 25. Using the same argument and the same mathematics as the case with three people, we have the number of total possible birthday combinations in a room of 25 is 365x365x. . . x365 twenty-five times. The number of ways no two can share a common birthday is 365x364x363x. . . x341. The quotient of these two numbers is 0.43 and 1 - 0.43 = 0.57. In other words, in a room of twenty-five people there is a better than 50-50 chance that at least two will have a common birthday. Interesting, no? Amazing what mathematics and in particular what probability theory can show.

So for those of you whose birthday is today as you are reading this article, or will be having one shortly, happy birthday. And as your friends and family are gathered around your cake to sing you happy birthday, be glad and joyful that you have made another year—and don’t forget the birthday paradox. Isn’t life grand?

See more at Cool Math Sites and Cool Math Ebooks

Joe is a prolific writer of self-help and educational material and an award-winning former teacher of both college and high school mathematics. Under the penname, JC Page, Joe authored Arithmetic Magic, the little classic on the ABC’s of arithmetic. Joe is also author of the charming self-help ebook, Making a Good Impression Every Time: The Secret to Instant Popularity; the original collection of poetry, Poems for the Mathematically Insecure, and the short but highly effective fraction troubleshooter Fractions for the Faint of Heart. The diverse genre of his writings (novel, short story, essay, script, and poetry)—particularly in regard to its educational flavor— continues to captivate readers and to earn him recognition.

Joe propagates his teaching philosophy through his articles and books and is dedicated to helping educate children living in impoverished countries. Toward this end, he donates a portion of the proceeds from the sale of every ebook. For more information go to http://www.mathbyjoe.com


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