You are here:

Advanced Math/Probability question....sorta

Advertisement

Expert: - 2/25/2010


Question
Mr. Wilson,
I've read (in several sources) that it only takes 23 people in the same room before the probability is better than 50% that at least two of them will share the same birthday.  
Is there a mathematical formula for that, or do you just have to add up all the potential probabilities.  Every person has a 1 in 365 chance that his birthday is the same as someone else, I guess.  And when there are 23 people in the room, there are all sorts of combinations...it wouldn't be 23 factorial times 1/365, would it?
Anything you can share about this would be greatly appreciated.
Warm regards,
David Gardner

Answer
We need to check every pair of people in the room.
Given only 2 peoole, say A and B, there is only 1 combination - AB.
Given 3 people, say A, B, and C, we need to check AB, AC, and BC - 3 choices.
Given 4 people, say A, B, C, and D, we need to check AB, AC, AD, BC, DB, and CD - 6 choices.
Given 5 people, we need to check AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE - 10 choices.

So this is what we have so far
2 1
3 3
4 6
5 10

As can be seen, the choices with each additional person is the number of people already present.
Thus, a 6th person would be 10+5=15 choices.

The formula for this is n(n-1)/2.  Thus, 2 is 2*1/2 = 1, 3 is 3*2/2 = 3, 4 is 4*3/2 = 6,
5 is 5*2/2 = 10, ... 20 is 20*19/2 = 190, 21 is 21*20/10 = 210, 22 is 22*21/2 = 231,
and 23 is 23*22/2 = 253.

Thus, there is no guarantee that two people will have the same birthday,
but with 23 people, that gives 253 possible pairs of people.

The chance of having no matching birthdays can also be approximated by A choose B.
We have A days (365) and 23 people (B).  For them all to be independent, is would be
365 choose 23, or 365!/(23!342!).

Suppose there were only 8 days in a year and we had 4 people.  That seems a bit silly,
but just go along with it.  For them all to be independent, there would be 8*7*6*5/4*3*2*1
ways to do it.  That is 70 ways.  To get the total ways they could have birthdays,
take 8*8*8*8 = 4096.  As can be seen, it is almost sure that there would be two with
the same birthday since there there are only 70 ways out of 4,096 total.

Thus, for 365 days and 23 people, it can be seen to have pretty good odds.



There may be only 23 people, but what they don't say when they tell this is
it takes two people, and there are 253 ways of making pairs of people.

Advanced Math

All Answers

Answers by Expert:

Ask Experts

Volunteer

Scott A Wilson

Expertise

I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can even tell you it takes me over 2,000 steps to go a mile, but is that relevant?

Experience

Experience in the area; I have tutored people in the above areas of mathematics for almost two years in AllExperts.com. I have tutored people here and there in mathematics since before I received a BS degree almost 25 years ago. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

Publications
My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

Education/Credentials
Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

Past/Present Clients
My clients have been students at OSU, people nearby, friends with math questions, and several people every day on the PC, and you're probably make one more.

©2012 About.com, a part of The New York Times Company. All rights reserved.