**Document ID**: 06_07_99_1

**Date Received**: 1999-06-07 **Date Revised**: 2000-01-04 **Date
Accepted**: 2000-01-06

**Curriculum Topic Benchmarks**: M7.4.2, M7.4.3, M7.4.6, M7.4.11,
M7.4.15

**Grade Level**: \[9-12\] High School

**Subject Keywords**: probability, exponents, counter-intuitive result

**Rating**: \*\*\*

## Matching Birthdays

**By**: Phyllis A Salmons, Embry-Riddle Aeronautical University, 461
Brentwood Drive, Daytona Beach FL 32117 e-mail:
phyllis.salmons@msfc.nasa.gov;salmonsp@cts.db.erau.edu

**From**: The PUMAS Collection http://pumas.jpl.nasa.gov

©1999, California Institute of Technology. ALL RIGHTS RESERVED. Based on
U.S. Gov\'t sponsored research.

An interesting and well-known riddle in the area of probability is "The
Birthday Problem." You can find information about this problem in many
introductory books in probability, probability and statistics, and math
tricks books for laymen. This problem falls in the category of problems
called counter-intuitive \-- meaning that you may be surprised at the
answer.

You can gather data and participate easily in any of your classes or
group activities. Here's **the problem: In a group with people of any
age, what are the chances that two people in the group have the same
birthday (disregarding the year of birth)?**

There are 365 days in a year and we won't consider the complication of
leap year -- it wouldn't make that much difference. It seems likely that
the number of people in the room would have to be substantial to have a
promising likelihood of two people matching the month and day of birth,
wouldn't you think?

Calculating the probability of at least two of n people having matching
birthdays provides a good calculator exercise for students and an
interesting challenge for beginning computer programmers.

**Assumptions:**

1\. It is not leap year.

2\. There are no twins in the room.

3\. Disregard the year of birth.

**Definitions:**

n = the number of people in a room

P(2) = the probability that at least two people have matching birthdays

P(D) = the probability of no matches

**Theory:** Using this terminology, probability theory states

P(2) = 1 -- P(D).

Also, if an event is made up of several sub-events, then the
probabilities are multiplicative.

That is, the probability of the event D can be found by multiplying the
sub-events( a,b,c,\...) leading to event D

P(D) = P(a) ^.^ P(b) ^.^ P(c) ^.^ \...

In order for event D to occur, sub-event a must occur AND sub-event b
must occur AND sub-event c must occur AND so forth. Multiplying the
probability of the sub-event will give us the probability of event Z
occurring. The probability of the each sub-event is less than one. Each
time you multiply numbers less than one, the result is even smaller yet.

**Procedure:** Let's calculate P(D).

P(D) = the number of ways that nobody's birthday matches, divided by the
number of ways that birthdays can be given to everyone in the room.

The denominator of P(D) is 365n, since each person was born on one of
the 365 days in the year. In the numerator, we must consider how many
ways birthdays can be distributed to n people so that there are no
matches. The first person can have any one of the 365 days for a
birthday without restriction (with 365 ways out of 365 days). The second
person can only have a birthday on one of the 364 days left (364 ways
out of 365 days), and the third person must be born on one of the
remaining 363 days (363 ways out of 365 days), and so forth.

**Calculate the chances of two people having the same birthday:**

This means for two people, the probability that they have different
birthdays is

P(D) = [(365) (364)]{.underline} = [(365)(364)]{.underline} = 0.9973

\(365\) (365) 3652

and: P(2) = 1 -- 0.9973 = 0.0027

Therefore, with only two people (who are not twins) in the room, there
is a 0.27% chance that they have the same birthday (that is, of course,
if they are not twins!) Well, we knew that given only two people, it was
highly unlikely that they would have the same birthday, didn't we?

**In a room containing three people, calculate the chance that at least
two of them have the same birthday:**

For three people, the calculation looks like this

P(D) = ([365)(364)(363)]{.underline} = [(365)(364)(363)]{.underline} =
0.9918

(365)(365)(365) 3653

and: P(2) = 1-0.9918 = 0.0082

With three people in the room, there is a 0.82% chance that two of them
have the same birthday. Not much better really. But what happens as the
number of people in the room grows? Just continue in the same fashion
for more people.

**In a room containing n people, calculate the chance that at least two
of them have the same birthday:**

For n people in the room,

P(D) = [(365)(364)(363)....(365 -- (n-1))]{.underline}

365n

and: P(2) = 1 -- P(D).

# Activities:

1\. Given 365 possibilities for each person, what is the probability
that at least 2 out of the n people match? Use a calculator or write a
computer program to confirm the following:

  ----------------- ----------------- ----------------- -----------------
  **N**             **Probability of  **N**             **Probability of
                    at least one                        at least one
                    match**                             match**

  2                 0.0027            27                0.6269

  3                 0.0082            28                0.6545

  4                 0.0164            29                0.6810

  5                 0.0271            30                0.7063

  6                 0.0405            31                0.7305

  7                 0.0562            32                0.7533

  8                 0.0743            33                0.7750

  9                 0.0946            34                0.7953

  10                0.1169            35                0.8144

  11                0.1411            36                0.8322

  12                0.1670            37                0.8487

  13                0.1944            38                0.8641

  14                0.2231            39                0.8782

  15                0.2529            40                0.8912

  16                0.2836            41                0.9032

  17                0.3150            42                0.9140

  18                0.3469            43                0.9239

  19                0.3791            44                0.9329

  20                0.4110            45                0.9410

  21                0.4437            46                0.9483

  22                0.4757            47                0.9548

  23                0.5073            48                0.9606

  24                0.5383            49                0.9658

  25                0.5687            50                0.9704

  26                0.5982                              
  ----------------- ----------------- ----------------- -----------------

2\. Test this theory in each of your classes and clubs at school and in
the community. Record the number of people in the room and the results
of the birthday poll. Did any of the groups buck the odds?

**Questions**

1\. What is the minimum number of people needed to have at least a 50/50
chance of at least one match? A 75% chance? A 90% chance?

2\. Are there other circumstances where predictions can be made from a
surprisingly small sample of people? (What about the media predictions
from polls taken on election day?)

3\. How many people would be needed in the group to be CERTAIN that two
people will have the same birthday? Compare this with the number of
people which gives a probability of 90%. Interesting, don't you think?

**For more information:**

Mosteller, Frederick. *Fifty Challenging Problems in Probability*. Dover
pubs, ISBN 0-486-65355-2)

Paulos, John. Innumeracy: *Mathematical Illiteracy and Its
Consequences*. pp 35-37.

The Birthday Problem, http://silver.sdsmt.edu/\~rwjohnso/module0.htm