The Birthday Problem



There is a problem in mathematics relating to birthdays.   Since a year has 366 days (if you count February 29), there would have to be 367 people gathered together to be absolutely certain that two of them have the same birthday.

Now, if we were content to be just fifty percent sure, how many people are needed to be in the room?   A good guess might be 183 since that is half of 366.   The surprising answer is that there need only be 23 people present!   Stated differently, half of the time that twenty-three randomly selected people are gathered together, two or more of them will share the same birthday (an explanation is given in my letter below).

This is a delightful activity to do with students in your classroom.   If your class size is 23 or more, you have at least a 50% chance that two students will share a birthday.

The Problem: In a group of n people, what is the probability that two of them will have the same birthday?

Here is a letter that I wrote to Marilyn vos Savant about the Birthday Problem in Statistics (She published an edited version of this letter in November 1997 in her column in Parade magazine):



                                         
                                                                                      John Handley High School
                                                                                      P.O. Box 910
                                                                                      Winchester, VA  22604
                                                                                      August 3, 1997

Ask Marilyn
PARADE
711 Third Avenue
New York, NY  10017

Dear Ms. vos Savant:

     I am a mathematics teacher and eagerly look forward to your column each week.  
I often use items from your column and from your books as material in my classes.  
One of my colleagues assigns her first year computer science students to write a 
program which simulates your famous "Game Show" problem, and every year you are 
proven correct!  I have always agreed with the mathematical answers that you have 
given and marveled at your explanations.  You explain things in layman's terms 
which are often better understood by my students than my mathematical proofs.  

     
     But in a recent column (8/3/97), I believe that you gave an incomplete answer 
to the question posed by Robert Shearn.  Your logic was correct (you cannot reason 
"that if there's a 50% chance that at least two out of 23 people will share the same 
birthday, there must be a 100% chance that at least two out of 50 will!").  But while 
that type of extrapolation is incorrect, it is a coincidence that there is a 97% 
chance that at least two people out of a group of 50 will share a common birthday, 
and I believe it is that to which Mr. Shearn is referring.  Of course, since there 
are 366 days on which people could have birthdays, there would have to be 367 people 
gathered in order to be absolutely 100% certain that at least two of them would have 
the same birthday.

     In mathematics, this is called "The Birthday Problem."  I use it along with your 
"Let's Make A Deal" problem and several other problems when I teach probability.  I 
try to motivate this with my students by using the same approach you used in your 
response to Mr. Shearn -- by examining the probability that everyone in the group 
would have a different birthday.  This probability along with the probability that at 
least 2 people have the same birthday must equal 1 (or 100% if you are using percentages).

     For example, take a group of three people.  The number of ways that any three 
dates (representing their birthdays and allowing for repetition) may be chosen is 
(366)x(366)x(366).  In order for all of them to be born on different days, the first 
could be born on any of the 366 dates, the second could be born on any of 365 dates 
(she can not be born on the one day that the first person was born), and the third 
person could be born on any of the remaining 364 dates.  So, all the ways that this 
could occur is (366)x(365)x(364).  And therefore, the probability that no two people 
would have the same birthday out of a group of three people is the number of ways 
that their birthdays could be different divided by the total number of ways in which 
their birthdays could occur or    (366)(365)(364) / (366)(366)(366).


The probability that at least two people would have the same birthday out of a 
group of three is 

             (366)(365)(364)
     1  -  ---------------------    =  .00818  or .8%
             (366)(366)(366)  


Similarly, it can be shown that out of a group of 23 people, the probability that 
at least two of them would have the same birthday is

             (366)(365)(364)(363)...(344)
     1  -  --------------------------------------   =  .506 or 50.6%.       
             (366)(366)(366)(366)...(366)


And it can be shown that out of a group of 50 people, the probability that at least 
two of them would have the same birthday is

              (366)(365)(364)(363)...(317)
     1  -  --------------------------------------   =  .97 or 97%.
              (366)(366)(366)(366)...(366)

     I have written two computer programs in C++ to illustrate the birthday problem 
and I have enclosed them in this mailing.  The first program, called MARILYN.EXE, 
computes the theoretical probabilities using the techniques described above for groups 
ranging in size from 2 to 100.  I have included that printout below.  The second program, 
called BIRTHDAY.EXE, is a simulation of the problem allowing you to choose the size of 
the group and the number of times that you wish to run the program.  It determines the 
number of times that at least two people have a matching birthday using pseudorandom 
numbers (this is similar to the technique our computer students use in verifying your 
"Game Show" problem).

     As an application of the Birthday Problem, would you predict whether any two of the 
41 different Presidents of the U.S. have the same birthday?  The theoretical probability 
that at least two of them would be born on the same day is 90.25%.  Upon an examination 
of their birthdays, we find that the eleventh President of the United States, James K. 
Polk, was born on November 2, 1795, and that the twenty-ninth President, Warren G. 
Harding, was born on November 2, 1865.  Of course, if we use all 42 Presidents, there is 
another pair of Presidents with the same birthday - the 22nd and the 24th Presidents 
(Grover Cleveland and himself).

     I hope the explanations make sense.  Thank you for getting me focused on the new 
school year - I start my 30th year of teaching in less than two weeks!

                                                                                           Sincerely,

                                                                                           David H. Pleacher

 








 
The Probability of At Least Two People in a Group Having The Same Birthday
(Computer Printout)

  Group Size    Probability        Percent Chance 
      2           	0.002732            0.273224%
      3           	0.008182            0.818179%
      4           	0.016311            1.631145%
      5           	0.027062            2.706214%

      6           	0.040354            4.035364%
      7           	0.056086            5.608555%
      8           	0.074139            7.413856%
      9           	0.094376            9.437597%

     10          	0.116645           11.664541%
     11           	0.140781           14.078078%
     12           	0.166604           16.660431%
     13           	0.193929           19.392876%

     14           	0.22256             22.255971%
     15          	0.252298           25.229786%  
     16           	0.282941           28.294139%
     17          	0.314288           31.428821%

     18          	0.346138           34.613822%
     19          	0.378295           37.829535%
     20           	0.41057             41.056964%
     21           	0.442779           44.277895%

     22          	0.474751           47.475065%
     23           	0.506323           50.632301%
     24           	0.537346           53.734643%
     25           	0.567684           56.768437%

     26          	0.597214           59.721412%
     27           	0.625827           62.582733%
     28           	0.65343             65.343023%
     29           	0.679944           67.994376%
 
     30           	0.705303           70.530341%
     31           	0.729459           72.945887%
     32           	0.752374           75.237356%
     33           	0.774024           77.402396%

     34           	0.794399           79.439884%
     35          	0.813498           81.349841%
     36          	0.831333           83.133326%
     37           	0.847923           84.792343%

     38           	0.863297           86.329729%
     39           	0.87749             87.749047%
     40           	0.890545           89.054476%
     41           	0.902507           90.250708%

     42           	0.913428           91.342842%
     43           	0.923363           92.336286%
     44           	0.932367           93.236668%
     45           	0.940497           94.049746%

     46           	0.947813           94.781335%
     47           	0.954372           95.437233%
     48           	0.960232           96.023162%
     49           	0.965447           96.544714%

     50           	0.970073           97.007307%
     51           	0.974161           97.416145%
     52           	0.977762           97.77619%
     53           	0.980921           98.092141%

     54           	0.983684           98.368416%
     55           	0.986091           98.609142%
     56           	0.988182           98.81815%
     57           	0.98999            98.99898%

     58           	0.991549           99.154876%
     59           	0.992888           99.288803%
     60           	0.994034           99.40345%
     61           	0.995012           99.501245%

     62           	0.995844           99.584371%
     63           	0.996548           99.654778%
     64           	0.997142           99.714201%
     65           	0.997642           99.764177%

     66           	0.998061           99.806058%
     67           	0.99841             99.841031%
     68           	0.998701           99.870132%
     69           	0.998943           99.894261%

     70           	0.999142           99.914195%
     71           	0.999306           99.930606%
     72           	0.999441           99.944068%
     73           	0.999551           99.955071%

     74           	0.99964             99.964032%
     75           	0.999713           99.971304%
     76           	0.999772           99.977184%
     77           	0.999819           99.981922%

     78           	0.999857           99.985725%
     79           	0.999888           99.988768%
     80           	0.999912           99.991192%
     81           	0.999931           99.993117%

     82           	0.999946           99.99464%
     83           	0.999958           99.995841%
     84           	0.999968           99.996784%
     85           	0.999975           99.997522%

     86           	0.999981           99.998098%
     87           	0.999985           99.998545%
     88           	0.999989           99.998891%
     89           	0.999992           99.999157%

     90           	0.999994           99.999362%
     91           	0.999995           99.999519%
     92           	0.999996           99.999639%
     93           	0.999997           99.999729%

     94           	0.999998           99.999798%
     95           	0.999999           99.99985% 
     96           	0.999999           99.999889%
     97           	0.999999           99.999918%

     98           	0.999999           99.99994%
     99           	1		99.999956%
     100          	1		99.999968%

 

 

     These last probabilities are not really equal to one - they are just rounded to 
one because I only took the decimal representation out to six places.

     Isn't it interesting to see that in groups of 55 or more, you have a 99% chance 
of having at least two people with the same birthday?  And yet, to be 100% certain, 
you would need a group size of 367!


Some references for The Birthday problem include: Paulos, John Allen, Innumeracy, Hill and Wang, New York, 1988, p. 27. Johnson, John, "The Birthday Problem Explained", The Mathematics Teacher, N.C.T.M., Reston, VA, January, 1997, pp. 20-22.



Here is Marilyn vos Savant's column from the November 23, 1997 PARADE magazine:





Send any comments or questions to: David Pleacher