######
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: