Three Share a BirthdayDate: 09/17/1999 at 16:06:46 From: Kay McGraw Subject: Probability In my classroom one year I discovered a very rare situation as we were graphing our birthdays (month/day). In a class of 24, I had 3 students sitting one behind the other with exactly the same birthday. As I do this activity every year, I would like to be able to tell them the probability of that occurrence and let them know how truly unusual this is. No one, including retired and current teachers, seems to know how to combine the fact that 3 out of 24 students have the same birthday with the fact that they are one behind the other in a room of 24. Many of us would appreciate it if you could send me the formula, and while I won't explain it in detail to my class of 9-year-olds, I will give them the chances of that happening again. I can't begin to tell you the combinations different staff members have suggested. One that might make sense is 365*365*24*23*22, but it seems too large an answer. I realize that the probability of having a birthday on any given date in a year is not a constant, but I'm willing to ignore that. (The date is Jan. 21, if you're curious.) Thank you very much for any assistance that you might give me. Kay McGraw Date: 09/17/1999 at 17:37:18 From: Doctor Anthony Subject: Re: Probability With 23 people in a class there is a 50% probability of at least two sharing a birthday. The probability of at least three in a class of 24 sharing a birthday is calculated as follows. First find the probability that no two share a birthday and then that exactly two share a birthday. Add these probabilities and subtract from 1 and we have the probability of three or more sharing a birthday. If no two share a birthday, then the first person can choose a birthday in 365 ways, the second person in 364 ways, the third in 363 ways, and so on. If there are n people in the room, the probability that no two share a birthday is: 365 * 364 * 363 * ... * (365-n+1) P(365,n) --------------------------------- = -------- 365^n 365^n Setting n = 24, we get: P(365,24) P(none sharing a birthday) = --------- 365^24 = 0.4616557 Next we find the probability of exactly 2 sharing a birthday. We can choose 2 from 24 in C(24,2) = 276 ways. These two can choose their birthday in 365 ways and then all the other 22 must have different birthdays. 276 * 365 * 364 * ... to 23 factors Probability = ----------------------------------- 365^24 276 * P(365,23) = --------------- 365^24 = 0.37256428 So the probability of none or two sharing a birthday = 0.4616557 + 0.37256428 = 0.83422 So the probability of 3 or more sharing a birthday = 1 - 0.83422 = 0.16578 approx. probability = 1/6 As you can see this is not such an unusual occurrence for a class of 24. Unfortunately, there is not enough information to complete the question about the probability of three students with the same birthday sitting one behind the other in a room of 24. We would need to know how the desks are arranged: 6 x 4, 8 x 3, or (highly improbable) 1 x 24 seating arrangements would have different solutions. - Doctors Anthony and TWE, The Math Forum http://mathforum.org/dr.math/ Date: 01/15/2001 at 11:30:25 From: John McGregor Subject: The Birthday Problem - a mistake? Dr. Anthony: In your reply to a question about "Three share a birthday" given on July 19/99, there seems to me to be an error. Essentially you say that the probability that 3 or more share a birthday equals the probability that 2 or more share a birthday minus the probability that exactly 2 share a birthday: P(>=3 share) = p(>=2 share) - p(exactly 2 share) This is fine as far as it goes, but what you then actually calculate for the third term is: p(exactly ONE SET of 2 share) so that your resultant p(>=3 share) is an overestimate since it also includes the probability that MORE THAN ONE SET of 2 share. regards John Date: 01/18/2001 at 13:17:57 From: Doctor Rick Subject: Re: The Birthday Problem - a mistake? Hi, John. Dr. Anthony may not have seen your comments, but I've been working on this problem since you sent it in. You're quite right! The partition into: 1. none share a birthday 2. exactly two share a birthday 3. three or more share a birthday is not correct. A correct partition is: 1. none share a birthday 2. one pair shares a birthday 3. two pairs share different birthdays 4. three pairs share different birthdays : 1+N/2. N/2 pairs share different birthdays 2+N/2. three or more share a birthday The probability that no two people out of N share a birthday is, as Dr. Anthony had it, is: P_N(0) = P(365,N)/365^N where P(N,M) is the number of permutations of N things taken M at a time. The probability that exactly one pair share a birthday, and all the others are different, is (as Dr. Anthony showed): P_N(1) = 365*C(N,2)*P(365-1,N-2)/365^N The probability that exactly two pairs share a birthday is: P_N(2) = C(365,2)*C(N,2)*C(N-2,2)*P(365-2,N-4)/365^N That's because you -- choose 2 birthdays, one for each pair: C(365,2) -- choose two people for the first pair: C(N,2) -- choose two people for the second pair: C(N-2,2) -- choose all different birthdays for the remaining N-4 people: P(365-2,N-4) You can calculate the probability for exactly K pairs in the same way: P_N(K) = C(365,K)*C(N,2)*C(N-2,2)*...*C(N-2K+2,2)*P(365-K,N-2K)/365^N This can be simplified by noting that the successive C(...,2) terms have factorials that cancel in one denominator and the following numerator. The simplified formula is: P_N(K) = C(365,K)*P(N,2K)*P(365-K,N-2K)/(2^K*365^N) The probability that at least three people out of N will share a birthday is 1 minus the sum of P_N(0) and P_N(K) for K=1 through N/2. I haven't gotten this into closed form, but using a spreadsheet, I found the following: Number of pairs with same birthday At least Exactly People 0 1 2 3 Total trio 1 trio ===================================================================== 2 0.99726 0.00273 1 0 0 4 0.98364 0.01630 2.2E-05 0.99997 2.9E-05 2.9E-05 6 0.95953 0.03998 0.00033 3.0E-07 0.99985 0.00014 0.00014 8 0.92566 0.07239 0.00151 8.4E-06 0.99958 0.00041 0.00040 10 0.88305 0.11162 0.00437 6.1E-05 0.99911 0.00088 0.00083 12 0.83297 0.15530 0.00984 0.00025 0.99837 0.00162 0.00140 14 0.77689 0.20084 0.01877 0.00079 0.99733 0.00266 0.00227 16 0.71639 0.24562 0.03183 0.00198 0.99591 0.00408 0.00326 18 0.65308 0.28713 0.04936 0.00427 0.99407 0.00592 0.00438 20 0.58856 0.32319 0.07125 0.00818 0.99175 0.00824 0.00558 22 0.52430 0.35207 0.09694 0.01429 0.98892 0.01107 0.00680 24 0.46165 0.37256 0.12545 0.02309 0.98551 0.01448 0.00796 26 0.40175 0.38403 0.15541 0.03499 0.98150 0.01849 0.00900 28 0.34553 0.38643 0.18523 0.05012 0.97683 0.02316 0.00987 30 0.29368 0.38021 0.21323 0.06834 0.97146 0.02853 0.01053 The column labeled "At least trio" is the probability that at least 3 people share the same birthday; it is calculated as 1 minus the sum of columns 0 through N/2. I omitted columns 4 through 15 so the table could fit on one line. The last column, "Exactly 1 trio", is the probability that exactly 3 people share the same birthday, and all others have different birthdays. This is a sanity check: if it were greater than the "At least trio" column, there would be something wrong. It is calculated as: P_N(1 trio) = 365*C(N,3)*P(364,N-3)/365^N The conclusion is that the probability of at least 3 people out of 24 sharing the same birthday is much less than the roughly 1 in 6 that we previously estimated. It is more like 1 in 70. The probability of finding two pairs is much higher, about 1 in 8, accounting for most of the 1/6. It is more likely even to find THREE pairs than to find one trio! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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