Prof. Templeton Posted August 9, 2008 Report Share Posted August 9, 2008 There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet. How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ). Quote Link to comment Share on other sites More sharing options...
0 TwoaDay Posted August 9, 2008 Report Share Posted August 9, 2008 20 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 why doesn't 20C2 or C(20,2) or Combination(20,2) or 20 choose 2 or however you want to say it is 190, ie, there are 190 unique pairs (380 if in either order, but that doesn't matter, so 190) of people from within 20 people. In each of those 190 pairs, if person A has a set birthday, and person's B chance of having the same birthday is 1/365, so each pair has a 1/365 chance of getting a match. With 190 pairs present, there is a 190/365 chance of a match happening, which is a little over 52% 20 is less than 23 20 work? I think I can prove it: I believe you made a arithmetic mistake, and a minor logical one: First, the arithmetic 1 - (364/365)20_C_2 = 1 - (364/365)190 = 1-0.593770541 = 0.406229459 < .5 < .52 (what you gave) Second, there is a minor logical error. If you do the problem one person at a time under the premise that no two people have the same birthday then you get the following sequence (already brought up by someone else): First person__second____third___fourth__etc 365_______365-1___365-2__365-3_______365-(n-1) ------__*___-------- * -------- * -------- * ... *------------ 365________365____365____365_________365 The numbers subtracted here are for each day they cannot use because it has been occupied. If you subtract the final product of all of these values from 1, you get the probability of at least 2 of the people having the same birthday. This is slightly more accurate that what you gave, though the deviation is minimal (the raw difference between the two is between 0 and 0.009891937 overall). Using combinations assumes independence in the 364/365 probability and the sequence does not. This causes the error. As proof that there is an error: It is understood that (if we neglect leap years) with 366 people there is a 100% probability that there is an overlap because there are only 365 days. Using your method the probability is: 1 - (364/365)366_C_2 = 1 - (364/365)66795 = 1 - 2.6004E-80 < 1 As a matter of fact using combinations mistakenly suggests that no matter how many people are in the room, there is never a 100% probability. Using the sequence (and combinations because the error is minimal). The probability of there being 2 people with the same birthdays is just below 50% with 22 people (0.475695308) and just over at 23 people (0.507297234). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 9, 2008 Report Share Posted August 9, 2008 I believe it's 20 as 19+18+17...+1=190 which is the probability that 2 people in a group of 20 will have the same birthday. This also happens to be 52% of 365. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 10, 2008 Report Share Posted August 10, 2008 20. I read somewhere a long time ago that if you had 20 people in a room, chances are that at least 2 of those people would have the same birthday. I cant imagine how this would be plausible given the odds but that is what it said! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 11, 2008 Report Share Posted August 11, 2008 There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet. How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ). This is not intuitive it is strictly mathematical and the answer is suprising Looking at the probability of NOT having a birthday match for person 2 it is 364/365, person 3 it is 363/365 and so on. multiplying these fractions gives the probability of NOT having a birthday match for X number of people. Using a spreadsheet at 23 people there is at least a 50/50 chance. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 11, 2008 Report Share Posted August 11, 2008 why doesn't 20C2 or C(20,2) or Combination(20,2) or 20 choose 2 or however you want to say it is 190, ie, there are 190 unique pairs (380 if in either order, but that doesn't matter, so 190) of people from within 20 people. In each of those 190 pairs, if person A has a set birthday, and person's B chance of having the same birthday is 1/365, so each pair has a 1/365 chance of getting a match. With 190 pairs present, there is a 190/365 chance of a match happening, which is a little over 52% 20 is less than 23 20 work? I think I can prove it: With 20 random people you will have an average of .5 birthday matches but sometimes you will have 2 or 3 matches and this is being worked into your calculation of average matches. Since multiple matches do not count any more than one the number of people needed to get at least a 50/50 chance of having at least one is 23. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 11, 2008 Report Share Posted August 11, 2008 why doesn't 20C2 or C(20,2) or Combination(20,2) or 20 choose 2 or however you want to say it is 190, ie, there are 190 unique pairs (380 if in either order, but that doesn't matter, so 190) of people from within 20 people. In each of those 190 pairs, if person A has a set birthday, and person's B chance of having the same birthday is 1/365, so each pair has a 1/365 chance of getting a match. With 190 pairs present, there is a 190/365 chance of a match happening, which is a little over 52% 20 is less than 23 20 work? I think I can prove it: I am wondering why it is not the same as well. I think your approach is a solution to n * (n - 1) = 365 and that puts n right at about 19.6. However I have more faith in the solution using the fractions 364/365...... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 12, 2008 Report Share Posted August 12, 2008 There is a 50/50 chance, you would only need one other person to be in the room. there is a 50% chance that they either do or do not have the same birthday. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 12, 2008 Report Share Posted August 12, 2008 No math or anything used here...I've just heard this before. It's 25. If not, then I got misinformation. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 13, 2008 Report Share Posted August 13, 2008 14 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 15, 2008 Report Share Posted August 15, 2008 (edited) 20 -- first person has no chance number 2 has 1:365 number 3 has 1:365 for first one, and 1:365 for second..so his chance is 2:365, and the total is 3:365 number 4 1:365........................................................................... .................................6:365 The 20th person adds in 19:365 making it a total of 192 out of 365 which is more than 183/365 The probability of 50:50 does not mean that the next person has a 50:50 chance...it means that within the sample there is a total probability of a situation existing. Edited August 15, 2008 by AmyL Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted August 19, 2008 Author Report Share Posted August 19, 2008 The probability of 50:50 does not mean that the next person has a 50:50 chance...it means that within the sample there is a total probability of a situation existing.20 -- first person has no chance number 2 has 1:365 number 3 has 1:365 for first one, and 1:365 for second..so his chance is 2:365, and the total is 3:365 number 4 1:365........................................................................... .................................6:365 The 20th person adds in 19:365 making it a total of 192 out of 365 which is more than 183/365 20 is not the answer Out of 20 people there are 20*19/2=190 ways to select a pair from the pool. With 365 days in a year, the probability of one pair having the same birthday is 1/365 and likewise the probability of 1 pair NOT having the same birthday is 1-1/365 or 0.99726. So if we figure in each of the 190 pairs not having the same birthday seperately, than P(of 190 pairs not having the same B-day)= 0.99726^190 = .593739. Therefore the probability of one of those 190 pairs having the same birthday is 1-0.593739=.406261 So 20 people would give a 41% chance and not at least 50%. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2008 Report Share Posted August 19, 2008 You must have 367 people in a room to guarantee people having the same birthday, so 184 (half a person is still a person) people would give you a 50% likelihood of sharing the same birthday Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 19, 2008 Report Share Posted August 19, 2008 20 is not the answer Out of 20 people there are 20*19/2=190 ways to select a pair from the pool. With 365 days in a year, the probability of one pair having the same birthday is 1/365 and likewise the probability of 1 pair NOT having the same birthday is 1-1/365 or 0.99726. So if we figure in each of the 190 pairs not having the same birthday seperately, than P(of 190 pairs not having the same B-day)= 0.99726^190 = .593739. Therefore the probability of one of those 190 pairs having the same birthday is 1-0.593739=.406261 So 20 people would give a 41% chance and not at least 50%. Yes, now I see how we can get the exact answer! (1-1/365)^(n(n-1)/2) = .5 n(n-1)log(1-1/365) = log .5 n(n-1) = 505.3040 n = 22.98454 Just one problem, how do we account for a leap year? Do we change 365 to 365.2425 (365 + 1/4 - 1/100 + 1/400)? If we use this number n is equal to 22.992 Quote Link to comment Share on other sites More sharing options...
0 Prime Posted August 19, 2008 Report Share Posted August 19, 2008 That's a tough math. I've got an equation: ((364!)/(364-x)!)/365x = 1/2 Not sure how to solve it, I used spreadsheet finding: 23 people is the smallest number for which probability of at least 2 people sharing b-day is over 50% (50.7297%). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 20, 2008 Report Share Posted August 20, 2008 There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet. How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ). Here is a variation I thought of: Let's say the problem was re-worded to ask exactly how many people do you need in a room to make this a 50:50 chance. Do you have enough information to obtain an exact calculation? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2008 Report Share Posted August 23, 2008 The answer is as follows: 12 Factorial + 366 Factorial = Number of people. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2008 Report Share Posted August 23, 2008 This is a classic probability puzzle that we went over when I took probablility in college. The answer is in fact 23, for reasons given by the origianl poster. There's a funny version of this though. On the first day of a college probability class, the instructor presented this problem, and said, "It's too bad that there are only 15 people in the class, since I could demonstrate this if we had 23." A student said, "I'll bet you $50 that there are two people in this class with the same birthday." The instructor insisted that this was a bad bet with such a small group, and took him up on it. He started going around the class and asking students for their birthdays, stopped before finding a match, and started laughing instead. He gave $50 to the student. The student didn't know the birthdays of any of his fellow students. How did he win the bet? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2008 Report Share Posted August 23, 2008 The student didn't know the birthdays of any of his fellow students. How did he win the bet? there were twins in the class? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 24, 2008 Report Share Posted August 24, 2008 Yep. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 3, 2008 Report Share Posted November 3, 2008 Assume that there are n students, and I'm one of them. Probability of me to share my birthday with a friend is (n-1)/365. Since there are n students, each has a (n-1)/365 probability, sum is n(n-1)/365, but as hand shaking we should divide it by 2 ---> n(n-1)/730. This is needed to be equal to 1/2, then n(n-1)=365. If we give 20 to n, n(n-1)/730=0.52 is the probability for 20 students. I don't forget that this is the probability, unless each students other than me have all different birthdays. However, if two of them share their birthdays, this doesn't reduce the odds, because anyhow, we are looking for such a pair. I see that 23 is the accepted answer for this question, but where is my error? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 3, 2008 Report Share Posted November 3, 2008 Ah I just did this the other day in my calculus class. I don't have the work on me, however I will in a few hours. Will reply soon =] Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted November 3, 2008 Author Report Share Posted November 3, 2008 Assume that there are n students, and I'm one of them. Probability of me to share my birthday with a friend is (n-1)/365. Since there are n students, each has a (n-1)/365 probability, sum is n(n-1)/365, but as hand shaking we should divide it by 2 ---> n(n-1)/730. This is needed to be equal to 1/2, then n(n-1)=365. If we give 20 to n, n(n-1)/730=0.52 is the probability for 20 students. I don't forget that this is the probability, unless each students other than me have all different birthdays. However, if two of them share their birthdays, this doesn't reduce the odds, because anyhow, we are looking for such a pair. I see that 23 is the accepted answer for this question, but where is my error? I think the key to this question is to determine how many people do not share the same birthday. The answer can not be 20 if you look at it like thisOut of 20 people there are 20*19/2=190 ways to select a pair from the pool. With 365 days in a year, the probability of one pair having the same birthday is 1/365 and likewise the probability of 1 pair NOT having the same birthday is 1-1/365 or 0.99726. So if we figure in each of the 190 pairs not having the same birthday seperately, than P(of 190 pairs not having the same B-day)= 0.99726^190 = .593739. Therefore the probability of one of those 190 pairs having the same birthday is 1-0.593739=.406261 So 20 people would give a 41% chance and not at least 50%. Quote Link to comment Share on other sites More sharing options...
0 Prime Posted November 3, 2008 Report Share Posted November 3, 2008 (edited) Assume that there are n students, and I'm one of them. Probability of me to share my birthday with a friend is (n-1)/365. ... For starters, that estimate of probability is incorrect. Say, instead of 365, there were only two possible values -- "0" and "1". Let number of people be n=3 and the first one holds "1". Your formula for the probability of matching first man's number against the remaining two yields: P = 2/2 = 1 (or certainty). Note, how for more than 3 people the formula would produce probability value of greater than 1. Actually all combinations between two people are "00", "01", "10", "11". Four combinations in all. In three of those at least one other person holds the same number ("1") as the first person we set aside. For the resulting probability 3/4. A popular technique to calculate that kind of probability is to estimate the probability that none of the others has the same number (birthday). It makes it easier because you can view that probability as a series of dependent probabilities, which you can multiply. Thus for each of the n-1 individuals the probability not to share your b-day is 364/365. The probability that none of n-1 shares your b-day is (364/365)n-1. Then, since the probability of someone sharing you b-day or not is 1, the probability of at least one person of the remaining n-1 sharing your b-day is 1 - (364/365)n-1. The OP, however, asked to find when it is 50% of having at least any two people (not just you) in the room to have the same b-day. The simplest solution that someone already have posted here is to estimate the probability that no two people have the same b-day. So when there is one person in the room, the second walking in has a 364/365 chance not matching. The third person now must mismatch two b-days at the probability 363/365, the fourth person 362/365, and so on... The overall probability for n people is a product (364/365)*(363/365)* ...*(366-n)/365. Edited November 3, 2008 by Prime Quote Link to comment Share on other sites More sharing options...
0 Guest Posted November 4, 2008 Report Share Posted November 4, 2008 (edited) Thanks to Prof.Templeton and Prime for detailed explanations. I simulated such a probability for 1 million of times each, and found that: People count Probability 20 ---- % 41.13 21 ---- % 44.38 22 ---- % 47.46 23 ---- % 50.76 46 ---- % 94.83 At least 23 people are needed for % 50. I didn't try more but I think %100 is not possible. Edited November 4, 2008 by nobody Quote Link to comment Share on other sites More sharing options...
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Prof. Templeton
There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet.
How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ).
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