Birthday Probability, Class of 25Date: 11/14/2002 at 16:55:21 From: C.J. John Subject: Birthday probability A class consists of 25 students. Find the probability that no two students will have the same birthday, assuming that all days in a year are equally likely birthdays and ignoring leap years. Date: 11/15/2002 at 21:24:30 From: Doctor Fwg Subject: Re: Birthday probability Dear C.J. This problem, in one form or another, seems to come up frequently, and the answer almost always seems astounding when first encountered. There are several other ways to solve this problem but the one I am listing here is the easiest to understand, I think. You can also find other related solutions in our archives - search using the keyword birthday - and in the Dr. Math FAQ: The Birthday Problem http://mathforum.org/dr.math/faq/faq.birthdayprob.html In any case, each person can compare his/her birthday with every other person, but only once. For example, if there are six people in a room, the first can compare his/her birthday with five others, the second can compare with four others, etc. So, for six people, there are 15 possible combinations like this. For n people there are n(n - 1)/2 combinations. Now, assuming 364 days in a year (ignoring leap years), the approximate probability of a birthday match between any two people is (1/364), but the probability of no match is [1.0 - (1/364)], or (363/364). So, the probability of two "no matches" in a row is: (363/364)(363/364) = (363/364)^2 = 0.9945. Now, if there are n combinations of possible matches, the probability of n "no matches" in a row is (363/364)^n. For 25 people, n = 300. So, for this case, the probability of no birthday matches for these 25 people (assumming no twins or triplets, etc.) is (363/364)^300 = 0.4381. You can see that in this case (of 25 people) the chances are better than even that at least two of these people will have the same birthday. So, if you were to survey, say, thousands of different rooms, each holding 25 randomly selected people (with no twins, triplets, etc.), you would find that about 56% of these rooms would have at least two people with the same birthday. I hope this has been helpful. With Best Regards, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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