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# Birthday probability problem

The probability that at least 2 people in a room of 30 share the same birthday. Created by Sal Khan.

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• The total # of possible outcomes of different birthday is n!/(n-k)! as presented.
I am ok with that.
But what happens to combination rule (where order is not important) - using the combination rule I thought this # would be n!/k!(n-k)!
What happens to the k!
I know we end up using permutation where order is important - but why order is important in this problem? I thought order is NOT IMPORTANT so we should use combination, NOT permutation. •   Hi site2n2. Order is important in this problem because we care about treating each person as a distinct person. For example, if there were three people in the room, namely Alice, Bob, and Eve, then assuming we have one possibility where Alice was born on 8/18, Bob was born on 4/13, and Eve was born on 8/10, we want to count all the permutations of these three dates since it's very possible for Alice to be born on 8/10, Bob to be born on 8/18, and Eve to be born on 4/13 instead. Notice that n!/(n-k)! treats the k objects as distinct (and hence does not count permutations of the k objects as only 1 possibility) while nCk = n!/(n-k)!*k! divides by k! and counts permutations of the k objects as 1 possibility.
• I can see that this is a permutation: n!/(n-k)! but I don't really understand then why he divides by 365^30 as this is not in the permutation ula we have learnt? Can anybody explain this? •   Jennifer,
This is not a permutation problem.
Matthew,
This is not a combination problem.
The birthday problem should be treated as a series of independent events. Any one person’s birthday does not have an influence on anybody else’s birthday (we will assume no twins, triplets, etc. are in the same class). In finding the probability that no one has the same birthday, Sal took the probability that each member of the class, in turn, did not have the same birthday as all the other members of the class before them. It would be like finding the probability that in five separate rolls of a six-sided die you did not get the same number (which would be (6/6)*(5/6)*(4/6)*(3/6)*(2/6) . There are no combinations or permutations there. Likewise there is no combination or permutation in the thirty terms of (365/365)*(364/365)*(363/365)* … (338/365)*(337/365)*(336/365) for the birthday problem. Sal only wanted to simplify the numerator of that series of numbers. Looking at just the numerator (the denominator becomes 365^30), Sal has 365*364*363*… 338*337*336. Another way to write this is 365!/335!, which is also 365!/(365-30)! which looks like the n!/(n-k)! permutation formula. Just because it looks like the permutation formula does not mean we are using the permutation formula.

I wish Sal had explained that the factorials he was using was not an attempt to apply the permutation formula. This is fooling a number of people.
• Shouldn't we keep in mind leap years? Meaning, the probability that 1 person was born on a specific day is 1/366, not 1/365? •  Except that that wouldn't be the correct probability if we were to consider it. Leap years only happen every fourth year (more or less), so the probability that a person was born on February 29 is 1/1461 and the probability of every other day would be 4/1461. Since the events aren't all equiprobable in this new system, calculating the probability that lots of people all have different birthdays is far more complicated, which is why people traditionally ignore leap years for a reasonable estimate.
• How would this calculation change if we wanted to know the probability of at least 3 people out of 30 share the same birthday? • I know how to do the problem, but just thnking about the answer, it seems really high. There are 30 people in a class, 360ish days in a year and if you think about it like 360 spots for people to stand in and the 30 kids need to stand in a spot, it'd seem really hard for 2 people to stand in the same spot given ALL the possible spaces there are. I know this logic isn't "holeproof" but still, it seems that the probability should be like much less than such a high number like 70%. • Could you instead just work out the probability that one person has a birthday on one day and then times it by the probability that another person has a birthday on that day. • May I ask for example the number of people is more than 365? what equation will be used instead? •  The birthday probability problem is trivial if the number of people is greater than 365, as then there is a 100% chance that 2 people share a birthday.   