- 16 Mar '05 14:24If one takes a group of 64 individuals, and asked them for their birthdates (day and month, not year), the chances of there being at least one match (at least two persons born on the same day of the same month), is 99% (actually about 995/1000).

"Logic" dictates that is should be 64/365 (or 1 in 6ish). Note, I say "logic".

Now, can a mathematically inclined person please explain to me in layman terms, why this is? (the 99% thing, not the "logic" thing!)

Oh, also, please will anyone else share any "weird" coincidences of a mathematical nature? (Not "One time, when my aunt was at the store, you will NEVER believe who she saw, buying tooth paste, no less"

alcra - 16 Mar '05 15:04 / 3 edits

OK, here is with some simplifications (366 equally probable birthdays):*Originally posted by Alcra***If one takes a group of 64 individuals, and asked them for their birthdates (day and month, not year), the chances of there being at least one match (at least two persons born on the same day of the same month), is 99% (actually about 995/100 ...[text shortened]... NEVER believe who she saw, buying tooth paste, no less"**

alcra

The probability of having 'at least one double' = 1- the probability of having no doubles at all. Which is

for one student: 366/366

for two students: 366*365/366*366

for three students: 366*365*364/366*366*366]

etc.

for n students:366*365* .... (366-n)/366^n

or, writtten otherwise: 366! / [(366-n)! * 366^n]

for 64 students : 366!/(362! * 366^64) = 0.29% chance of no doubles,

or 99.71 % chance of at least one double

Another example: you expect more than 50% of the classes with 23 students to have at least one double birthday - 19 Mar '05 10:42

For large k, the following approach might be useful:*Originally posted by THUDandBLUNDER***Ever tried working it out for at least k people with the same birthday?**

Let Dk(x1,...,xk) be a polynomial which takes the value zero if x1 = x2 = ... = xk, and is not zero otherwise (eg the sum over i,j dinstinct of (xi-xj)^2). Let Pk be the product (over all subsets of size k of your set of people) of Dk. Then the probability you're looking for is prob(Pk = 0). This looks a bit ugly, but since Pk is a polynomial in independent random variables with a known distribution (assume uniform, to first approximation) it should be possible to deal with as long as you have a computer handy. - 21 Mar '05 11:35

Well, *that* was easy 'nuff!**Originally posted by Mephisto2***The probability of having 'at least one double' = 1- the probability of having no doubles at all. Which is*

for one student: 366/366

for two students: 366*365/366*366

for three students: 366*365*364/366*366*366]

etc.

for n students:366*365* .... (366-n)/366^n

or, writtten otherwise: 366! / [(366-n)! * 366^n]

for 64 students : 366!/(362! * 366^64) = 0.29% chance of no doubles,

or 99.71 % chance of at least one double