# Coincidences

Alcra
Posers and Puzzles 16 Mar '05 14:24
1. Alcra
Lazy Sod
16 Mar '05 14:24
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 995/1000).

"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"ðŸ˜‰

alcra
2. 16 Mar '05 15:043 edits
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"ðŸ˜‰

alcra
OK, here is with some simplifications (366 equally probable birthdays):

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
3. 16 Mar '05 18:163 edits
Ever tried working it out for at least k people with the same birthday? ðŸ™„
4. Acolyte
19 Mar '05 10:42
Originally posted by THUDandBLUNDER
Ever tried working it out for at least k people with the same birthday? ðŸ™„
For large k, the following approach might be useful:

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.
5. Red Matrix
21 Mar '05 11:35
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
Well, *that* was easy 'nuff!