Understanding the Birthday Paradox

I have a question: 6 people, one movie being advertized…3 people having the same birthday…and same birthday show on advertisement at the same time. What would the ‘chances’ be? This was an actual event.

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I asked a question on 4-1-09 about the ‘birthday paradox’ and an actual event. The reason I would like to know the ‘odds’ of that happening, is because it was one event out of four similar events. Any suggestions on where I can find some answer to the ‘odds’ other than here?

[…] Statistics: Combinations & permutations, Birthday Paradox, Bayes’ Theorem, […]

[…] Then I thought I should at least leave you with something. So I’ll introduce you to the concept of the birthday paradox. Basically, given a group of 23 people, what’s the percentage chance that any 2 persons in that group share the same birthday? (hint: it’s higher than you think. Here’s a less clinical explanation of the paradox than the one from Wikipedia.). […]

Hi Ashton, you might want to ask your math teacher to see if you’ve covered the necessary topics in class. You’ll probably need statistics and combinatorics.

all my brothers and sisters(not by both same parents)have the birthday of 2or 16

[…] The math gets somewhat complicated, but you can check it out in more detail: Understanding the Birthday Paradox and Wikipedia’s page. […]

The explanations given are all approximations, in order to get an exact result you follow the start to Appendix A, but don’t attempt to simplify with e^x. The solution is actually fairly simple, for n possibilities (days in the year) and k events (people at the party) we get a probability of:

1 - (P(n-1,k-1)/(n^(k-1))).
Where P(n,k) is the number of ways to pick k elements from a set of n, or n!/(n-k)!.

This will give an exact solution, the probability of finding two people with the same birthday from a crowd of 23 is more accurately: 50.7297234%

I hope that this makes sense, if it doesn’t, look at the page on combinatorics and/or think about the fact that (1-(j/n)) = ((n-j)/n) with reference to Appendix A.

Aren’t there 366 possible birthdays? (feb 29)

this is really interesting!! im doing a math project on this !! nice topic!!

@angelina: Awesome, glad you liked it!

the way i intuitively see the 50% is like this,

imagine throwing 23 point blobs of paint at a calender on the wall. Then move all the dates that hit into a tidy ~5x5 square in the corner.

Now throw another 23 blobs of paint at the wall.

To me, it is almost inconceivable that no paint blobs will now touch the dates in the 23 blob square in the corner (50/50 maybe)

[…] are savvy with Hamming’s error correcting code or not, listen to Kalid Azad when he presents Understanding the Birthday Paradox posted at BetterExplained in which he explains the Birthday Paradox from […]

I got it after third heading…HAHAHAHAHAHA!!!

Thank you.
Well detailed and nicely structured guide for a really misinterpreted problem.

[…] maksymalnej ilości kombinacji (bez dzielenia). Ta ilość prób wg ogólnego wzoru (wyprowadzenie tutaj w załączniku B) to tylko ok. 1,18 * sqrt(N) gdzie N oznacza ilość wszystkich możliwych […]

The Birthday Paradox in Clojure and Incanter…

The Birthday Paradox is an interesting little problem in probability theory. To quote Wikipedia:

“[...] the birthday problem, or birthday paradox pertains to the probability that in a set of randomly chosen people some pair of them will have t...