September 8th, 2008
This problem came up in conversation recently, and I realized I don't have the mathematical chops to figure it out. It's about the probability of birthdays...
Answer these questions, assuming the distribution of birthdays is even (which it actually isn't):
1) what is the probabiility that person A's birthday is January 1?
>> you'd say "one in 365". Right?
2) what is the probability of my birthday being the same as Person A's?
>> is it the same? 1:365? Because my birthday is a fixed date, so the question is really the same as #1. Or is it?
3) what is the probability that two people chosen at random will have the same birthday?
>> Are the odds different in this scenario? Here I don't know the birthday of person A, or person B. Yet this seems like it would have the same probability as #1 and #2, since you're just asking if Person B has Birthday A, where Birthday A is arbitrarily chosen by assigning Person A.
4) Considering the questions above, in a group of 3 people, what is the probability that two of them will have the same birthday? What about a group of 10, 20, ...
n people?
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September 8th, 2008 at 4:12 pm
1-3 are the same question, and yes, it’s 1:365 if birthdays are evenly distributed. Which, world-wide, I’m guessing is probably a fair assumption. It doesn’t matter what date you’re refering to - January 1st, your birthday or person B’s birthday - it’s 1:365 every day (assuming a 365 year).
4 is the birthday problem: probability that at least 2 have same birthday in a room of n people is 365! : (365^n*(365-n)!)
Note that n must be less than 366 for this formula. At 366, 2 people have to have the same birthday, so the probability is 1:1.
You don’t have to take my word for any of this though:
http://en.wikipedia.org/wiki/Birthday_problem
September 8th, 2008 at 4:15 pm
Whoops - I meant the probability that everyone in the room has a different probability from each other was that formula. So the probability that at least 2 people share a birthday is 1 minus 365!/(365^n*(365-n)!)
See… wikipedia is way better than randomly posters to your blog.
September 8th, 2008 at 4:17 pm
Whoops - I meant “I meant the probability that everyone in the room has a different BIRTHDAY from each other…”
I wish there was an edit button…
September 10th, 2008 at 11:58 am
I started figuring this problem out offline, and here is my reasoning:
consider a group of 3 people: A, B, and C
number the days of the year, 1 - 365
The combinations can be expressed as a sequence, like:
1,1,1
1,1,2
1,1,3
1,1,4
1,1,5
…
365,365,364
365,365,365
There are 365^3 different combinations.
of those, there are 365 values for x where x=A=B=C (all three have the same birthday)
there are 365 values for A=B, where C is different; and for each value x=A=B, there are 365 values of C. But one of those cases was the one where A=B=C, which we have already counted above. This means the pattern A = B != C offers 365*364 more unique combinations; that’s 365*364 = 132860
The same pattern exists for
A = C != B
B = C != A
these are exclusive sets; there can be no overlap, since each pattern has an inequality and an equality which are not repeated in any of the other patterns.
So the total number of combinations where two people in a group of 3 have the same birthday is:
((365*364) * 3) + 365
= 398945
out of the possible cominbations:
398945/48627125
which is approximately 1 in 122
Thus in a group of 3, it is more likely that you’ll find 2 identical birthdays than if it was a group of 2. Makes sense.
@BO, does your equation result in the same answer?
1 - (365!/(365^n*(365-n)!))
I’m not sure, because my calcualtor craps out when I attempt to display (365!)