There are 16 people at my work and 3 of us have the same birthday. What are the odds of that happening?

Question

There are 16 people at my work and 3 of us have the same birthday. What are the odds of that happening?

Answer 1

Slim

Answer 2

Higher than you think.

It's called the birthday paradox.  Among other things, it says that you only need 23 people in a group for the probability of any two people having the same birthday to be more than 50%.  It's obviously less for 3 people in 16 though...

I couldn't work out how to work it out exactly for 3 people as opposed to just two, but a quick simulation tells me it's about 4%, or about 25:1 against.

Answer 3

The two answers given so far seem to be trying to find the probability that exactly 3 of 16 employees have the same birthday. This isn't correct, because it ignores the supersets -- that is, it ignores the probabilities of there being more than 3 with the same birthday (which also satisfies the condition)  and that there may be more than one set of 3 or more.

(And the math on one of the previous answers is incorrect and the other, well, bizzare.)

The general solution to the birthday problem is simple and straightforward in that it is really based on finding the probability of no duplicates, so it's a nice easy sequence. The complement is the sum of ever other conceivable situation -- a single pair of people with the same birthday, 3, 4, or more with the same birthday, multiple pairs, triples, quads, and so on. And of course, some overlap so that a quad is also a pair of pairs, ad nauseum.

So it seems to me that the easiest thing to do is to find the usual complement to the probability of unique birthdays, and then to reduce it by the probability associated with pairs of employees having the same b'day, with the caveat to not reduce it by quads, sextets, etc (not that any of the latter will amount to much).

The probability of 2 or more of 16 people with the same birthday is .283604.

As suggested by asldas, the probability of exactly 2 of a group of 16 having the same birthday is:

365/365 * 1/365 * (364/365)^14 = .002636

(The first doesn't matter, the second same as first, and rest different from the first.)

There are 16C2 = 120 such combinations of 16 (only the first two count, everyone else automatically falls into the latter 14). The probability of 1 or more such combinations can be found with the binomial formula:

= 1 - b(x=0, n=120, p=.002636) = .271521

Subtracting: .283604 - .271521 = .012083

There is a .012083 probability that, in a group of 16 people, there are 1 or more subgroups of 3 or more that have the same birthday.

[If anyone asks about 4 or more with the same birthday, you will be hunted down ...]

Answer 4

P= 16C3 * 1 * 1/365 * 1/365 * 364P13/ 365^13

= 0.326562276%

P stands for permutation or (364!/(364-13)!)

odds are about 1:306 of happening and probabilty is 0.32656%

hope this helps

Answer 5

this is how ya find out.

http://en.wikipedia.org/wiki/Bayes_formu...

personaly, I would find the office accountant and ask them.

Answer 6

Note I won't solve for the case of the leap year, as it over complicates and isn't significant enough to change our numbers.

The odds of this event could be summed by 1 * (1-event 1) * (1-event 2)... these events will be described below

1 comes from the fact you much have a birthday, it's a starting point.

Event 1 - The odds of 15 people not having person ones birthday are (355/356)^15/100.

Event 2 - The odds of 14 people not have the same birth day as either person one or two is (355/356)^14.

Thus the total probability of this event of at least 3 people sharing the same birthday are 1*(1-(355/356)^15)*(1-(355/356)^14)... which equals 0.00159546428 or about .16%

Some notes on these calculations. First this is the probability of at least 3 people of 16 sharing birth days, it includes the event that 4 could have the same birthday etc. Second this method uses the concept of 1 - unfavorable events = favorable events. Third if you want to leave the math for a moment consider birth dates aren't completely random. Consider there are certain times of the year when conception is more likely, such as new years.