All mathematic interessting magicians know, that the probability that two of 23 by random selected persons have the same birthday (not necessarily the same year) is more than 50% and in a group of about 55 persons it is more than 99%.

But what is the probability that a group of n persons consists of:

a) exact three

b) at least three persons

who have the same birthday?

And how to calculate the result?

Angelo

Just a quick guess here that might have some holes in the reasoning:

a)

Three people can be taken in nC3 ways from among n people.

Those three have the same b-day with probability 1 * 1/365 * 1/365

The rest have to have different b-days which they all have with probability

364/365 * 363/365 * ... * (364 - (n-4) )/365

So the probability of exactly three having the same b-day is

nC3 * 364! / (365^(n-1) * (367-n)!)

b)

You already know how to find out the probability of exactly zero having the same b-day from the classic version of the problem.

Use the same reasoning as in a) to find out the probability of exactly two having the same b-day.

The compliment to the sum of those two probabilities will be the probability of at least three having the same b-day.


I have not tried the above suggestions with any numbers yet to see if they hold water, so it's very possible that some tweaking needs to be done.

/Tomas

Congratulations, Tomas!

Very good!

b) is difficult, because we have to subtract not only one couple with the same day, but different numbers of couples (different days) - so I think we will not find a nice formula depending on n for those problem


I solved a) in a different way - I show it the next time when I am here!

Angelo

All bets are off in triplets walk into the room.

Greg

Just a bump of this to say that when I finally solved b), I now realize that my answer to a) is very underestimated. Can you see why?

/Tomas

According to Wolfram|Alpha, the probability for exactly 3 people is roughly 0.82%.
http://tinyurl.com/q53d2jc

I'd imagine the probability for at least 3 people would be at least roughly 0.82%.

Here's a good post on understanding the math behind the Birthday Paradox:
http://betterexplained.com/articles/unde......paradox/

Scott, how many people are there in the group you are working with there? Do you mean if there are only 3 people in the room? The question in a) is regarding n people and exactly three share birthday. My answer above is very wrong.

I'll pose problem c) which can be used to solve b):

c) In a group of n people, what's the probability of exactly two sharing the same birthday?

That one I did wrong the first time I tried to solve it.

/Tomas

Sorry, Thomas, I was thinking he was asking about 2 that share the same birthday in a group of 3. Ignore my previous post.

I'll take a try at part c.

Probably a nicer way to notate this, but:

nC2*1/365 * [(366-n)+1]/365*[(366-n)+2]/365*[(366-n)+3]/365*...*[(366-n)+(n-1)]/365

Could you give some of the reasoning behind that formula? For 88 people I got a probability of 0.48892 for exactly 2 people having the same birthday. What do you get with that formula?

Maybe I should give a spoiler as to what I missed the first time, as I think you are missing that too here?

/Tomas

Too lazy to figure out the answer with 88, but it sure seems like you're getting a higher answer than I would.

My reasoning is this: Let's assume n=4 for now with persons A, B, C, D. We have 4C2 = 6 possible pairings i.e. AB, AC, AD, BC, BD, CD. We want only one pair to have a match. Let's say we want AB to match. The probability of that is 1/365. But we also don't want C to match either A or B. So the probability of that is 1/365*364/365. Now we don't want D to match either AB or C so the probability of that happening is 1/365*364/365*363/365. So basically we want to multiply those probabilities together times the number of possible pairs.

My formula is just the generalization of the above. Seems right to me, but what am I missing?

What I was missing at first was that "exactly two" needs to mean "every set of people that share birthday contains exactly two people" to be able to solve b). So in your four people case you you can actually have two sets of two people sharing birthdays, or just one set. You have calculated it for one set, I think.

AB c d
AC b d
AD b c
a BC d
a BD c
a b CD

But here are the ones I initially missed:

AB CD
AC BD
AD BC

where you actually have two pairs sharing birthday with eachother, but not with any other pair.

I admit it's probably not clear what is meant by "exactly two", but this is the way that needs to be interpreted for b).

This means that a) is probably also very hard to calculate. Is it a success if seven triplets share birthday? Is it a success if one triple share birthday and a lot of pairs share birthday? I assume "yes" to these questions.

/Tomas

Ah, if that's what you mean then yes, I misunderstood your intention. I think I'm right though for the problem of exactly two matching birthdays only in a group of n people. I'll have to think about this other question.

Turning to my favorite method, looking up the answer, Wolfram|Alpha says that you need 88 people for there to be a greater than 50% chance of at least 3 people sharing the same birthday.
http://tinyurl.com/k9hrz4h

This is a tricky problem. In the classic version, you only have to eliminate the possibility that no people share a birthday. Once you do that, the only remaining possibility is that 2 or more people share a birthday.

With this version, you not only have to eliminate the possibility that no two people share a birthday, but also all the pair possibilities. For, say, a group of 10 people, you have to eliminate the possibility of 1 pair sharing a birthday with everybody else having different birthdays, AND 2 pairs of people sharing a birthday with everyone else having different birthdays, AND 3 pairs of people..., and so on up to 5 pairs.

Calculating this for 88 people would involve doing all this up to 44 pairs!

Wow, Scott! I had no idea Wolfram Alpha automatically would give the answer for at least 3. I did this manually and got the same result. As for 99% I got 168 people, which Wolfram Alpha confirms using your way of input.

/Tomas

There's a great discussion of the birthday problem concerning at least 3 people sharing a birthday at this page: http://mathforum.org/library/drmath/view/56650.html

Interestingly, the first answer is actually incorrect, but still worth reading because you get an idea of how easy it is to be led astray in probability. Further down, in a post dated January 18, 2001, is a more detailed and correct answer, and it explains the math every step of the way.

If you have the Wolfram|Alpha app, you can try entering the formula as below. It doesn't seem to work on the website.

1-((P(365,88)/365^88)+(Sum[(C(365,K)*P(88,2K)*P(365-K,88-2K))/(2^K*365^88), {K, 1, 44}]))

This calculates the probability of at least 3 people (as opposed to exactly 3 people) sharing the same birthday among 88 people. If you want to work out a different number of people, replace all the 88s with that number, and the 44 at the end with a number equal to half the number of people (rounded down, if needed).

For example, if you want to run 23 through the above formula, replace all the 88s with 23s, and change the 44 at the end to 11 (because 23 / 2 = 11.5, which rounded down is 11).

Here's the one I used that calculates the compliment: http://www.wolframalpha.com/input/?i=sum......o+88%2F2
The term regarding that none have the same birthday can be ignored when you reach this many people, as it becomes small quickly.

/Tomas