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Probability of 3 birthdays on the same day
By Kelvin Porter (T3141) on Monday, October 9, 2000 - 10:50 am:

Firstly let me say that I am a Maths teacher. This does not mean that I am NOT a Maths student as well. My first love is pure Maths and therefore do not feel that confident when dealing with complex probability problems. Recently our year 7 maths group, which contains 21 pupils, had three birthdays on the same day. The group were asked by their friendly Geography teacher to ask me what is the probability of this happening. I have given it some thought and have an answer. I wondered if anyone else would like to give me their ideas ?

By Dave Sheridan (Dms22) on Monday, October 9, 2000 - 11:28 am:

This is a classic problem. Normally a slight variant is phrased, especially for school classes. Suppose you have 23 pupils - what's the probability that at least two will have the same birthday? Most people are then surprised to hear that it's just over 50%.

How do we work this out? We must assume that birthdays are equally likely to be on each day of the year (if not, it's more likely that they'll pile up on one day) and that different people have independent birthdays (seems reasonable unless their parents were conspiring together...) but we'll also ignore leap years although it's not difficult to add them too (by considering 4 years instead of 1).

We want 3 people to have the same birthday. Does it matter what the day is? Not at all. Any choice of day will do.

What's the probability that 3 specific people having their birthday on this one day and everyone else not having their birthday on this day? That's easy, since we have independence - we multiply probabilities.
(1/365)3(364/365)18
ie 3 people must have this day and 18 people must not have this day.

Now we must consider how we choose these 3 specific people. How many ways are there of choosing 3 people from 21? This is a combinatoric problem, and hopefully you've seen combinations and permutations before. If not, ask about them. But the answer is 21!/3!18!

Finally, we need to choose the day. Any day will do, so there's 365 choices.

So, for a specific day, with 3 specific pupils we know the probability. But this is true for every combination of day and pupils, so we must multiply the number of ways this happens together giving a final answer of
365×(1/365)3(364/365)18×21!/3!18!
which is about 1% (I think - quick bash on the calculator anyway)

From this, can you see how to answer the more complicated question, "at least two people share the same birthday"?

-Dave

By Kelvin Porter (T3141) on Wednesday, October 11, 2000 - 12:08 pm:

Dave,
Thanks for replying but I think your answer is too high. I think that your calculation has included within it the possibility that all the other 18 pupils have their birthday on the same day as each other. My original problem intended that the other 18 pupils all have different birthdays from each other; not just different from the 3. My apologies for not making this crystal clear at the start. Any thoughts?

By The Editor:

That will mean that instead of (364/365)18 we will have (364/365)×(363/365)×(362/365)×(361/365)×(360/365)×... = 364!/(346!×36518), as for each person there will be one less day on which their birthday could be.
It's not going to be a huge difference.