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The Birthday Sharing Problem

In a class of 23 children what is the probability that at least two children share the same birthday? cake

The easiest way to solve this problem is to work out the probability that NO two children share the same birthday. That is what we call the 'complementary' event to the one we are interested - it's like the 'opposite'. The probability of any event is always a number between 0 and 1. If you know the probability p of the complementary event then the probability of the event you are interested in is just 1-p.

 Imagine we line the children up. Once we know the first child’s birthday there are 364 ‘unused’ other days for the second child’s  birthday to be different. So the probability that the second child’s birthday is different is 364/365.

 Now we consider the third child. For its birthday to be different to both the first and second child it must fall on one of the 363 unused birthdays, so the probability that its birthday is different from the other two is 363/365. This means that the probability that all three birthdays are different is 364/365 times 363/365.

 Now consider the fourth child. For its birthday to be different to each of the first three children it must fall on one of the 362 unused birthdays, so the probability that its birthday is different from the other three is 362/365. This means that the probability that all three birthdays are different is 364/365 times 363/365 times 362/365.

 Carrying on this argument for 23 children you find that the probability that none of the children share the same birthday is:

birthdays formula

This works out to approximately 0.49.
cake2

So the probability that at least two children share the same birthday is 1 minus 0.49 which is equal to 0.51.  This is just better than a one in two chance.

You can use the same method to calculate the probability that at least two children share the same birthday in a class of size n (where n is any number). For example, in a class size of 40 (which was the typical class size when I was a kid) the probability is 0.9.  It is therefore not surprising that there were invariably kids with the same birthday in almost every class I ever knew.

The fact that people underestimate this kind of probability so badly is often used by people to trick you in various ways. Bookmakers use it to trick you out of money by offering ‘odds’ that look more generous than they really are. Magicians and self-proclaimed psychics use it to fool you into thinking they can predict incredible events, which actually turn out to be very likely indeed.

Finally, it is worth pointing out that we have made some simplifying assumptions in the above calculations.  We have ignored leap year birthdays (which makes the probability lower by a tiny fraction) but set against this we have ignored the fact that in practice birthdays are not spread evenly throughout the year; in fact if you take into account this ‘non uniform’ distribution of birthdays (far more birthdays occur in May in England than in October even though both months have 31 days), it turns out that you need just 20 children in a class for the probability of finding a duplicate birthday to be about 0.5, in other words about 50% of any classes of 20 children will have children with duplicate birthdays.

Why do people get it so wrong? Probably because they think in terms of someone else sharing the same birthday as themselves. That probability is fairly low. It is the same reason why people massively underestimate the probability of events that they think are incredibly rare. Read this example to see what I mean.

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Norman Fenton


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