Birthday Betting

Here's another Q&A from my Betting Surgery column in the "unpublished" final edition of The Winning Report...

Q. Please can you settle an argument with a neighbour? He claims that, in a room of 26 people, the chances are at least two will have the same birthday. I can’t believe this is right. Can you settle our dispute and show how the answer is worked out? There’s a tenner riding on this.
Michael Waters, Blackpool

A. We can settle your bet, but you won’t like the answer! Your neighbour is quite right. The probability of two or more people from 26 sharing the same birthday is 63 per cent, or almost 2/1 in favour.

The way to work this out is to establish the probability of no birthdays matching. The chance of the second person’s birthday differing from the first is 364/365, the chance of the third person having a different birthday again is 363/365 and so on, until the chance of the 26th person not having a matching birthday is 340/365.

All of these conditions must be met for no-one to have matching birthdays – so for the bet to fail you must multiply these fractions together. That gives a figure of 0.37. Subtracting this from one gives a probability of 0.63, or 63 per cent, that two or more people WILL have matching birthdays.

As a matter of interest, with 23 people in the room the chances of a match are just over even; with 34, the chances are 80 per cent. Why not try offering this as a bet at the next party you go to? You might just win your tenner back!