Thursday, July 26, 2007

A little bit of maths fun!

While we were coming home in the car the other day Ally, my daughter, asked me a maths question. She'd been looking at all the small change that sits in the cup holder of my car and wondered aloud how many coins you needed for it to be probable that two came from the same year. Now for those non-UK readers, we went decimal in 1970 (?), I don't remember exactly but I know I was at secondary school. That means the earliest date for any coin was going to be 1970 and the latest 2007, which gives a nice beginning and end for this problem.

Now I can't remember exactly how to do the probability maths (it's 30 years since I did any!) so someone may need to help and that's where the fun comes in if you're mathematically inclined. As I remember it the way we'd work this out is to calculate the probability of two, then three etc, coins coming from different years and then subtracting that from 1 to get the probability that they'd be the same. did we use something called a Laplace transform to do this? Once this number dropped below 0.5 then the probability that they were the same year was greater than the probability that they were from different years. Or something like that. Anyway, if I'm right then I think the answer is 15. But I'm not at all convinced that I'm anywhere near right!

I do seem to remember than we were asked to do the same thing to calculate the probability of two people sharing a birthday the answer was 23. In other words, if you have a group of 23 people or more, then it's more probable that two of them share a birthday than that they all have different birthdays. If that's true then 15 seems a little high for the coin question given that there are 365 days to choose from for the birthday thing!

If there's a mathematician out there who can do the maths properly and knows they've done it properly, I'd love to know the answer. I'd be even more thrilled to discover that I'd got it right in the first place!

1 comment:

Richard Pool said...

Aha! I think I've found the solution:

You take the probability that the second is different from the first and the third different from the first and second and so on and multiply these probabilities. When this falls below 0.5 it's more probable that they are the same year.

so it's:

(36/37)*(35/37)... until the answer is <0.5

That gives 8 as the answer.

Now to put it to the test...