@exitthelemming (Samuel West) posted an interesting question today, which took me back to early grammar school maths – and an answer I’ve wanted to publicly show off ever since!
His question was:
Q: My World Cup sticker album has spaces for 638 stickers. How many packets of 5 must I buy before it\’s more than 50% likely I get a swopsy?
So, on the face of it, many people might think that it would be a number approaching 319 (half the number of stickers), however this is not the case.
Before I show the answer though, there is an important caveat – the answer to this question assumes that the 638 stickers are evenly distributed in all the packets, and that the manufacturers have not pulled the usual trick of having a few rare stickers to bump up sales of the packets. So, the maths (and many thanks to Rob Williams – my superb maths teacher from Davenant Foundation Grammar School in the 70’s) is:
The probability of the first sticker that you have is 638/638 or 1. It has to be unique.
For the second sticker to be different it must be any 1 of the 637 remaining possibilities, so that is a probability of 637/638 or 99.8433% (to 4 decimal places).
Likewise for the third sticker to be unique – 636/638 or 99.6865% (to 4 dp again).
So you can carry on with this 635/638, 634/638…
So this gives us the probability of any sticker being different from all the preceding stickers.
So the probability of the first three stickers being unique is the product of all three probabilities – 1 times 637/638 times 636/638 (or 1 * 99.8433 * 99.6865) which is 99.5303%.
Although the spreadsheet above was a new one for me, the one that Rob Williams taught us (although in those days it was pen, paper and slide rule!) shows that if you have 23 people in a room, then there is a greater than 50% chance that 2 of them share a birthday. So that means in any school (provided classes are randomly distributed by birthday) every other class should have 2 children sharing a birthday.