Thread: Christmas present probability

THere are 11 people in my family that will be at the Christmas dinner/tree event. We are putting everyone's name on a piece of paper and putting it into a bowl. What is the probability that nobody will draw their own name from the bowl. If they draw their own name, they'll of course need to switch with somebody else. I'm not just interested in the actual probability, but also how to approach this problem.

Thanks.

This is a well-known problem: a permutation of the numbers from 1..n which does not leave any element fixed is a 'derangement' and you can find a Wikipedia article at http://en.wikipedia.org/wiki/Derangement and a Mathworld article at http://mathworld.wolfram.com/Derangement.html

The Mathworld one says take the derangements and divide them by the permutations to get the probability that nobody is buying their own present. This is 1/e which is .37. My Aunt and Brother each got their own at first by the way. I think its interesting that once you get to 4 and higher for the number of points, the probability is still 1/e.

This is good stuff, thanks for the info.