Probability question: names in hats

[robvalue]

You're right Chas, but the problem is that the combinations don't all produce the same probability.

I think the solution involves both counting combinations and being able to count up how many there are for each probability. I got stuck into it at one point, but I didn't quite get there.

It's the "not your own name" that throws the wrench in, and the probability for each person changes depending on whether their number has already come up or not.

I did the example for 4 people to show this. There's 2 legal combos, but they have different probs:

For 4 players:

2, 3, 1, 4 = 1/3 * 1/3 * 1/2 = 1/18
or
3, 1, 2, 4 = 1/3 * 1/2 * 1/2 = 1/12

Total = 5/36