Probability question: names in hats

[robvalue]

Hehe

I tried several approaches. I simplified the puzzle down to just 3 people, so that the calculation is easy. Then up to 4 people, to see if there was some sort of iterative formula. I didn't find one, although there could be.

The annoying thing is that each pick changes some of the probabilities...

Like if player 1 picks the name of player 2, player 2 then has 8/9 chance of picking a player from 1 to 9. But if player 1 picks a higher number, player 2 has a 7/8 chance, because his own name is effectively not for grabs. And so on. They have a ripple effect through each other which I found very hard to put into a general formula.

Maybe there is a simple way of doing it that I just haven't thought of. But it seemed like a total disaster.

Usually, with this kind of thing, you get an easily countable number of branches on a probability tree, with the same probability. But they don't work like that here, as far as I can see.

You could also approach it in reverse: find the probability that one of the first 9 take player 10's name. This will then be (1-probability it is left).

So probability player 1 takes it is 1/9, since he can't take his own. But then the chances of player 2 depends again on whether his name as already been picked or not...