Probability question: names in hats

[robvalue]

Yes, that's exactly what I'm saying.

Once we enforce the rules of the game, there is a probability of 1/2 that player 1 picks number 3. He can pick either 2, or 3.

(It can be made explicit as 1/3 + 1/3 * 1/2 = 1/2)

Once he has picked that number, player 2 is left with numbers 1 and 2. He can't pick his own number, so he must pick number 1. And then player 3 must pick number 2.

So half the time, player 1 picks number 3. Every time that happens, we always get 3, 1, 2

But if player 1 takes the number 2 instead (the other half of the time), there are two equally likely possibilities:

2, 3, 1

And

2, 1, 3

So those two share the probability. They get 1/4 each.

This is a very nonstandard puzzle!