Probability question: names in hats

[Whateverist]

Whateverist:

It seems like you have the same solution paths as me for 3 and 4 people, but you're assuming the probability of each combination is the same. Sadly, they are not. So simply counting them isn't enough. That's what makes this problem harder.

For example, with 3 people, it must be 2, 1, 3

Player 1 can pick either name 2 or name 3. He can't pick his own. So the probability he picks name 2 is 1/2.

You can write it out explicitly including his own number, but you get the same result, as if it wasn't in there:

Prob(pick 2 on first try) + prob(pick own number)*prob(pick 2 on second try)

=1/3 + 1/3 * 1/2

=1/3 + 1/6

=1/2

So now player 2 must pick number 1. The numbers left are 1 and 3. So it's a straightforward 1/2.

Combining these gives 1/2 * 1/2 = 1/4

Similarly for the two lines for 4 people.

But if in deed the probability is 1/4 you must be able to demonstrate 4 distinctly different drawing sequences in which the first two people draw any name but their own and where only one of those results in the third one drawing his own name (or else some multiple of those numbers like 2 out of 8).  I've found just the three: 1b,2a,3c ; 1c,2a,3b ; and, 1b,2c,3a .. where the numbers are assigned to the people by virtue of the order in which they pick and the letter with the same ordinality represents the name slip of each person.  

I think I can convince you these are the only three drawing sequences which satisfy the condition of the problem that none but the last person can keep their own name slip.  Here goes.

As you say, the first person can only keep person two's name slip "b" or person 3's name slip "c". 

If he draws person 2's name slip there are two ways that can go ->  1b,2a,3c  and  1b,2c,3a
If he draws person 3's name slip there is only one way that can go ->   1c,2a,3b  since reversing the last two results in 2b which isn't allowed.

What other possible drawing meets the conditions of the problem?  Since there only are three possibilities to consider the probability is 1/3 that the last guy gets stuck with his own name tag when three people play.