Probability question: names in hats

[Whateverist]

There are only 3 possible sequences which don't violate the rules that you can't pick your own number (except for the last guy):

2, 1, 3: probability 1/4

2, 3, 1: probability 1/4

3, 1, 2: probability 1/2

That's it. That's all the ways it can happen. They don't all have equal probabilities. And only the first one satisfies the criteria of player 3 drawing his name.

All I've done is extracted these 3 sequences from the possible 6 permutations of 1, 2, 3 by excluding the ones where player 1 or player 2 take their own number.

I've excluded 1, 2, 3

1, 3, 2

And

3, 2, 1

These three can't happen.

Wait, are you saying that the 3,1,2 sequence is twice as likely to occur as either the 2,1,3 or the 2,3,1 sequence?  Lets see about that.

A:  1,2,3
B:  1,3,2
C:  2,1,3
D:  2,3,1
E:  3,1,2
F:  3,2,1

These at least have an equal chance of occurring if we ignore the requirement that sequences A,B & F require a redraw.  Now do we know whether the rejected slip is put back in before the redraw?  (Does any of this really matter?  If so, I'm beginning to hate this problem.)  Without knowing about what to do with the rejected slip it is hard to know how to proceed from here.

But setting that concern aside, I don't see how you can say sequence E is twice as likely to occur as either sequence C or D.  The way I analyze the experiment/game is this:

N players place a strip of paper with their name on it into a hat.  (Let's assume their names are as distinct as they are.)
They then take turns drawing, but only keep the first strip which does not contain their own name.  Mis-draws are returned to the hat.
The only player who might be stuck with his own name strip in this game is the last to draw.
The question is: what is the probability the last person indeed draws his own name?

By my analysis, a game with n players can take many more than n draws but never more than n allowable draws.  The sample space is therefore limited to the sequences composed only of allowable draws.  These, I believe, are equally likely without regard to the number of unallowed draws which might occur in the course of the game.  I think this is where we disagree.  Is that so?