Probability question: names in hats

[Chas]

Can anyone confirm that these are all the combinations in which all but the last person can draw any name but their own when there are 3 people total and for when there are 4 people total?

Let 1 = first person to draw, and a= his name slip; Let 2 = second person to draw, and b = his name slip; and so on.

Then for three people, the combinations in which all but the last person can draw any name but their own:

1b,2a,3c     1c,2a,3b             Of these three combinations there is only 1 in which the last person draws his name, 
1b,2c,3a                              P(last person is left with his won name) = 1/3 for 3 people

For four people, the combinations in which all but the last person can draw any name but their own:

1b,2a,3d,4c     1c,2a,3b,4d     1d,2a,3b,4c             Of these there are only two combinations  
1b,2c,3a,4d     1c,2a,3d,4b     1d,2c,3a,4b             in which the last person draws his name, 
1b,2c,3d,4a     1c,2d,3a,4b     1d,2c,3b,4a             P(last person is left with his won name) = 2/11 for 4 people
1b,2d,3a,4c     1c,2d,3b,4a     


It would be no fun to type up all the combinations I got for 5 people,  but I count 51 combinations in which all but the last person draws a person's name other than their own.  Of these I count 9 of them in which the last person draws his own name.  (But I am much less confident in these totals.)  If correct that would make P(last person is left with his won name) = 9/51, or if you like, 3/17 for 5 people

Again, if anyone can confirm my count of combinations I would be obliged.  Rob, you were certainly correct about this being one hell of an unwieldy problem.  I wonder if it will allow for an elegant solution.  I've got nothing so far.

Hmmmm, I miscounted the number of legal picks for 4 people, so my elegant solution is wrong.