Probability question: names in hats

[Whateverist]

HA! I just worked out the proper solution. I'd neglected to put probability of n being selected into the formula correctly. Funnily enough adding it simplifies the formula hugely. Thus:

1.                   9/10 - 1/10 * 1/9 = 8/9
2. 8/9 * (8/9 * 8/9 - 8/9 * 1/9 * 1/8) = 7/9
3. 7/8 * (7/9 * 7/8 - 7/9 * 1/8 * 1/7) = 6/9
4. 2/3 * (2/3 * 6/7 - 2/3 * 1/7 * 1/6) = 5/9
5. 5/9 * (5/9 * 5/6 - 5/9 * 1/6 * 1/5) = 4/9
6. 4/9 * (4/9 * 4/5 - 4/9 * 1/5 * 1/4) = 3/9
7. 1/3 * (1/3 * 3/4 - 1/3 * 1/4 * 1/3) = 2/9
8. 2/9 * (2/9 * 2/3 - 2/9 * 1/3 * 1/2) = 1/9
9.                           1/9 * 1/9 = 1/(9*9)

1/81 = .01234567890123456789012345678901234567890123456789...

It's still approximately 1%, thus my original guess was pretty damn close. This time I'm SURE it's correct, due to its eloquence.

But please show what your procedure (which I'm not sure I completely understand*) returns as an answer for the two cases for which the answer is known: 3 players and 4 players.


*I believe you mean that the probability for just one person playing would be 8/9 and for two, 7/9.  But we can ignore these results as noise since we understand the game doesn't work at all for 1 or 2 persons.  But I admit I'm not at all sure what you are showing here.

Is it just a procedure for finding the probability for ten people .. or can you stop it at any point to find the answer for less people?  If that is the way it works then your probability for three people, 6/9, would actually be the probability that the last person does not draw his own name .. and that agrees with my reckoning.  But your result for four players in the fourth line does not agree with what I find based on an elaboration of the actual sample space.