My previous calc is obvously wrong, I think ford's is too.

2 persons:
If there are two other people, there's a 50% chance that the person with it's seat free gets in first, followed by a 100% that the last person doesn't get its seat. If the person with the seat stolen enters first, it's a 50% chance/risk that he takes the remaining persons seat and 50% that he takes fords. This amounts to 25% chance for the last person to sit in its own seat.

3 persons:
The tree of possibilities for three persons is no that much harder, as all but one probabilities lead to the the calc above. I will call the person with the stolen seat Soviet from now on, and any other will be called Xenu.

1/3 that Soviet enters first -> 1/3 that Soviet takes fords seat and 2/3 that Soviet takes one of the others
I.e. a 1/9 chance that the last person will get bound to get its own seat after this event, and 2/9 resulting in the case above with 25%.

2/3 that Xenu enters first -> Also leads to the case above.

P = 8/9*0.25 + 1/9 -> 1/3

4 persons:

1/4 Sov -> 1/4 takes ford, 3/4 takes someone elses leading to the probabilty above
3/4 Xenu -> probability above

P = 15/16*1/3 + 1/16 = 3/8

5 persons:

1/5 Sov -> 1/5 takes ford, 4/5 takes someone elses
4/5 Xenu

P = 24/25*3/8 + 1/25 = 217/625

Hmmm, iterative solution?

P(n) = ((n^2-1)/n^2) * P(n-1) + 1/n^2
P(n) = ((n^2-1) * P(n-1) + 1)/n^2

Edit: this doesn't take into account the chance that ford takes his own seat and it didnt really count him as a person either, easily fixed.

P(n) = 1/n + ((n-1)/n) * S(n)
S(1) = 0
S(2) = 1/4
S(n>=3) = ((n^2-1) * S(n-1) + 1)/n^2