If the drunk somehow chooses his correct seat, the probability is 100% as everyone else will find their correct seat. However, if the drunk does not, everyone who has a correct seat still will find their seat accept for the one person who's seat has been taken by the drunk, in which case he will take the remaining seat that the drunk left behind. He either gets the correct seat or he doesn't... 50%.
That's somewhat what I said...although, if the second person to enter is the owner of the seat that the drunk took, they will sit in a random seat. If the third person to enter is the owner of the seat that the second person sits in, he too will sit somewhere else. Either way, the last person has a 50% (or 1/2) chance of getting his seat.
What I did originally, was make up a random number of seats (50), and I sat the drunk in seat one (and made his seat, seat #50). And I simply put the second person (who's seat was seat #1), in seat 2. If this pattern continued, the last person would not be able to sit in their seat (seat 49, because they will be left to sit in seat 50, the drunk's seat.) But that's only if the pattern continues, if say the second to last person sits in seat 50 instead of 49, then the last person to enter is free to take their own seat.
So no matter how you look at it (unless the drunk takes his own seat), it's a 50% or 1/2, chance.
of course, you could just assume that there are only three seats (3 people).
The drunk is asisgned to seat #1, the second person, seat #2, and the last #3.
The drunk sits in seat #2: the second person either sits in seat #1 or #3 (which leaves the last person with whatever's left (could be his seat, or not)
The drunk sits in seat #3: the second person sits in his own seat, seat #2, and the last person is forced into the first seat. (in this case, the outcome is 100% that he will sit in the wrong seat.)
So, I guess it's not exactly 50%...hmm. does that mean that the odds are about 1/3 that he will sit in his correct seat?
or...is it a...3/2 chance...? No that doesn't make sense either...agh! This is much more complicated than I thought it would be...
The answer is: it's either 50% or 0%. There isn't one correct answer.
(I'll make diagrams if I must.)
There is only one answer. It's 50%.
You can't say "If the drunk takes the last person's seat, then the chances are 0%, but if he takes his own seat, then it's 100%, therefore there isn't one correct answer." That logic is faulty. That's like saying, "If I win the lottery, then my chances of winning are 100%, but if I lose, then my chances are 0%." You want to look at the overall probability.
Anyways, providing a person will always take their own seat if it's available, there will always be 2^(n-1) seating combinations where n = number of seat/people. Half of those combinations will always end up with the last person in the correct seat.
If there are 2 seats, there are two combinations with one leaving the last person in the correct seat. 50%
If there are 3 seats, there are four combinations with two leaving the last person in the correct seat. 50%
If there are 50 seats, there are 562949953421312 seating combinations with 281474976710656 combinations leaving the last person in the correct seat. 50%
It's a math problem, not a riddle.
Anyways, sorry for turning this riddle (which was meant to be fun) into a math lesson.