Probability question: names in hats

[Whateverist]

This is a question someone posed to me many years ago. It may be well known, and the answer might be on the internet. I would ask that if anyone does go look it up, that they please don't spoil it for everyone else (and me) by posting the solution here.

I have as yet been unable to solve this, not that I've been trying constantly! I put a few hours into it here and there, and I felt I was coming close but came up empty. Here is the question:

There are ten people, who each write their name on a piece of paper. These are then all put into a hat.

Each of the ten people, in turn, select a name from the hat using the following rule: (the order of the people is not important)

1) They select a piece of paper at random from those remaining in the hat.
2) If the name is not their own name, they keep the piece of paper.
3) If the name is their own name, they pick again randomly, and then return their name to the hat.

The question is: what is the probability that the tenth person is left with their own name in the hat?

Drawing a tree diagram will drive you insane! It's the "putting back your own name" that really makes this a tough puzzle. Regular probability and combination tricks don't apply as neatly.

Sorry if I'm covering familiar ground but I don't want to read any other responses until I try this myself.

Seems to me that there is some chance that every person who went before might have drawn the last person's name.  So when you come to the ninth person there is very little chance either name left in the hat belongs to the tenth person.  But I see no reason to think the probability is zero.  Perhaps this problem calls for a straight up calculation of probability rather than simply logic.  

If you want to know what the probability is of an 80% free throw shooter making all three free throws after getting fouled on a long attempt you would just multiply 8/10 • 8/10 • 8/10 to get 512/1000, so slightly better than 50/50.

So the probability that the last guy gets his own name would be the product of the probabilities of each preceding person not drawing his name. The probability for the first person not to have wound up with the last person's name would be 8 out of nine possible. (I'm excluding his own name since that isn't a possible keeper.) Continuing in this way we get:

Person 1: 8/9
Person 2: 7/8
Person 3: 6/7
.
.
.
Person 9: 1/2

So the probability of the last guy drawing his own name would be the product of all these probabilities =8/9 • 7/8 • ... • 2/3 • 1/2

Since the numerator of each person's probability is cancelled by the denominator the next person's probability the product reduces to 1/9, which is the answer to the problem.