Which simple puzzles have fooled professional mathematicians? Although I'm not a professional mathematician by training, I felt I should have easily been able to answer straight away the following puzzle:

Three men go to a shop to buy a TV and the only one they can afford is £30 so they all chip in £10. Just as they are leaving, the manager comes back and tells  the assisitant that the TV was only £25. The assistant thinks quickly and decides  to make a quick profit, realising that he can give them all £1 back and keep £2.
So the question is this: If he gives them all £1 back which means that they all  paid £9 each and he kept £2, wheres the missing £1?
3 x £9 = £27 + £2 = £29...??

Well, it took me over an hour of thinking before I finally knew what the correct answer to this puzzle was and, I'm embarrassed.
It reminds me of the embarrassement some professional mathematicians must have felt in not being able to give the correct answer to the famous Monty Hall problem answered by Marilyn Vos Savant:
http://www.marilynvossavant.com/articles/gameshow.html

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?
Yes; you should switch.

It's also mentioned in the book: The Man Who Only loved Numbers, that Paul Erdos was not convinced the first time either when presented by his friend with the solution to the Monty Hall problem.
So what other simple puzzles are there which the general public can understand yet can fool professional mathematicians?
 A: I guess this well known von Neumann anecdote fits the description of the question: 
The following problem can be solved either the easy way or the hard way.
Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown?
The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.
When this problem was posed to John von Neumann, he immediately replied, "150 miles."
 "It is very strange," said the poser, "but nearly everyone tries to sum the infinite series."
 "What do you mean, strange?" asked Von Neumann. "That's how I did it!"
A: Along the same lines as the Monty Hall Problem is the following (lifted from Devlin's Angle on MAA and quickly amended):

I have two children, and (at least) one of them is a boy born on a Tuesday. What is the probability that I have two boys?

Read a fuller analysis here.
A: I'm not sure if this qualifies as I don't have any reports of this actually tricking any mathematician, but it's a good problem.  So, at the risk of violating the criteria, the following is Robert Connelly's "Say Red" (taken from Gardner's "Fractal Music, Hypercards and more", Chapter 14):

The banker shuffles a standard deck of 52 cards and slowly deals them face up.  The dealt cards are left in full view where they can be inspected at any time by the player.  Whenever the player wants, he may say "Red."  If the next card is red, he wins the game, otherwise he loses.  He must call red before the deal ends, even if he waits to call on the last card.  What odds should the banker give to make it a fair game, assuming that the player adopts his best strategy on the basis of feedback form the dealt cards?  The player must announce the size of his bet before each game begins.

A: How about the Two envelopes problem?
A: Perhaps not what you were aiming at, but have a look at this. Fabio Massacci provides a counterexample for a lower bound proved by Cook and Reckhow (that's the same Cook from Cook's theorem), and also referred to in several other papers (Section 5 of the paper).
A: I don't know if this qualifies as simple.
A train line from an airport to the Cantor Hotel operates in the following manner. There is a station at each ordinal number, and every station is assigned a unique ordinal. The train stops at each station, in order. At each station people disembark and board, in order, as follows:
i) if any passengers are on the train, exactly $1$ disembarks, and then
ii) $\aleph_0$ passengers board the train.
Station $0$ is at the airport, and the Cantor Hotel is at station $\omega_1$, the first uncountable cardinal. The train starts its journey empty. $\aleph_0$ passengers board the train to the Cantor Hotel at the airport (station $0$), and off it goes. When the train pulls into the Cantor Hotel at station $\omega_1$, how many passengers are on it?
A: Erm... I don't even understand the solution, but there is this one:
An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1 km per second (so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc). Will the ant ever reach the end of the rope?
A: Every few years the Daily Mail newspaper publishes a story of three people in the same family having the same birthday, featuring a different family each time the story is dusted off and recycled.
One one occasion in 2010, they used this version of the story, asking the mathematician Roger Heath-Brown what the odds were of this happening.
Professor Heath-Brown stated that the odds were 48,627,125 to one, ($365^3$ to one). The odds are of course 133,225 to one ($365^2$ to one), not allowing for leap years.
This means that there are usually several births each year that provide the Daily Mail with opportunities for new versions of the story.
A: OK, here's one which had me stumped and when I knew the answer, I was pretty embarrassed. Even more so when I told my mother and she immediately saw the solution. And she's no mathematician.

A music lover lives in a town which has two concerts every weekend. One is in the North of the town, the other in the South. The music lover lives in the centre of town and decides to take the bus to the concert. There's a bus to the North and one to the South every hour. The music lover just decides to take the first bus he sees and go to that concert. At the end of the year, he notices that he only went to the concert in the South. How come?

Here's the answer:

 The clue is that there's a bus every hour, but I didn't specify at what time in the hour the buses leave for the North and South. Imagine that the bus for the South leaves only 5 minutes before the one for the North. Then, it is more likely that if the music lover arrives at a random time, he will take the bus to the South. The odds are 1 to 11 that he takes the one to the North. If they leave even closer to one another, the probability will be even smaller.

