Some of you will have heard of this.

On the game show "Let's Make a Deal," host Monty Hall used to present three giant doors. Behind one was a prize like a car. Behind the other two were booby prizes like a goat.

The contestant was asked to pick a door. At that point, of course, the contestant's chance of picking the door that hid the car was 1 in 3. (Or 2 to 1 against, if you prefer.)

Whichever door was selected, Monty (who knew where the car was) would open one of the remaining two doors to reveal a goat. He'd ask then the contestant whether they wanted to stick with their original choice or switch to the single unchosen and unopened door, which must contain either a car or a goat.

Here's the problem: mathematical geniuses say that switching is a good idea because it raises the odds of winning the car from 1 in 3 to 2 in 3.

Others, including myself, cannot follow this reasoning. Wikipedia tries to explain (unsuccessfully in my view) at great length:

http://en.wikipedia.org/wiki/Monty_Hall_problem

Among the skeptics, by the way, are sophisticated engineers and logicians, who evidently are not sophisticated enough. The mathematicians claim, with equations, that switching changes the odds.

Obviously (?) eliminating one door increases the odds that your first choice was correct from 1 in 3 to 1 in 2. But how then does switching your choice boost the odds of winning to 2 in 3? You still have no idea where the car is!

I'm prepared to believe the math geniuses because they offer incomprehensible equations and I'm not very smart. They also say the answer is obvious and lament the gullibility of the non-mathematician public. But some of the dissenters also are smarter than I am.

Can anybody explain CLEARLY and with a minimum of math how seeing the contents of one of the three doors and then switching your choice (from one unknown quantity to another) increases the odds that you'll win the car?

Or do I have to sign up for an advanced course in probability theory?