I mentioned in a previous post that I don't fully "get" the solution to the Monty Hall problem.
An friend of mine, John, didn't accept that I couldn't understand this and sent me an email about it.
On thinking about it again I think I do get it now.
My problem was this: intuitively I think that once Monty shows you one of the wrong doors, you imagine that you are faced with two doors (a "stick or switch" choice) so your changes are 50:50. So why should you switch?!
But intuition is doing you a disservice here.
Your changes of getting the right door from the original 3 were 1 in 3. You will only have the right door one time in 3 on average. Your changes if you stick are not 50:50 - they are still 1 in 3!
The chances that one of the other doors is the right one are therefore 2 in 3. But since Monty has shown you a wrong door, your changes if you switch to the remaining door are therefore 2 in 3.
So my intuition that Monty showing you a wrong door has changed the odds of you being right (by either sticking or switching) to 50:50 is the mistake I was making.
If Monty gave you the opportunity to change your selection without opening another door, there would be no reason to do so: the odds would still be 1 in 3.
But since he knows the right door, and is therefore able to open a wrong door, he has changed the dynamic. He has compressed the 2 in 3 odds into a single door!
So your chances of moving OFF the right door (which you had all along) are 1 in 3.
But your changes of moving TO the right door are 2 in 3.
So you should move!
Thanks John: I think I get it now!