Dear Esteemed Community,
I am a reasonably bright person (though not a particularly good chess player). I have fancied puzzles, trivia, and useless knowledge since as far back as I can remember. I did well on the math part of the SATs and the math and logic parts of the GREs. Though I work as a liturgical musician, I keep sharp with the very sorts of puzzles presented in this forum.
AND I DON'T GET THE MONTY HALL PROBLEM. I've used the internet. I've worked it out. I've looked at it and stared and thought and contemplated. And it still seems that the odds are 1/2 once Monty shows the goat.
Usually, when I don't get a puzzle (and I don't get a lot of the tough ones), if I look at the answer, I am able to work it out, learn a new way of thinking, and move on and apply it.
Since I don't get the &^%$*^& Monty Hall problem, I certainly don't get this one with prisoners A, B and C. Can someone work this out for me?
I've been here: http://mathworld.wolfram.com/MontyHallProblem.html
and read (and reread, and rereread, etc. etc.) the explanation. And I get caught up here:
But after Monty has eliminated one of the doors for you, you obviously do not improve your chances of winning to better than 1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3 chance you will win the car (counterintuitive though it seems).
INDEED IT DOES SEEM COUNTERINTUITIVE. If he elimates the door, and asks if I want to switch, it is essentially the same as saying, "Ok, pick the door you REALLY want (the one you said first or the other one)." Obviously, two doors, one car, 1/2. Obviously???? Help!!
My problems are analogous with the A, B, and C prisoner problem originally posed. Please help me. I am agonizing over this and I am sure it is easier than this web page is making it seem.
I thank you all and am,