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hot_pastrami · 06-13-2003, 03:31 PM

Er, ok... let me try another way of explaining. This is fun, because it's challenging to explain it, and bends my brain like a pretzel.

I think we can agree that before the player selects a door, each one has a 1/3 chance of being the winning door. When a player selects a door, they basically divide the doors into two groups... the door they have chosen, and doors they have not chosen. The door they have chosen has a 1/3 chance of being a winner, and the doors they have not chosen, as a group, has a 2/3 chance of containing a winner.

So Monty opens one of the doors in the "doors the player did not choose" group. Now, even though you know what is behind one of the doors in that group, the "doors the player did not choose" group still has a 2/3 probability of containing the winning door, you just know not to choose the one that had it's contents revealed. So you go with the remaining door in that group, and use that 2/3 odds to your advantage.

Intuition wants us to break down the problem and say that after a door is opened, and we have one winning door and one losing door before us, we have a 50/50 chance. But it's better to think of the do-I-switch choice as "Was I right on my initial guess?" Chances are, you weren't. It's a 2/3 chance you were wrong, as a matter of fact.

Any better?

06-13-2003, 03:31 PM	#16
hot_pastrami I am meaty Join Date: Dec 2001 Location: Salt Lake City, UT Posts: 1,119	Er, ok... let me try another way of explaining. This is fun, because it's challenging to explain it, and bends my brain like a pretzel. I think we can agree that before the player selects a door, each one has a 1/3 chance of being the winning door. When a player selects a door, they basically divide the doors into two groups... the door they have chosen, and doors they have not chosen. The door they have chosen has a 1/3 chance of being a winner, and the doors they have not chosen, as a group, has a 2/3 chance of containing a winner. So Monty opens one of the doors in the "doors the player did not choose" group. Now, even though you know what is behind one of the doors in that group, the "doors the player did not choose" group still has a 2/3 probability of containing the winning door, you just know not to choose the one that had it's contents revealed. So you go with the remaining door in that group, and use that 2/3 odds to your advantage. Intuition wants us to break down the problem and say that after a door is opened, and we have one winning door and one losing door before us, we have a 50/50 chance. But it's better to think of the do-I-switch choice as "Was I right on my initial guess?" Chances are, you weren't. It's a 2/3 chance you were wrong, as a matter of fact. Any better? __________________ Hot Pastrami!