Ok so I googled the Monty Hall Problem and found lots of different explanations for why switching doors is the right thing to do, however none of them are based on sound mathematical principle.
For instance
Quote:
|
Let's say that you choose your door (out of 3, of course). Then, without showing what's behind any of the doors, Monty says you can stick with your first choice or you can have both of the two other doors. I think most everyone would then take the two doors collectively.
|
No matter which door you choose after one of the non-winning doors has been revealed, you still have "two doors collectively".
Quote:
|
You'll lose 2/3 of the time if you don't switch. If you do switch, then the prize is still available 2/3 of the time, and you'll make the wrong choice 1/2 the time, so the chance of losing when you switch is only (2/3)*(1/2) = 1/3. You've halved your chance of losing = doubled your chances of winning by switching.
|
This is also an erroneous statement, because "if you switch" the prize isn't available 2/3 of the time, it's only available half of the time because there are only two choices left.
Regardless of how many choices you started with, when given the option to switch it becomes an entirely new situation, one with only two choices and not three.
I'll keep looking for a better explanation, but at this point none of them have me convinced.