Quote:
Originally posted by vsp
But... as always, I don't grasp the causal relationship between the two choices.
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Part of the problem is that the scale is so small that intuition misleads us. Here's what basically happens in the three-doors scenario.... the contentant picks one door, then the host opens
all but one of the remaining doors. That remaining closed door was unopened because:
A) it contains the prize (likelihood: x-1/x), or
B) you selected the door that contained the prize, so an empty door was randomly opened from those that were left (likelihood: 1/x).
...where x=3 in the Let's Make a Deal context.
So imagine that there are not three doors, but a thousand (x=1000). The contestant picks a door, and like in the three-door game, Monty opens all of the other doors except one... 998 doors spring open to reveal goats. Now, like in the three door game, there are only two closed doors to choose from. But are the chances 50/50? No, because that fact that there are only two doors now
does not change the fact that your chances of guessing right were one in a thousand before those 998 doors opened. You can still bet that one of the 999 doors you
didn't pick is the winner (likelihood 999/1000, or 99.9%), you just happen to know what is behind 998 of them, so you know not to pick those.
Any better? I know my explanations aren't the greatest.
EDIT: I had the little formulas reversed in my A/B scenarios... oops