Quote:
Originally posted by hot_pastrami
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.
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It still doesn't entirely make logical sense to me. x can be 3, 1000 or 3.4985767E+12, but if (x-2) are eliminated before you make a choice, you're still choosing between two options in the end, one of which is an unknown and the other one of which is (998 losers + one unknown). You make a choice earlier in the game, of course, but that choice doesn't determine whether you win or lose.
If I was applying for a job with 999 others, and only two of us were brought back for a second interview (myself being one of them), I'd be a hell of a lot more confident after the reduction than before...