>>13008363>Why? The new odds I was right are 1/2. Stop acting like choosing twice gives you more info.You aren't "choosing twice", you're making two different choices (picking a door, switch or stay), or if you think about, one choice (whether or not to switch), since the initial "picking a door" isn't really a choice, in the sense that you can't do any better than chance. This also isn't the gambler's fallacy. The gambler's fallacy would be picking "stay" after "switch" has been correct 9 times in a row because there's 'no way "switch" could win again'.
Let me enumerate all the possible scenarios for three doors (though if you truly think there's an equal chance the door you initially picked and the remaining door in the 100-doors scenario are equally likely, then I don't think there's any helping you.)
All possible layouts are as follows, with equal probability
1 2 3
G G C
G C G
C G G
There's one car, and it can be in any of three positions. You pick one of three doors:
You pick door 1:
(G) G C - You win if you switch
(G) C G - You win if you switch
(C) G G - You lose if you switch
You pick door 2:
G (G) C - You win if you switch
G (C) G - You lose if you switch
C (G) G - You win if you switch
You pick door 3:
G G (C) - You lose if you switch
G C (G) - You win if you switch
C G (G) - You win if you switch
In 6 out of 9, or 2/3, of equally likely cases, switching makes you win. In 3 out of 9, or 1/3, of equally likely cases, staying makes you win. Your pick of a random door in the 100-door version doesn't get any better then "99% goat" just because they open a bunch of extra doors - there's no world in which they would ever open your door, or tell you the you've already won or you picked wrong. Imagine if there was no option to switch, they just opened the doors: obviously this doesn't make your pick better! But there's only one other door, and you know that if you've picked a goat (99% chance), then that door MUST hold a car (thus, 99% chance).
