I'm not saying this is incorrect, but some people try to explain the Monty Hall problem in terms of what the host knows and has to reveal. — Purple Pond
My question is, in this simulation of the Monty Hall show, with this extra random variation where the computer doesn't "know" which door contains the goat or car, but reveals a goat in one of the other doors by chance, is the probability still 2/3 that the other unopened door contains the car? Or does the probability change because of the extra random process? — Purple Pond
And here's the crucial part. If the computer, who randomly opens one of the other two doors you didn't choose, reveals the car, the program terminates right there, and starts again. If on the other hand, the computer opens one of the other doors, and by chance it is a goat, the program continues and you are given a choice to switch or not. — Purple Pond
Again - why are you biasing in favour of one door?
The goat likewise raises the probability of door 1. — Shamshir
Still 2/3. The door you chose retains the initial 1/3 chance and the other door must have a 1 - 1/3 chance, i.e. 2/3. — Michael
One door has zero chance and chances are split to 1/2.
Where's the indication that one door is more likely, assumptions that one chose wrong aside? — Shamshir
But one door has zero probability and the remaining two are equiprobable.
In theory the chances are 50/50 from beginning to end as the revealed dud is nonfactor; it only appears to be 2/3 if you focus on the dud with the assumption that you chose wrong, if you assume you chose right vice versa. — Shamshir
In practice, you might not win one game by switching.
But you practically can, you can win all 99 by not switching. — Shamshir
Without the assumption that you chose wrong, the three doors are equiprobable - and staying or switching you would win 33 games anyway, by the posited logic. — Shamshir
If we simply choose one and reveal them all it's 1/3.
If we remove the dud from the equation, it should be obvious you're limited to two options and it's 1/2.
If we by condition of the dud, assume we chose wrong, then yes, you can skew the chances to 2/3. — Shamshir
I don't think the initial probability of picking a random door is 1/3. Think about it. — Purple Pond
Six permutations, but the combination G+G is eliminated by the exposure of the car, as indicated above by the invisible strikes. One game in three my door wins, one game in three Monty's door wins, and one game in three is abandoned. So it's even odds. — unenlightened
I think I get it too now. If I get the option to switch, that means the game didn't restart. It's less likely I picked the right door initially. But, given that the game did not restart, it becomes more likely that I picked the right door. Not sure exactly how the mathematical operation looks though.
Monty hall problems are apparently difficult even if you know the basics. — Echarmion
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