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>= 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3. Your sen
by jasode 4y ago
>= 2/3. Just from this knowledge alone, you could decide that you will always switch, since the probability that the other boxes have the keys is 2/3.
Your sentence the particular way you worded it is not the correct mathematical model.
The player does not get to switch to BOTH OF THE OTHER 2 boxes as an alternative to just the 1st box. Therefore the 2/3rd probability doesn't apply.
Where the non-intuitive 2/3rds probability becomes the answer instead of 50/50 is the host's perfect knowledge of always choosing the door without the car.
- kasperni 4y ago> Your sentence the particular way you worded it is not the correct mathematical model. It wasn't really my sentence I just quoted the article. Nonetheless I disagree with you. The probability that the other boxes have the key is 2/3 and that is all that really matters. Opening a door doesn't change anything.
- basch 4y agoYou do get to switch to both other doors. One of the two remaining doors is a goat, and it is opened for you. Another way to phrase that is getting to pick both doors and the goat doesnt count against you. The Monty Hall problem distills to simply "would you like one or two doors, (if the prize is behind any door in the set you pick you win.)"