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"Hence in the Monte Hall Problem, there is no advantage to switching if Monte selects the unchosen door to open at random, and it happens to be a goat. That sit
by Psyonic 12y ago
"Hence in the Monte Hall Problem, there is no advantage to switching if Monte selects the unchosen door to open at random, and it happens to be a goat. That situation is consistent with the description, but perhaps not the 'feel' of the game show setup."
This isn't quite right. Obviously if there's a chance Monty might open the car, it screws up the gameshow, but it doesn't actually change the odds in the situation that he opened a door with a goat.
The knowledge in Monte's head doesn't change the odds.
To simulate this properly, you'll need to change the problem statement somewhat... you'll either have to throw out cases where he opened the door with the car, or state the problem such that we're only look at cases where he randomly opened a door with the goat. But you'll still get 2/3.
- Psyonic 12y agoI may be wrong about this, actually. Guess that goes to show you that precise terminology matters here. This python gist appears to simulate the situation correctly, and shows that it becomes 50/50. https://gist.github.com/ecdavis/da8f67258860e9f35620 https://gist.github.com/ecdavis/da8f67258860e9f35620 That said, I'm not sure I completely agree with this solution just throwing out cases where he opens the winning door, without replacement. I feel like at this point we're calculating a different problem where we haven't precisely defined what it is we're measuring, so you could tweak things to get either answer.
- PalmerEldritch 12y ago"But you'll still get 2/3.". No you don't, you get a 1/2. If you consider only those cases where Monty reveals a goat (and discard those where he reveals the car) then 50% of the time you've already picked the car