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I'm quoting the following from https://www.sciencenews.org/article/when-intuition-and-math-probably-look-wrong https://www.sciencenews.org/article/when-intuitio

by cuu508 12y ago

I'm quoting the following from https://www.sciencenews.org/article/when-intuition-and-math-probably-look-wrong https://www.sciencenews.org/article/when-intuition-and-math-... """Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests. On the other hand, suppose that you are looking for a male beagle puppy. You want a puppy that has been raised with a sibling for good socialization but you are afraid it will be hard to select just a single puppy from a large litter. So you find a breeder who has exactly two pups and call to confirm that at least one is male. Then the probability that the other is male is 1/3. In the scenario of Mr. Smith, you’re randomly selecting a child from his two children and then noticing his sex. In the puppy scenario, you’re randomly selecting a two-puppy family with at least one male."""

sago 12y ago
The first example isn't very good. I see what they're trying to do there, to collapse the ambiguity about which child is the boy. But if you work from the set of possible scenarios just purely from those that match the description, without assuming any methods or probability distributions on the way Mr Smith selected the child to go walking with, you get 1/3. I advise students to always work this through from the combinations of possible scenarios. The deeper truth of the vos Savant story is that verbal descriptions are ambiguous, and the even deeper truth is that probability calculations are highly volatile on information. Eventually disagreements come down (among people who can competently do the calculations) to arguments about what such and such a phrase means, or what underlying probability distribution it implies. 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. In science, it is wise to be aware of this. If you have probability calculations that are so volatile based on interpretation, then you probably want to avoid trusting the results, no matter how careful your hermeneutic.
- 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 ago
    I 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

by cuu508 12y ago

sago 12y ago

The first example isn't very good. I see what they're trying to do there, to collapse the ambiguity about which child is the boy. But if you work from the set of possible scenarios just purely from those that match the description, without assuming any methods or probability distributions on the way Mr Smith selected the child to go walking with, you get 1/3. I advise students to always work this through from the combinations of possible scenarios. The deeper truth of the vos Savant story is that verbal descriptions are ambiguous, and the even deeper truth is that probability calculations are highly volatile on information. Eventually disagreements come down (among people who can competently do the calculations) to arguments about what such and such a phrase means, or what underlying probability distribution it implies. 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. In science, it is wise to be aware of this. If you have probability calculations that are so volatile based on interpretation, then you probably want to avoid trusting the results, no matter how careful your hermeneutic.

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 ago
  I 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