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> Not only is Marilyn's interpretation grammatically valid I would consider both interpretations equally valid. If anything, in law we have the "rule of the la

> Not only is Marilyn's interpretation grammatically valid I would consider both interpretations equally valid. If anything, in law we have the "rule of the last antecedent" (phrase/clause modifies nearest antecedent noun) so that the hater's interpretation could be more correct if based only on this rule. This language is ambiguous: "host ... opens another door, say #3, which has a goat." The two options are Marilyn's interpretation of the door always containing a goat (invariant version), or that one door he opens (say #3) just happens to contain a goat (hater's version) but it could have contained a car. As suggested by someone below, a much clearer formation could have said "host opens a door with a goat, say door #3" to avoid this particular ambiguity. There's also the issue of usage of "which" signifying a non-restrictive non-essential clause.

hulium 3y ago
If the the door contains a car, then it doesn't matter whether you switch or not. If it is a goat you get the problem as intended. You still have the intended problem as a subproblem, so this is just a distraction and not a serious objection.
- tga_d 3y ago
  If a goat was revealed randomly instead of intentionally, there is no advantage to switching. See the "Monty Fall Problem" in this paper already linked elsewhere in the discussion: https://web.archive.org/web/20230706235720/https://probability.ca/jeff/writing/montyfall.pdf https://web.archive.org/web/20230706235720/https://probabili... To use the 100 door extension some people find helpful: If Monty always reveals the 98 doors that don't contain goats, then 99% of the time (every time you picked a non-car door), the other door will have the car. If Monty is opening a random set of 98 doors, then 98% of the time Monty will reveal the car, 1% of the time you'll have picked the car on the first guess, and 1% of the time the car is in the other door. When you're in those 2% of cases where no car is revealed, you have a 50/50 shot of being in either of the 1% states where that happens.

hulium 3y ago

If the the door contains a car, then it doesn't matter whether you switch or not. If it is a goat you get the problem as intended. You still have the intended problem as a subproblem, so this is just a distraction and not a serious objection.

tga_d 3y ago
If a goat was revealed randomly instead of intentionally, there is no advantage to switching. See the "Monty Fall Problem" in this paper already linked elsewhere in the discussion: https://web.archive.org/web/20230706235720/https://probability.ca/jeff/writing/montyfall.pdf https://web.archive.org/web/20230706235720/https://probabili... To use the 100 door extension some people find helpful: If Monty always reveals the 98 doors that don't contain goats, then 99% of the time (every time you picked a non-car door), the other door will have the car. If Monty is opening a random set of 98 doors, then 98% of the time Monty will reveal the car, 1% of the time you'll have picked the car on the first guess, and 1% of the time the car is in the other door. When you're in those 2% of cases where no car is revealed, you have a 50/50 shot of being in either of the 1% states where that happens.