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See my cousin comment[0] about this, I think you're getting tripped up with the idea of the host's "knowledge" mattering. That's a distraction here, what's impo

by Cushman 12y ago

See my cousin comment[0] about this, I think you're getting tripped up with the idea of the host's "knowledge" mattering. That's a distraction here, what's important is the probabilities. Two game shows, two hosts, both reveal a goat. The random host revealed a goat with 1/3 probability; the canonical host revealed a goat with 1 probability. We know that both hosts revealed a goat. We also know (for the purposes of this problem) that they had different likelihoods of doing so before the fact. Does it make intuitive sense to you that the prior probability of an event could affect how you react to it? That's all that's going on here. [0] https://news.ycombinator.com/edit?id=8664771 https://news.ycombinator.com/edit?id=8664771

skizm 12y ago
The premises of the riddle in both cases is that the host reveals a goat. There is no randomness. The fact that there was a chance the host might show a car doesn't matter. He showed a goat. The probability of the host showing a goat in both cases is 1. I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.
- Cushman 12y ago
  > The probability of the host showing a goat in both cases is 1. Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat. Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?
  - deleted 12y ago
    [deleted]
- dllthomas 12y ago
  There are two separate things here. First, the premise of the riddle. Second, the procedure of the game show (as fictionalized within the riddle). If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.
  - skizm 12y ago
    The question is phrased ad-hoc. We know the host picked a goat in both scenarios. Decision time. Do we switch or no? It didn't ask: "On an infinite timeline where the host might not always pick the goat, should we switch and if so when?" Which I think is what op was trying to say.
    - Cushman 12y ago
      No. Even in the case of a single trial, it matters. If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point. If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch. Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.
      - dllthomas 12y ago
        "If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point." 1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.
        Cushman 12y ago

by Cushman 12y ago

skizm 12y ago

The premises of the riddle in both cases is that the host reveals a goat. There is no randomness. The fact that there was a chance the host might show a car doesn't matter. He showed a goat. The probability of the host showing a goat in both cases is 1. I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.

Cushman 12y ago
> The probability of the host showing a goat in both cases is 1. Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat. Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?
- deleted 12y ago
  [deleted]
dllthomas 12y ago
There are two separate things here. First, the premise of the riddle. Second, the procedure of the game show (as fictionalized within the riddle). If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.
- skizm 12y ago
  The question is phrased ad-hoc. We know the host picked a goat in both scenarios. Decision time. Do we switch or no? It didn't ask: "On an infinite timeline where the host might not always pick the goat, should we switch and if so when?" Which I think is what op was trying to say.
  - Cushman 12y ago
    No. Even in the case of a single trial, it matters. If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point. If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch. Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.
    - dllthomas 12y ago
      "If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point." 1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.
      - Cushman 12y ago