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Here is a simulation in Clojure showing that the host's knowledge is important. (def doors {0 :goat 1 :goat 2 :prize}) (def choose (rand-nth 3)) (

Here is a simulation in Clojure showing that the host's knowledge is important. (def doors {0 :goat 1 :goat 2 :prize}) (def choose (rand-nth 3)) (defn game [] (let [choice (choose) remaining (dissoc doors choice) monty (rand-nth (vec (keys remaining)))] {:choice (when (= (get doors monty) :goat) (get doors choice)) :monty (get doors monty) :switch (when (= (get doors monty) :goat) (first (vals (dissoc remaining monty))))})) So Monty is choosing a random door that you have not picked. When he picks the prize, the game is over and you lose. (def ten-thousand-games (repeatedly 10000 game)) (frequencies (map :choice ten-thousand-games)) ;; {:goat 3309, :prize 3307, nil 3384} (frequencies (map :switch ten-thousand-games)) ;; {:goat 3307, :prize 3309, nil 3384}

Jtsummers 4y ago
The problem statement doesn't permit him to choose the prize. He is explicitly stated as having revealed the goat. Given that, do you switch? Even if he could have revealed the prize (due to lack of knowledge or care), he does not reveal the prize so why are we considering that circumstance?
- antris 4y ago
  >The problem statement doesn't permit him to choose the prize Doesn't disallow it either. That's the problem: if you describe the problem without all necessary details, it doesn't work like the Monty Hall Problem is supposed to. If you fill in the details yourself (host never showing the prize in this case) then of course the problem becomes complete and then works like Monty Hall Problem again.
  - Jtsummers 4y ago
    The statement is that he revealed the goat, in what world could that be interpreted as him revealing the prize? "But he could have" but he didn't. "But does he always" but he did. What he always does or could do in an alternative scenario are irrelevant, you are given one scenario: He reveals the goat. Don't overthink it, it's not hard, switch and double your odds or stick with it and keep the 1/3rd chance.
    - antris 4y ago
      If the door the host chooses is random, and the door you choose first is random, then it's 50/50, because there's only two doors left and neither door choice was made with any information of what's behind the door. Your chances improve from 1/3 to 1/2 when the host opens the empty door, but whether you stay on the first door or not doesn't matter.
      - Jtsummers 4y ago
        If the host randomly selects the goat or not doesn't change the odds. You should switch. The statement is that he does reveal the goat. Why or how is irrelevant. One of three doors is selected. One of the other two is opened, revealing a goat. Should you switch to the unopened door? Yes, it doesn't matter how that goat was revealed, it could've kicked the door open itself. The odds are better to switch.
        antris 4y ago
        You're the one who's saying the odds change when the door is opened. I'm the one saying that the odds remain the same (or rather, increase to 50/50) if the doors picked are random. If the host knows where the prize is, the odds change when the host picks a door, because we know the hosts strategy. If it's just random, then the odds don't change and changing the door doesn't matter, because each unopened door has the same chance of revealing the prize (1/3 before a door is revealed, 1/2 after). Look up "other host behaviors" on the Wikipedia article, especially the table with "Ignorant Monty", if you don't believe me.
        Jtsummers 4y ago
        > You're the one who's saying the odds change when the door is opened. No, I said the player's odds of winning improve if they switch (to the unopened door). If the host randomly revealed a goat or deliberately revealed the goat is immaterial. A goat was revealed, that's all that matters. The player has 3 doors to choose from. 1 is selected, 2 are not. That means the other two have a combined 2/3rds chance of holding the prize. The selected door has a 1/3rd chance of holding the prize. A goat is revealed behind one of the two unselected doors. The remaining closed, but unselected, door has a 2/3rds chance of being the one with the prize. The selected door's odds of being the winning door remains 1/3rd after the reveal. The player's odds of winning improve if they switch.
- jdminhbg 4y ago
  > The problem statement doesn't permit him to choose the prize. I agree, but this subthread is discussing a misinterpretation of the problem statement, and asking the question about whether switching still matters if the host is also picking randomly. This is showing that it does in fact matter whether the host is picking deliberately or randomly.
  - Jtsummers 4y ago
    OP for the thread: > I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. [emphasis mine] It doesn't matter if it was at random or deliberate. It matters that a non-prize (goat in most versions) is revealed. The question is about one very specific scenario, even in what antris was originally given: The goat is revealed. In that scenario, you switch. The OP misunderstood the problem because they forgot (as many others have in this discussion) that it's about one scenario, not about all times. If Monty selects a door at random and reveals the prize, then it's not this same scenario anymore so the odds of that situation are necessarily different over all scenarios. And if we include all possible scenarios (really, two) then the odds of winning drop, but only because you have a 100% chance of losing when he reveals the prize (or 100% chance of winning and the odds increase). This is the fundamental misunderstanding in many of the comments. If I tell you, "We select, each day, either a prize or a tiger to put behind this door at random. Today it's the tiger. Do you open it?" You'd be a fool to open it. And over all scenarios you'd have a (if chosen uniformly) 50/50 chance of getting the prize or the tiger. But if today I tell you, "It's the prize" should you open it? There's a 100% chance it's the prize, I've told you it's the prize. Now we can get into adversarial situations, whether I'm honest or not. But that's diving into something entirely different and makes it impossible to consider the problem statement because we'll be chasing nonsense after a few minutes. I've just told you, 100% chance it's the prize today. You should open it. The Monty Hall problem is the same. It is one scenario, a goat is revealed. The reason or mechanism of that reveal is irrelevant. He could know it, he could not know it. He could have revealed the goat by dumb luck. Or maybe because it started kicking at the door and was too obvious not to reveal. But today, for this game show he revealed a goat. So your odds are better if you switch. And even antris' original statement presented that scenario, they just answered a different question.
mathgeek 4y ago
Can you elaborate on why you feel this proves that the host's knowledge is important in deciding on whether to switch after a goat is revealed from one of the two doors you haven't chosen? As far as I'm aware (but it's been a while), the host's knowledge is only important up to the point where the first door to be revealed contains a goat. All scenarios where the host reveals a prize on the first door to be opened are discarded for the sake of the problem statement (when dealing with the host being ignorant).
- colinmhayes 4y ago
  So if the host can reveal a prized door your odds are 1/3 prize, 1/3 goat, 1/3 redo. So each door has a 50% chance of having the prize after Monty reveals a goat door. Basically the game is just reduced to having 2 doors instead of your choice vs. all other doors.
  - deleted 4y ago
    [deleted]
- jdminhbg 4y ago
  Sure. This simulation shows the scenario OP described in his misunderstanding of the MHP, which is that he didn't realize the host was picking with foreknowledge of which doors were which. So if the host doesn't know that and is himself simply picking a random door (and 1/3 of the time, revealing the prize), then switching does not improve your odds.
  - ALittleLight 4y ago
    Foreknowledge has absolutely no effect. Again, go back to the problem I posed where Monty reveals a random door and has revealed a goat. Should you switch or not? The correct answer is that you should. Monty's knowledge has no effect on the problem or the player's choices. What has an effect is the information the player learns from seeing the revealed door. If the player sees a goat behind the revealed door - that is the relevant information. Whether Monty was surprised or not by the revealed goat is irrelevant.

Jtsummers 4y ago

The problem statement doesn't permit him to choose the prize. He is explicitly stated as having revealed the goat. Given that, do you switch? Even if he could have revealed the prize (due to lack of knowledge or care), he does not reveal the prize so why are we considering that circumstance?

antris 4y ago
>The problem statement doesn't permit him to choose the prize Doesn't disallow it either. That's the problem: if you describe the problem without all necessary details, it doesn't work like the Monty Hall Problem is supposed to. If you fill in the details yourself (host never showing the prize in this case) then of course the problem becomes complete and then works like Monty Hall Problem again.
- Jtsummers 4y ago
  The statement is that he revealed the goat, in what world could that be interpreted as him revealing the prize? "But he could have" but he didn't. "But does he always" but he did. What he always does or could do in an alternative scenario are irrelevant, you are given one scenario: He reveals the goat. Don't overthink it, it's not hard, switch and double your odds or stick with it and keep the 1/3rd chance.
  - antris 4y ago
    If the door the host chooses is random, and the door you choose first is random, then it's 50/50, because there's only two doors left and neither door choice was made with any information of what's behind the door. Your chances improve from 1/3 to 1/2 when the host opens the empty door, but whether you stay on the first door or not doesn't matter.
    - Jtsummers 4y ago
      If the host randomly selects the goat or not doesn't change the odds. You should switch. The statement is that he does reveal the goat. Why or how is irrelevant. One of three doors is selected. One of the other two is opened, revealing a goat. Should you switch to the unopened door? Yes, it doesn't matter how that goat was revealed, it could've kicked the door open itself. The odds are better to switch.
      - antris 4y ago
        You're the one who's saying the odds change when the door is opened. I'm the one saying that the odds remain the same (or rather, increase to 50/50) if the doors picked are random. If the host knows where the prize is, the odds change when the host picks a door, because we know the hosts strategy. If it's just random, then the odds don't change and changing the door doesn't matter, because each unopened door has the same chance of revealing the prize (1/3 before a door is revealed, 1/2 after). Look up "other host behaviors" on the Wikipedia article, especially the table with "Ignorant Monty", if you don't believe me.
        Jtsummers 4y ago
        > You're the one who's saying the odds change when the door is opened. No, I said the player's odds of winning improve if they switch (to the unopened door). If the host randomly revealed a goat or deliberately revealed the goat is immaterial. A goat was revealed, that's all that matters. The player has 3 doors to choose from. 1 is selected, 2 are not. That means the other two have a combined 2/3rds chance of holding the prize. The selected door has a 1/3rd chance of holding the prize. A goat is revealed behind one of the two unselected doors. The remaining closed, but unselected, door has a 2/3rds chance of being the one with the prize. The selected door's odds of being the winning door remains 1/3rd after the reveal. The player's odds of winning improve if they switch.
jdminhbg 4y ago
> The problem statement doesn't permit him to choose the prize. I agree, but this subthread is discussing a misinterpretation of the problem statement, and asking the question about whether switching still matters if the host is also picking randomly. This is showing that it does in fact matter whether the host is picking deliberately or randomly.
- Jtsummers 4y ago
  OP for the thread: > I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. [emphasis mine] It doesn't matter if it was at random or deliberate. It matters that a non-prize (goat in most versions) is revealed. The question is about one very specific scenario, even in what antris was originally given: The goat is revealed. In that scenario, you switch. The OP misunderstood the problem because they forgot (as many others have in this discussion) that it's about one scenario, not about all times. If Monty selects a door at random and reveals the prize, then it's not this same scenario anymore so the odds of that situation are necessarily different over all scenarios. And if we include all possible scenarios (really, two) then the odds of winning drop, but only because you have a 100% chance of losing when he reveals the prize (or 100% chance of winning and the odds increase). This is the fundamental misunderstanding in many of the comments. If I tell you, "We select, each day, either a prize or a tiger to put behind this door at random. Today it's the tiger. Do you open it?" You'd be a fool to open it. And over all scenarios you'd have a (if chosen uniformly) 50/50 chance of getting the prize or the tiger. But if today I tell you, "It's the prize" should you open it? There's a 100% chance it's the prize, I've told you it's the prize. Now we can get into adversarial situations, whether I'm honest or not. But that's diving into something entirely different and makes it impossible to consider the problem statement because we'll be chasing nonsense after a few minutes. I've just told you, 100% chance it's the prize today. You should open it. The Monty Hall problem is the same. It is one scenario, a goat is revealed. The reason or mechanism of that reveal is irrelevant. He could know it, he could not know it. He could have revealed the goat by dumb luck. Or maybe because it started kicking at the door and was too obvious not to reveal. But today, for this game show he revealed a goat. So your odds are better if you switch. And even antris' original statement presented that scenario, they just answered a different question.

mathgeek 4y ago

Can you elaborate on why you feel this proves that the host's knowledge is important in deciding on whether to switch after a goat is revealed from one of the two doors you haven't chosen? As far as I'm aware (but it's been a while), the host's knowledge is only important up to the point where the first door to be revealed contains a goat. All scenarios where the host reveals a prize on the first door to be opened are discarded for the sake of the problem statement (when dealing with the host being ignorant).

colinmhayes 4y ago
So if the host can reveal a prized door your odds are 1/3 prize, 1/3 goat, 1/3 redo. So each door has a 50% chance of having the prize after Monty reveals a goat door. Basically the game is just reduced to having 2 doors instead of your choice vs. all other doors.
- deleted 4y ago
  [deleted]
jdminhbg 4y ago
Sure. This simulation shows the scenario OP described in his misunderstanding of the MHP, which is that he didn't realize the host was picking with foreknowledge of which doors were which. So if the host doesn't know that and is himself simply picking a random door (and 1/3 of the time, revealing the prize), then switching does not improve your odds.
- ALittleLight 4y ago
  Foreknowledge has absolutely no effect. Again, go back to the problem I posed where Monty reveals a random door and has revealed a goat. Should you switch or not? The correct answer is that you should. Monty's knowledge has no effect on the problem or the player's choices. What has an effect is the information the player learns from seeing the revealed door. If the player sees a goat behind the revealed door - that is the relevant information. Whether Monty was surprised or not by the revealed goat is irrelevant.