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I don't think the host's knowledge is important. Consider the alternative where: 1. You pick a door. 2. The host reveals a different door at random. The hos

I don't think the host's knowledge is important. Consider the alternative where: 1. You pick a door. 2. The host reveals a different door at random. The host has no knowledge. 3. The door the host revealed is a goat. Should you switch or stay?

jjbinx007 4y ago
The puzzle doesn't work without the host always revealing a booby prize. In the example you gave it's just random chance. In the Monty Hall puzzle it works out that statistically you should swap.
- ALittleLight 4y ago
  You should also swap in the example I gave.
  - antris 4y ago
    Nope. That version is listed here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... "Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car. Switching wins the car half of the time. Swapping doesn't affect the outcome, it's a 50/50 whether you swap or not.
abirch 4y ago
I would switch. The question is what happens if the door is the prize?
tgb 4y ago
This is what I thought at first but simulating it out, it’s not true. The strategy of switching doors works 2/3rds of the time but only if you allow picking the opened door (when it has the prize). That’s easy to see since it works whenever the prize isn’t behind your first choice. Conditioning on the opened door revealing a goat gives that strategy only 50% chance. After all, the opposite situation happens 1/3rd of the time and has a 100% success rate (since you can see the car), so this other case (opened door has a goat) has to have a 50% chance to add up to the unconditioned success rate. It’s still unintuitive to me though.
- icambron 4y ago
  The choice happens after the door is opened. If in this variant you are allowed switch to the revealed prize door, you win 100% of the time the prize turns out to be behind the door they open. If you aren't allowed, you lose 100% of the time that happens. So that new "branch" doesn't affect any decision you could make. But given they do open a door with a goat, the probability is the same as the regular Monty Hall problem. So the parent's point stands: it's not the host's knowledge that makes the game "work". He's leaking information by not opening your door, thus "concentrating" the 2/3 chance it wasn't behind your door into the one remaining door.
  - tgb 4y ago
    It’s not the same. Try stimulating it out if my argument above didn’t convince you. WLOG if you pick door 0 out of 0,1,2 and host picks door 1 always, then the the relevant value is in numpy: correct = numpy.random.choice(3, 10000) (correct[correct != 1] == 0).mean() This gives 0.49 or similar when I run it.
    - Jtsummers 4y ago
      Then you aren't simulating the problem statement.
    - icambron 4y ago
      Hmm, you are right. If the host chooses randomly, we get this tree: - 1/3 it is behind your door - 1 host picks a goat - switching always loses -> 1/3 chance of losing - 2/3 it is behind one of the other two doors - 1/2 host picks the goat - switching wins always -> 2/3 * 1/2 = 1/3 chance of winning - 1/2 host picks the prize - result undefined -> 1/3 chance of ending up here So you get 1/3 wins, 1/3 losses, 1/3 undefined. So given the goat is revealed, you have a 1/2 chance to win by switching. On reflection, this makes sense. The host is indeed leaking information by not be able to pick the around the prize. If the prize is not behind your door, his choice is constrained. So his revealing it supplies information about that constraint. If he is not constrained, then you are both picking doors at random. I'm sold.
tomonl 4y ago
This reminds me of the two envelopes problem https://en.wikipedia.org/wiki/Two_envelopes_problem https://en.wikipedia.org/wiki/Two_envelopes_problem
GistNoesis 4y ago
If the host has knowledge what matter is whether the host has the freedom to not open a second door (which they often have in TV shows where time is of the essence), and deprives you of a second round. If the host has this freedom, and he is adversarial and assumes you know the monty-hall paradox, then the only time where he will offer you to show a second door is if you have guessed right to make you change.
banannaise 4y ago
In this version it doesn't matter. #3 eliminates the 1/3 chance that the host reveals the prize by accident, so you're left with the 1/3 chance that you chose the prize, or the 1/3 chance that you didn't choose the prize and the host also didn't choose the prize.
cupofpython 4y ago
if it's opened randomly then it doesn't matter, it is 50-50. You have to account for the situation where the host might reveal the prize in his random pick, and by not doing so you learn that your choice is more likely to be the prize than initially thought. If you picked the prize P(A = 1/3), then the random open is going to be Goat P(1). if you didnt pick the prize P(B = 2/3) then the random open is going to be Goat P(1/2) or Prize P(1/2) P(Goat reveal) happens (1/3)(1) + (2/3)(1/2) = 2/3 times P(Prize reveal) happens (1/3)(0) + (2/3)(1/2) = 1/3 times P(you picked prize, given goat reveal) = (1/3)(1) / (2/3) = (1/2)
- Jtsummers 4y ago
  You don't have to account for the chance of him revealing the prize because he does not reveal the prize, he reveals the goat. So switch because it doubles your odds. Or don't, you're a free person. But it would be foolish.
  - colinmhayes 4y ago
    If he's guaranteed to reveal the goat then he knows which door has the prize. This situation he isn't guaranteed to reveal a goat, he just did it this time by chance.
    - Jtsummers 4y ago
      It doesn't matter if he's guaranteed to reveal the goat or not. He did reveal the goat, that's the scenario that's been described. So given that he did reveal the goat, should you switch? The answer is yes, from the information given, you should switch. Everything else is overthinking.
      - colinmhayes 4y ago
        Whether he reveals the goat by luck or because he knew which door had the prize is the entire point of the problem.
jdminhbg 4y ago
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.
afiori 4y ago
This is indistinguishable from the host having secret knowledge; without knowledge the host would open the door with the prize a third of the times.
- ALittleLight 4y ago
  Yes, that's the point, that it is indistinguishable. The host's knowledge doesn't matter. The host might know or not know. What matters is that the host has revealed a goat.
  - afiori 4y ago
    But the only way to always reveal a goat is to have knowledge of the doors.

jjbinx007 4y ago

The puzzle doesn't work without the host always revealing a booby prize. In the example you gave it's just random chance. In the Monty Hall puzzle it works out that statistically you should swap.

ALittleLight 4y ago
You should also swap in the example I gave.
- antris 4y ago
  Nope. That version is listed here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... "Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car. Switching wins the car half of the time. Swapping doesn't affect the outcome, it's a 50/50 whether you swap or not.

abirch 4y ago

I would switch. The question is what happens if the door is the prize?

tgb 4y ago

This is what I thought at first but simulating it out, it’s not true. The strategy of switching doors works 2/3rds of the time but only if you allow picking the opened door (when it has the prize). That’s easy to see since it works whenever the prize isn’t behind your first choice. Conditioning on the opened door revealing a goat gives that strategy only 50% chance. After all, the opposite situation happens 1/3rd of the time and has a 100% success rate (since you can see the car), so this other case (opened door has a goat) has to have a 50% chance to add up to the unconditioned success rate. It’s still unintuitive to me though.

icambron 4y ago
The choice happens after the door is opened. If in this variant you are allowed switch to the revealed prize door, you win 100% of the time the prize turns out to be behind the door they open. If you aren't allowed, you lose 100% of the time that happens. So that new "branch" doesn't affect any decision you could make. But given they do open a door with a goat, the probability is the same as the regular Monty Hall problem. So the parent's point stands: it's not the host's knowledge that makes the game "work". He's leaking information by not opening your door, thus "concentrating" the 2/3 chance it wasn't behind your door into the one remaining door.
- tgb 4y ago
  It’s not the same. Try stimulating it out if my argument above didn’t convince you. WLOG if you pick door 0 out of 0,1,2 and host picks door 1 always, then the the relevant value is in numpy: correct = numpy.random.choice(3, 10000) (correct[correct != 1] == 0).mean() This gives 0.49 or similar when I run it.
  - Jtsummers 4y ago
    Then you aren't simulating the problem statement.
  - icambron 4y ago
    Hmm, you are right. If the host chooses randomly, we get this tree: - 1/3 it is behind your door - 1 host picks a goat - switching always loses -> 1/3 chance of losing - 2/3 it is behind one of the other two doors - 1/2 host picks the goat - switching wins always -> 2/3 * 1/2 = 1/3 chance of winning - 1/2 host picks the prize - result undefined -> 1/3 chance of ending up here So you get 1/3 wins, 1/3 losses, 1/3 undefined. So given the goat is revealed, you have a 1/2 chance to win by switching. On reflection, this makes sense. The host is indeed leaking information by not be able to pick the around the prize. If the prize is not behind your door, his choice is constrained. So his revealing it supplies information about that constraint. If he is not constrained, then you are both picking doors at random. I'm sold.

tomonl 4y ago

This reminds me of the two envelopes problem https://en.wikipedia.org/wiki/Two_envelopes_problem https://en.wikipedia.org/wiki/Two_envelopes_problem

GistNoesis 4y ago

If the host has knowledge what matter is whether the host has the freedom to not open a second door (which they often have in TV shows where time is of the essence), and deprives you of a second round. If the host has this freedom, and he is adversarial and assumes you know the monty-hall paradox, then the only time where he will offer you to show a second door is if you have guessed right to make you change.

banannaise 4y ago

In this version it doesn't matter. #3 eliminates the 1/3 chance that the host reveals the prize by accident, so you're left with the 1/3 chance that you chose the prize, or the 1/3 chance that you didn't choose the prize and the host also didn't choose the prize.

cupofpython 4y ago

if it's opened randomly then it doesn't matter, it is 50-50. You have to account for the situation where the host might reveal the prize in his random pick, and by not doing so you learn that your choice is more likely to be the prize than initially thought. If you picked the prize P(A = 1/3), then the random open is going to be Goat P(1). if you didnt pick the prize P(B = 2/3) then the random open is going to be Goat P(1/2) or Prize P(1/2) P(Goat reveal) happens (1/3)(1) + (2/3)(1/2) = 2/3 times P(Prize reveal) happens (1/3)(0) + (2/3)(1/2) = 1/3 times P(you picked prize, given goat reveal) = (1/3)(1) / (2/3) = (1/2)

Jtsummers 4y ago
You don't have to account for the chance of him revealing the prize because he does not reveal the prize, he reveals the goat. So switch because it doubles your odds. Or don't, you're a free person. But it would be foolish.
- colinmhayes 4y ago
  If he's guaranteed to reveal the goat then he knows which door has the prize. This situation he isn't guaranteed to reveal a goat, he just did it this time by chance.
  - Jtsummers 4y ago
    It doesn't matter if he's guaranteed to reveal the goat or not. He did reveal the goat, that's the scenario that's been described. So given that he did reveal the goat, should you switch? The answer is yes, from the information given, you should switch. Everything else is overthinking.
    - colinmhayes 4y ago
      Whether he reveals the goat by luck or because he knew which door had the prize is the entire point of the problem.

jdminhbg 4y ago

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.

afiori 4y ago

This is indistinguishable from the host having secret knowledge; without knowledge the host would open the door with the prize a third of the times.

ALittleLight 4y ago
Yes, that's the point, that it is indistinguishable. The host's knowledge doesn't matter. The host might know or not know. What matters is that the host has revealed a goat.
- afiori 4y ago
  But the only way to always reveal a goat is to have knowledge of the doors.