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Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be th

Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.

crooked-v 6y ago
> If the host had just picked at random The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.
- fouronnes3 6y ago
  This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.
  - samatman 6y ago
    No. If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge. Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
  - dllthomas 6y ago
    Simulate it. I thought as you think. I wrote a simulation to show I was right. I was wrong. I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
    - Arnavion 6y ago
      Specifically, it's the difference between var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); and var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); (The difference is in the calculation of `montys_choice`.)
    - IggleSniggle 6y ago
      I think you’ve simply misread what they are saying. They are saying that the situation in which Monty opens a door and it reveals the grand prize, are the set of cases that we do not care about because we have already lost, and can thus discard them. This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats. Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch. Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.
      - Arnavion 6y ago
        There is a simulation in the sibling to your comment, ready to be pasted in your browser console.
        IggleSniggle 6y ago
        That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situations are irrelevant because they have no bearing on the fundamental question of whether or not to switch doors when Monty shows you a goat. Effectively, we discard all outcomes when Monty shows you the car, making it an identical “game” as when Monty simply does not ever choose the car. As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.
  - Enginerrrd 6y ago
    They're not the same if the host chooses randomly. Check it out: WLOG assume the car is in door A (The rest will be the same by symmetry.) You pick A, B, or C. The host picks one of the other two doors to show you. The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB. Since we're not in a universe where the host chose a car, we eliminate BA and CA. We are left with four possibilities: AB, AC, BC, CB. AB = You chose the correct door. (probability 1/4) AC = You chose the correct door. (probability 1/4) BC = You chose the wrong door. (probability 1/4) CB = You chose the wrong door. (probability 1/4) Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door. This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.
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kgwgk 6y ago
Imagine there are two contestants and each one selects a door. The host opens the remaining one and it happens to show goat. Do you think they will both increase their odds of winning by switching?
- basch 6y ago
  Can both select the same door? If the answer is no, you've changed the parameters of the game, because the person who chooses second can only pick from two doors. The subset of games where they close different doors is smaller than all possible games.
  - deleted 6y ago
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  - kgwgk 6y ago
    And if they could have selected the same?
    - diggernet 6y ago
      If both contestants choose the same door, and the host shows a goat, then yes, they both would increase their odds by switching.
      - kgwgk 6y ago
        And if they could have chosen the same but didn’t? The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)
        deleted 6y ago
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        the_af 6y ago
        If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other. As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.
mcphage 6y ago
> Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same. If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.
EricMausler 6y ago
To make it more obvious, assume there are 100 doors with 99 goats. And after you select a door, the host opens 98 goats doors

Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.

crooked-v 6y ago

> If the host had just picked at random The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.

fouronnes3 6y ago
This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.
- samatman 6y ago
  No. If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge. Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.
- dllthomas 6y ago
  Simulate it. I thought as you think. I wrote a simulation to show I was right. I was wrong. I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.
  - Arnavion 6y ago
    Specifically, it's the difference between var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); and var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins); (The difference is in the calculation of `montys_choice`.)
  - IggleSniggle 6y ago
    I think you’ve simply misread what they are saying. They are saying that the situation in which Monty opens a door and it reveals the grand prize, are the set of cases that we do not care about because we have already lost, and can thus discard them. This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats. Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch. Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.
    - Arnavion 6y ago
      There is a simulation in the sibling to your comment, ready to be pasted in your browser console.
      - IggleSniggle 6y ago
        That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situations are irrelevant because they have no bearing on the fundamental question of whether or not to switch doors when Monty shows you a goat. Effectively, we discard all outcomes when Monty shows you the car, making it an identical “game” as when Monty simply does not ever choose the car. As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.
- Enginerrrd 6y ago
  They're not the same if the host chooses randomly. Check it out: WLOG assume the car is in door A (The rest will be the same by symmetry.) You pick A, B, or C. The host picks one of the other two doors to show you. The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB. Since we're not in a universe where the host chose a car, we eliminate BA and CA. We are left with four possibilities: AB, AC, BC, CB. AB = You chose the correct door. (probability 1/4) AC = You chose the correct door. (probability 1/4) BC = You chose the wrong door. (probability 1/4) CB = You chose the wrong door. (probability 1/4) Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door. This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.
  - deleted 6y ago
    [deleted]

kgwgk 6y ago

Imagine there are two contestants and each one selects a door. The host opens the remaining one and it happens to show goat. Do you think they will both increase their odds of winning by switching?

basch 6y ago
Can both select the same door? If the answer is no, you've changed the parameters of the game, because the person who chooses second can only pick from two doors. The subset of games where they close different doors is smaller than all possible games.
- deleted 6y ago
  [deleted]
- kgwgk 6y ago
  And if they could have selected the same?
  - diggernet 6y ago
    If both contestants choose the same door, and the host shows a goat, then yes, they both would increase their odds by switching.
    - kgwgk 6y ago
      And if they could have chosen the same but didn’t? The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)
      - deleted 6y ago
        [deleted]
      - the_af 6y ago
        If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other. As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.

mcphage 6y ago

> Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same. If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.

EricMausler 6y ago

To make it more obvious, assume there are 100 doors with 99 goats. And after you select a door, the host opens 98 goats doors