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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

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.
      - Arnavion 6y ago
        >it simulates a situation where we count the times when Monty picks the car. No, it does not. >> if (montys_choice === car_door) { i -= 1; continue; } This discards the game where Monty chose a car. >> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); This makes Monty never choose a door with a car. Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.
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
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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.
    - Arnavion 6y ago
      >it simulates a situation where we count the times when Monty picks the car. No, it does not. >> if (montys_choice === car_door) { i -= 1; continue; } This discards the game where Monty chose a car. >> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); This makes Monty never choose a door with a car. Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.

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
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