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>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

by 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.
        antris 4y ago
        From the article: The original problem: >The host acts as noted in the specific version of the problem. >Switching wins the car two-thirds of the time. The random door pick: >"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.
        cortesoft 4y ago
        I feel like this is just going around in circles, but let me take a stab at explaining the flaw in your reasoning. Let’s try to simplify the problem. Let’s assume there are three numbered doors: 1, 2, and 3. A car is randomly placed behind one door, and goats behind the other two. To make it easy, I will always guess door 1. In the ‘ignorant host’ scenario, we can just have the host always reveal door 2 (the math doesn’t change if he randomly selects which door to open or always opens the same door, so keeping it always door 2 makes it easier to understand). Ok, so we guess door 1, and the host opens door 2. 1/3 of the time, the host reveals a car. Obviously we would switch in this case, since we know exactly where the car is. 2/3 of the time, the host will reveal a goat. When this happens, 1/2 the time the car will be in door #1 (our door) and 1/2 the time the car will be behind door #3. If the car is not revealed behind door #2, we have a 50% chance whether we switch or not. With this random reveal style, we end up winning 1/3 (the chance of revealing the car behind #2) + (2/3 * 1/2) = 2/3rds chance of winning… which makes sense, because we get our original 1/3 chance plus the free 1/3 chance it is behind the revealed door. Crucially, though, it doesn’t matter if we switch or not. You always win if it is behind door #2 (because you are shown it) and win half the time if it is behind #1 or #3. Contrast this with the case of the host always revealing a goat. In that case, you again have a 1/3 chance of it being behind your door #1 choice. However, in this case, you can essentially guess that your choice was wrong (2/3rds chance) by switching your guess, which will make your guess correct if it is either in door #2 or #3 (since the host will reveal the one which it isn’t behind if it is one of 2 or 3.) In both the ignorant host and non-ignorant host situations, you end up with a 2/3rds chance of getting the car. However, in the ignorant host situation, you don’t need to switch unless the host shows the car in his reveal to get the 2/3rds odds. Of course, it never hurts to switch in either case, so always switching is not a bad strategy, just not necessary if the host knows nothing.
- em500 4y ago
  This has already been explained elsewhere in this thread, but here it goes. The original problem formulation in Parade magazine is as follows: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? Assumptions about the strategy or decision rule of the host are not irrelevant, but crucial. Suppose the hosts' decision rule (unknown to the candidate) is IF (candidate's initial choice is correct): open one of the remaining goat doors ELSE: don't open anything then it's of course not advantagous to switch. The problem with the original Parade formulation is that it only describes what the candidate sees and what the host actually did in one particular scenario, not what rules the hosts must obay. So you don't know if the hosts is playing adversarial games, but you can't rule it out either. That's why (as per linked article) in later formulations the "standard assumptions" (that the hosts will always open a door with a goat) are usually given explicitly (in which case the solution is also a lot more obvious).
  - Jtsummers 4y ago
    It's entirely possible the host is a cheat and the prize is always moved to whichever door you didn't select so you only can ever get a goat. So don't play the game. We can come up with a million what-ifs, but none have a basis in the problem statement so there's no point in them.
    - em500 4y ago
      The original Parade problem is fairly ambiguous and incomplete. Therefor there is no "correct solution". Other formulations typically include the conditions that the host will always open a door, and that door will always contain a goat as explicit assumptions. In that case the math and logic are pretty straightforward. But in the setting of an actual repeated game show, I think the more realistic case to keep the show interesting is that the host will mix up his strategy, i.e. sometimes he opens another door, sometimes he doesn't. If it's a long running daily or weekly show, if he always opens a door with a goat, the public will work out pretty quickly that it's advantageous to always switch.

by antris 4y ago

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.
      - antris 4y ago
        From the article: The original problem: >The host acts as noted in the specific version of the problem. >Switching wins the car two-thirds of the time. The random door pick: >"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.
      - cortesoft 4y ago
        I feel like this is just going around in circles, but let me take a stab at explaining the flaw in your reasoning. Let’s try to simplify the problem. Let’s assume there are three numbered doors: 1, 2, and 3. A car is randomly placed behind one door, and goats behind the other two. To make it easy, I will always guess door 1. In the ‘ignorant host’ scenario, we can just have the host always reveal door 2 (the math doesn’t change if he randomly selects which door to open or always opens the same door, so keeping it always door 2 makes it easier to understand). Ok, so we guess door 1, and the host opens door 2. 1/3 of the time, the host reveals a car. Obviously we would switch in this case, since we know exactly where the car is. 2/3 of the time, the host will reveal a goat. When this happens, 1/2 the time the car will be in door #1 (our door) and 1/2 the time the car will be behind door #3. If the car is not revealed behind door #2, we have a 50% chance whether we switch or not. With this random reveal style, we end up winning 1/3 (the chance of revealing the car behind #2) + (2/3 * 1/2) = 2/3rds chance of winning… which makes sense, because we get our original 1/3 chance plus the free 1/3 chance it is behind the revealed door. Crucially, though, it doesn’t matter if we switch or not. You always win if it is behind door #2 (because you are shown it) and win half the time if it is behind #1 or #3. Contrast this with the case of the host always revealing a goat. In that case, you again have a 1/3 chance of it being behind your door #1 choice. However, in this case, you can essentially guess that your choice was wrong (2/3rds chance) by switching your guess, which will make your guess correct if it is either in door #2 or #3 (since the host will reveal the one which it isn’t behind if it is one of 2 or 3.) In both the ignorant host and non-ignorant host situations, you end up with a 2/3rds chance of getting the car. However, in the ignorant host situation, you don’t need to switch unless the host shows the car in his reveal to get the 2/3rds odds. Of course, it never hurts to switch in either case, so always switching is not a bad strategy, just not necessary if the host knows nothing.
em500 4y ago
This has already been explained elsewhere in this thread, but here it goes. The original problem formulation in Parade magazine is as follows: Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? Assumptions about the strategy or decision rule of the host are not irrelevant, but crucial. Suppose the hosts' decision rule (unknown to the candidate) is IF (candidate's initial choice is correct): open one of the remaining goat doors ELSE: don't open anything then it's of course not advantagous to switch. The problem with the original Parade formulation is that it only describes what the candidate sees and what the host actually did in one particular scenario, not what rules the hosts must obay. So you don't know if the hosts is playing adversarial games, but you can't rule it out either. That's why (as per linked article) in later formulations the "standard assumptions" (that the hosts will always open a door with a goat) are usually given explicitly (in which case the solution is also a lot more obvious).
- Jtsummers 4y ago
  It's entirely possible the host is a cheat and the prize is always moved to whichever door you didn't select so you only can ever get a goat. So don't play the game. We can come up with a million what-ifs, but none have a basis in the problem statement so there's no point in them.
  - em500 4y ago
    The original Parade problem is fairly ambiguous and incomplete. Therefor there is no "correct solution". Other formulations typically include the conditions that the host will always open a door, and that door will always contain a goat as explicit assumptions. In that case the math and logic are pretty straightforward. But in the setting of an actual repeated game show, I think the more realistic case to keep the show interesting is that the host will mix up his strategy, i.e. sometimes he opens another door, sometimes he doesn't. If it's a long running daily or weekly show, if he always opens a door with a goat, the public will work out pretty quickly that it's advantageous to always switch.