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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 wa

by 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.

by antris 4y ago

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.