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Solution: Host doesn't know what is behind the doors either. Contestant then has a true 1/3 chance. QED I guess that defeats the game though.

by Roridge 17y ago

mooism2 17y ago
...is the wrong answer. If the host happened to pick the door with the car, you wouldn't get the option of switching. And the host does not reveal what is behind the door you chose.
- Roridge 17y ago
  I know... that's why I said it.
sprachspiel 17y ago
This is actually an interesting variation of the problem. If a random door opens but by chance happens to not have the prize, then the result is different.
- cynicalkane 17y ago
  But the GP is wrong. If a random door opens, then the odds are 1/2, just like the intuitionistic answer predicts. There's a 1/3 chance you picked right, a 1/3 chance you picked wrong and a wrong door was revealed, and a 1/3 chance you picked wrong and the right door was revealed. Throw the last third out of your sample space and you get that the odds are 1/2. This is what that forced Marilyn to admit the odds can be 1/2 on some readings of the problem, which was ambiguously phrased the first time it was printed. (But I think most people read the problem for the 1/3 solution and still came up with 1/2.)
  - Splines 17y ago
    That doesn't make any sense. The intentions of the host prior to opening the second door shouldn't have any impact on the odds of picking a winning door. I could be wrong, but I think of it this way: You have a choice between picking one door, or picking two doors. If you choose to pick two doors, the host will happen to open one of them to reveal the non-prize. Irrespective of the differences between the wording of my problem and the MH problem, how are the actions different between the two? Here's the flaw in your logic: There's a 1/3 chance you picked right, and a 2/3 chance you picked wrong. In the 2/3 chance you picked wrong, there is a 1/2 chance of the correct door being revealed. No matter if the correct door is shown or not (in this case, it isn't shown), it doesn't change your original odds.
    - cynicalkane 17y ago
      The intentions of the host pass you information that you can use to choose the correct door. The information is this: "There's a 2/3 chance I was forced to reveal this door; ergo, there's a 2/3 chance the other door is the correct one." If the host is choosing randomly, no information is passed. With regards to your last paragraph, the odds of the host picking wrong are either 1/2 or 0 depending on the intentions of the host. I'm a tiny bit annoyed that you think you're pointing out a "flaw" in my logic when you missed something so simple, but whatever. If you don't like my explanation, direct your attention to the OP's link to Wikipedia, where there are many more variations and explanations that you might like.
- eru 17y ago
  Search for "forgetful Monty Hall" (or so) if you want to know more about this variation.

Solution: Host doesn't know what is behind the doors either. Contestant then has a true 1/3 chance. QED I guess that defeats the game though.

by Roridge 17y ago

Solution: Host doesn't know what is behind the doors either. Contestant then has a true 1/3 chance. QED I guess that defeats the game though.

mooism2 17y ago

...is the wrong answer. If the host happened to pick the door with the car, you wouldn't get the option of switching. And the host does not reveal what is behind the door you chose.

Roridge 17y ago
I know... that's why I said it.

sprachspiel 17y ago

This is actually an interesting variation of the problem. If a random door opens but by chance happens to not have the prize, then the result is different.

cynicalkane 17y ago
But the GP is wrong. If a random door opens, then the odds are 1/2, just like the intuitionistic answer predicts. There's a 1/3 chance you picked right, a 1/3 chance you picked wrong and a wrong door was revealed, and a 1/3 chance you picked wrong and the right door was revealed. Throw the last third out of your sample space and you get that the odds are 1/2. This is what that forced Marilyn to admit the odds can be 1/2 on some readings of the problem, which was ambiguously phrased the first time it was printed. (But I think most people read the problem for the 1/3 solution and still came up with 1/2.)
- Splines 17y ago
  That doesn't make any sense. The intentions of the host prior to opening the second door shouldn't have any impact on the odds of picking a winning door. I could be wrong, but I think of it this way: You have a choice between picking one door, or picking two doors. If you choose to pick two doors, the host will happen to open one of them to reveal the non-prize. Irrespective of the differences between the wording of my problem and the MH problem, how are the actions different between the two? Here's the flaw in your logic: There's a 1/3 chance you picked right, and a 2/3 chance you picked wrong. In the 2/3 chance you picked wrong, there is a 1/2 chance of the correct door being revealed. No matter if the correct door is shown or not (in this case, it isn't shown), it doesn't change your original odds.
  - cynicalkane 17y ago
    The intentions of the host pass you information that you can use to choose the correct door. The information is this: "There's a 2/3 chance I was forced to reveal this door; ergo, there's a 2/3 chance the other door is the correct one." If the host is choosing randomly, no information is passed. With regards to your last paragraph, the odds of the host picking wrong are either 1/2 or 0 depending on the intentions of the host. I'm a tiny bit annoyed that you think you're pointing out a "flaw" in my logic when you missed something so simple, but whatever. If you don't like my explanation, direct your attention to the OP's link to Wikipedia, where there are many more variations and explanations that you might like.
eru 17y ago
Search for "forgetful Monty Hall" (or so) if you want to know more about this variation.