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My crack at it... 1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it. But there's a 2/3 chance it's behind one of th

by bhc3 12y ago

My crack at it... 1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it. But there's a 2/3 chance it's behind one of the other doors. So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose). You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B. So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance Make the switch every time.

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dperny 12y ago
I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.
- BerislavLopac 12y ago
  I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.
- chrismcb 12y ago
  One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.
ntucker 12y ago
Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.
javajosh 12y ago
Dammit you beat me to it! Although maybe I can contribute because I was going to state it thusly... Imagine a related Monty Hall problem, where you select a door, and then Monty immediately asks (without revealing anything), "Do you want to keep that door, or would you like to pick the other two doors?" Clearly you'd pick two doors instead of one. When monty opens a door and gives you the "choice to switch" he is making noise designed to make picking two doors look like picking one door.
- beernutz 12y ago
  Thank you. When you put it that way, it makes so much more sense to me. Opening the door does not change the odds. Why is that so counter intuitive?
  - javajosh 12y ago
    Glad it helped! Why it's counter-intuitive-ness is a really, really good question because like 99.9% of people I found this one hard to accept, too. All I can say is that it's an incredibly effective verbal/logical obfuscation that relies on synergistic choice of number of doors and the (in my view) totally bogus part about Monty opening one of the other doors. In fact, some people (in this very thread!) are still analyzing the problem as if the opened door represents new, 'very valuable' information! It is truly just slight of hand to make it seem like you're only picking one door when you are picking two doors.
    - spronkey 12y ago
      My theory on why it's such a difficult problem to understand is that it's too easy to get caught up in the real world problem, when this isn't actually a real world problem. In the "real world" problem there are additional variables to consider, such as whether or not the host would open a door containing a car, whether that would result in a win or a loss, whether the host would offer a switch in all cases, whether if so, there is any additional incentive for switching (which wouldn't adjust the probability of winning, so is a red herring, but "intuitively" it seems that if the host is trying to make you lose by trying to make you switch, probability should go down). When you try to conflate the real world problem, which actually has multiple different probabilities depending on exact scenario, into an answer to what's generally accepted as the "Monty Hall Problem", some of this gets in the way. In addition, there are two probabilities in the "Monty Hall Problem" - the first is whether your switch results in a win. The second is the probability that, if you switch, it was the correct choice.
taeric 12y ago
Alternatively, there was a 1/3 chance you picked the winning door, which is the only scenario where you lose if you swap.
ecdavis 12y ago
Fantastic explanation. I think this is what the article was trying to get across with the example of the doors but I felt that just confused the matter by talking about "shifting" the probability.
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pbreit 12y ago
What makes the answer most obvious is to play the game with 100 doors.
dllthomas 12y ago
Note that this depends on whether Monty's choice of door was informed by the location of the prize. If he picks randomly and just, this particular time, happens not to have revealed the car then it doesn't matter if you switch or not. Of course, might as well switch anyway in that case just in case you misunderstood...
- ecdavis 12y ago
  I don't think that's correct. The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.
  - vubuntu 12y ago
    I agree with the parent poster that the participant's awareness of whether the host acted randomly or made an informed decision is critical for the participant to decide whether set B's 2/3 probability shifted/concentrated into the one unopened door in set B or whether set B's overall probability got reduced. I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.
    - taeric 12y ago
      See my response in a sibling. The host revealing does matter, in that otherwise you don't get to act on the reveal. If the host just blasts it away as you said, then the probabilities don't change. If the host reveals, then there is a chance you don't even get to the swap before you lose. But if you do get to the swap, you have better odds of winning.

by bhc3 12y ago

deleted 12y ago

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dperny 12y ago

I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.

BerislavLopac 12y ago
I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.
chrismcb 12y ago
One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.

ntucker 12y ago

Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.

javajosh 12y ago

Dammit you beat me to it! Although maybe I can contribute because I was going to state it thusly... Imagine a related Monty Hall problem, where you select a door, and then Monty immediately asks (without revealing anything), "Do you want to keep that door, or would you like to pick the other two doors?" Clearly you'd pick two doors instead of one. When monty opens a door and gives you the "choice to switch" he is making noise designed to make picking two doors look like picking one door.

beernutz 12y ago
Thank you. When you put it that way, it makes so much more sense to me. Opening the door does not change the odds. Why is that so counter intuitive?
- javajosh 12y ago
  Glad it helped! Why it's counter-intuitive-ness is a really, really good question because like 99.9% of people I found this one hard to accept, too. All I can say is that it's an incredibly effective verbal/logical obfuscation that relies on synergistic choice of number of doors and the (in my view) totally bogus part about Monty opening one of the other doors. In fact, some people (in this very thread!) are still analyzing the problem as if the opened door represents new, 'very valuable' information! It is truly just slight of hand to make it seem like you're only picking one door when you are picking two doors.
  - spronkey 12y ago
    My theory on why it's such a difficult problem to understand is that it's too easy to get caught up in the real world problem, when this isn't actually a real world problem. In the "real world" problem there are additional variables to consider, such as whether or not the host would open a door containing a car, whether that would result in a win or a loss, whether the host would offer a switch in all cases, whether if so, there is any additional incentive for switching (which wouldn't adjust the probability of winning, so is a red herring, but "intuitively" it seems that if the host is trying to make you lose by trying to make you switch, probability should go down). When you try to conflate the real world problem, which actually has multiple different probabilities depending on exact scenario, into an answer to what's generally accepted as the "Monty Hall Problem", some of this gets in the way. In addition, there are two probabilities in the "Monty Hall Problem" - the first is whether your switch results in a win. The second is the probability that, if you switch, it was the correct choice.

taeric 12y ago

Alternatively, there was a 1/3 chance you picked the winning door, which is the only scenario where you lose if you swap.

ecdavis 12y ago

Fantastic explanation. I think this is what the article was trying to get across with the example of the doors but I felt that just confused the matter by talking about "shifting" the probability.

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pbreit 12y ago

What makes the answer most obvious is to play the game with 100 doors.

dllthomas 12y ago

Note that this depends on whether Monty's choice of door was informed by the location of the prize. If he picks randomly and just, this particular time, happens not to have revealed the car then it doesn't matter if you switch or not. Of course, might as well switch anyway in that case just in case you misunderstood...

ecdavis 12y ago
I don't think that's correct. The use of sets really clarifies things. Set B has a 2/3 chance of containing the car-hiding door. A fair coin or an RNG chooses which door in Set B to open. If the car is revealed, the game is over and you don't have an opportunity to switch. If a goat is revealed, Set B still has its 2/3 chance of containing the car-hiding door so you should switch to the remaining door in Set B.
- vubuntu 12y ago
  I agree with the parent poster that the participant's awareness of whether the host acted randomly or made an informed decision is critical for the participant to decide whether set B's 2/3 probability shifted/concentrated into the one unopened door in set B or whether set B's overall probability got reduced. I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.
  - taeric 12y ago
    See my response in a sibling. The host revealing does matter, in that otherwise you don't get to act on the reveal. If the host just blasts it away as you said, then the probabilities don't change. If the host reveals, then there is a chance you don't even get to the swap before you lose. But if you do get to the swap, you have better odds of winning.