The monty fucking hall problem.

What you are failing to account for in this particular scenario (i.e., the scenario in which the host doesn’t know where the car is) is that we also may discover a car behind the door that the host opens. This never happens in the original scenario and it is this new information which changes the probabilities from a 2/3 chance of winning by switching to a 1/2 chance of winning by switching.

Think of it this way, Carleas: If the host doesn’t know where the car is, then the host will open the door with the car behind it 33.333% of the time, right? – at which time the game is aborted – nobody wins and nobody loses. We simply start over.

Obviously then, this means that 33.333% of the times that the game is started, there is no opportunity for the contestant to switch to another door because the host opens the door with the car behind it immediately after the contestant selects one of the three doors. Thus, the game is aborted at this point.

So what happens the other 66.667% of the time that the game is started?

This happens:

50% of the time the car is behind the door that the contestant selects AND 50% of the time the car is behind the door that the contestant doesn’t select and that the host does not open.

Those are the only two possibilities.

This should make it plain that if the contestant then switches to that other door (the only door that is both still closed and that he didn’t select originally) he wins 50% of the time and he loses 50% of the time.

I think we might be in agreement, and just considering different sets of outcomes. I am excluding from my set of outcomes cases in which Monty reveals the car when we picks a door. The reason I’m excluding this is because it doesn’t affect the answer to the question, “Should you switch doors,” because when Monty picks the car you don’t have the option to switch doors (the game has already ended).

Basically, my claim is that, if you are the contestant, you’ve picked your door, and Monty revealed a goat, whether or not Monty knew that he was revealling a goat before the actual reveal doesn’t affect whether you should switch. In that limited set of situations, the game functions exactly as it would if Monty had known.
You can see this by considering a situation in which you don’t know whether or not Monty knew he was picking the goat. The probability that switching doors will get you a car in this case is still 66%. There is no difference due to Monty’s knowledge in my Fancy Diagram, and that Diagram fits the situation where Monty knows and always picks a goat, and the situation where Monty doesn’t know and nevertheless has picked a goat.

I’m afraid we don’t agree, Carleas, because the probability is 0.5, not 0.667, that the contestant will win by switching to the other door IF MONTY DOES NOT KNOW WHERE THE CAR IS UNTIL THE CAR IS REVEALED.

Only by assuming both that Monty knows where the car is and that Monty will not reveal the car when he opens a door is the probability of winning by switching 0.667

The reason that the probability of winning by switching goes up to 0.667 in the scenario in which Monty knows where the car is and will not reveal it when he opens one of the doors is because Monty knows where the car is and will not reveal it when he opens one of the doors. :slight_smile:

Think of it like this: The two doors that the contestant doesn’t select have a combined probability of being the lucky door of 0.667 which can be broken down like this: one of the two doors has a 1 x 0.667 probability of being the lucky door and the other has a probability of 0 x 0.667. When Monty opens one of the two doors, he is in effect telling the contestant which of the two doors has the 0 x 0.667 probability (obviously that is the door that he opens) and which has the 1 x 0.667 probability.

No such information is given to the contestant in the scenario in which Monty does NOT know where the car is.

One more attempt before I call it a day . . .

Suppose that there are 1,000 closed doors in front of the contestant instead of just 3. Goats stand behind 999 of those doors. A new car sits behind one of them.

The contestant picks one of the 1,000 doors but before it is opened, the host opens all but one of the other 999 doors and behind every freaking door stands a goat!

What are the odds of that happening???

Well, obviously it depends.

IF the host KNOWS which door the car is behind and does not want to reveal the car by opening that particular door, then the odds are pretty good that the host will open 998 doors in a row to reveal goats behind them. In fact, those odds are much better than just “pretty good.” They are 1 out of 1 (100%). The odds, then, that the goat is behind the remaining unselected door that the host didn’t open are almost as good; they are 999 in 1000.

Thus, in this scenario the contestant will win the car by switching to the other door 999 times out of a 1000.

IF OTOH the host does NOT know which door the car is behind, then the odds that he will consecutively open 998 doors with goats behind them out of 999 possible doors are 1 in 999. Not very good. In fact, pitiful. This means that the odds that the car will just happen to be behind the one door that the host doesn’t open of the 999 doors that he could have opened plus the door that the contestant originally selected are 1 in 1000 – which just so happens to be the same odds that the car is behind the contestant’s orgininally selected door.

Thus, in this scenario the contestant will win the car 1 out of 2 times by switching to the other door, which means that his odds of winning are the same either by switching or by not switching.

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Fucking Monty Hall problem. You’re right.

I’m still not sure what happens if you don’t know whether the host knows or not, only that a goat has been revealed. What then?

Ah smears, its true that I oftentimes use questionable tactics when voicing my support for rationality or probability, but this time around, I don’t think it really was that per say.

Not that I don’t do what you say, I certainly do it and this board is full of examples of that, but in this specific case I wasn’t so much supporting that position with emotional ranting insults so much as expressing my hateful distaste for the ridiculous bullshit that people claim about ‘not believing probability’

My comments were against his comments more than in *support of mine, rather, I wouldn’t just use anger to support my own points, just to decry someone elses clown nonsense.

My replies were not over someone not believing what I claimed, but over claiming some ridiculous horseshit themselves and acting like it wasn’t total see-through bullshit.

I guess its a subtle point and by decrying someone elses claims I’m naturally supporting my own claim about probability and the monty hall problem.

But really its not about the monty hall problem at all, other then to explain how he’s wrong, its about how he goes around making clown statements that everyone knows better about, and then makes some ridiculous comment about other people’s ignorance.

i could give an example where he said i had no way to know that thousands of women were suffering under sharia law, but the mass protests that stopped sharia in iraq by women seem enough to me.

In that same breath, the fucking card deck being wielded by someone who understands the concept should do the trick.

The problem with the monty hall problem is that its a problem in conceptual space, no one is wrong about the switching doors switching probabilities (except those who have expressed a wrong probability assessment i mean)

but in real life, with real human actors, the probabilities would of course be massively skewed one way or another. Monty is likely to switch strategies, lie, use psychological manipulation in some small way to effect the choices of the player.

Theres a lot of unknown human variants, that in real life, may turn the monty hall problem into an even more mind-bending game.

no game show is going to run this trick the same way forever, eventually the audience catches on, it’d make a lot more sense for monty to only somtimes open a door, open a door when the *player picks the right door, etc.

if you actually take into consideration the actions of two *human players, in real life, under real circumstances like an extended gameshow.

a lot of people somehow get tied into this big character analyization of monty-hall when thinking about the monty hall problem, but its gotta come down to a very specifically worded problem.

the monty hall problem voiced in slighty different ways can obviously have different anwsers.

I just thought i’d add that, as irrelevant as it might be.