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### Re: The monty fucking hall problem.

Posted: Fri Aug 08, 2008 8:05 am
Uccisore wrote:
someoneisatthedoor wrote:I understand the mathematical proof, I just place no stock in mathematical proofs concerning probability.

Likewise, a lot of people do this intuitively. As such, it isn't that they 'don't get' the answer, it's that they don't believe in it.

Be simple enough to test with 2 people and a deck of cards, wouldn't it? The stats aren't just numbers in space, if the example is right, switchng really should get you the car (or ace) 66% of the time.

You don't even need two people. You can do it yourself.

Just shuffle two deuces and an ace (the deuces represent the "goats" and the ace, the "car"), and deal them face down in front of you. Pick one of the three cards, being sure to keep it face down. Turn over one of the two cards that you didn't pick. Then:

A) if the card revealed is an ace (i.e., "the car"), start over because in the "Monty Hall problem," remember, the host never reveals the car when he opens one of two doors that weren't picked by the contestant. He always opens a door with a goat behind it.

B) if the card revealed is a deuce (i.e., "a goat"), then either keep your original card OR switch to the other face down card.

By switching, you'll switch to the ace close to 67% of the time. By keeping your original card, you'll keep the ace about 33% of the time . . . guaranteed.

If you don't believe it now, do the experiment and you'll believe it then.

### Re: The monty fucking hall problem.

Posted: Fri Aug 08, 2008 8:36 am
Or you can mimic the Monty Hall problem by using an entire deck of cards including the joker. Let the joker represent the car.

Deal all 53 cards face down. Choose one card, keeping it face down. Then turn over 51 of the other 52 cards. If the joker is revealed when any of the 51 cards are turned over, start over again because, remember, the game show host in the problem never reveals the car when he opens the door.

If you ever arrive at a situation where only the card that you originally picked and one other card remain unrevealed, then either keep the card that you originally picked or switch to the other card.

By switching to the other card, you'll switch to the joker 52 out of 53 times. By keeping the card you originally selected, you will have the joker 1 out of 53 times.

Put this way, the problem doesn't seem so difficult.

### Re: The monty fucking hall problem.

Posted: Mon Aug 11, 2008 7:31 am
Cyrene wrote:Ihate my threads degrading to this drivel. This is a testable prediction, people who don't believe simply can't wrap thier heads around it, as the same statistical results or close enough happens consistently in the real world WHEN TESTED. leave the BULLSHIT run of the mill arguments against science/statistical predictions until you can show that in the real* world they don't routinely come true.

Now you have me curious what kind of drivel you were hoping for. I mean, were you intending to just say "The odds are higher if you switch doors, and if you don't like it, %&#@ YOU!!" and just let the thread die with no responses at all? The thread seems to have gone the way anybody could have predicted, as far as I can see. Sometimes I think you like to have things to complain about.

### Re: The monty fucking hall problem.

Posted: Mon Aug 11, 2008 6:09 pm
I think Cyrene believes that as far as varying degrees of certainty go in epistemology, that the most consistent method of prediction involves the scientific method.

That seems like a pretty uncontroversial position.

I don't really see anyone arguing against it.

The most interesting thing about it, to me at least, is that in making the point that using a disinterested and mathematically sound method is the best way to have the best knowledge, he resorts to emotional outbursts.

It just goes to show that while we may have different ways of coming to our conclusions, none of us are immune to the effects of our own emotions.

That's probably also the reason why most people don't pick the right door.

### Re: The monty fucking hall problem.

Posted: Mon Aug 11, 2008 8:28 pm
Smears wrote:I think Cyrene believes that as far as varying degrees of certainty go in epistemology, that the most consistent method of prediction involves the scientific method.

That seems like a pretty uncontroversial position.

Since given your remark I assume you are a non-American, things like the evolutionist/creationist debate probably don't get much attention where you live.

I wish that there were no controversy in America, too, about the relative certainty of scientific method compared to other epistemic methods. Over here, as opposed apparently to the situation in your country, it's a big problem.

### Re: The monty fucking hall problem.

Posted: Mon Aug 11, 2008 8:34 pm
Relative certainty. Exactly.

### Re: The monty fucking hall problem.

Posted: Mon Aug 11, 2008 10:40 pm
You don't need cards at all. Check it:
From the initial choice, there are three possible outcomes: Goat1, Goat2, or Car.

If you pick Goat1, Monty reveals Goat2. There are now two outcomes, switch to Car, or stay with Goat1.

If you pick Goat2, Monty reveals Goat1. There are now two outcomes, switch to Car, or stay with Goat2.

If you pick Car, Monty reveals either Goat1 or Goat2. There are now two outcomes, switch to Goat2 or Goat1, or stay with Car.

Tally it up. Of the three initial outcomes, in two cases switching will get you the Car, in one case it will get you the Goat, i.e. in 66.666. . .% of cases, switching gets you a car.

(As a preemptive, in the case where you pick the Car, the fact that there are two goats doesn't change the odds, because the goats are functionally interchangeable here, and they do not represent additional outcomes.)

Why are people saying that it matters whether or not Monty knows where the car is? If Monty always opens a door after the first guess, and the door he opens reveals a goat, you still get the same amount of information, and you still have the same incentive to switch doors.

### Re: The monty fucking hall problem.

Posted: Tue Aug 12, 2008 8:52 pm
Carleas wrote:Why are people saying that it matters whether or not Monty knows where the car is? If Monty always opens a door after the first guess, and the door he opens reveals a goat, you still get the same amount of information, and you still have the same incentive to switch doors.

Because it matters, Carleas.

Assume for example that the contestant chooses Door 1. At this point there are six equally likely possible outcomes. Either:

A) The Car is behind Door 1 and the host opens Door 2 to reveal a Goat. If the contestant switches, she loses.
B) The Car is behind Door 1 and the host opens Door 3 to reveal a Goat. If the contestant switches, she loses.
C) The Car is behind Door 2 and the host opens Door 2 to reveal a Car. Game Over. No opportunity to switch.
D) The Car is behind Door 2 and the host opens Door 3 to reveal a Goat. If the contestant switches, she wins.
E) The Car is behind Door 3 and the host opens Door 2 to reveal a Goat. If the contestant switches, she wins.
F) The Car is behind Door 3 and the host opens Door 3 to reveal the car. Game Over. No opportunity to switch.

Since the host reveals the car in options C) and F) and the game presumably ends at that point (or at the very least is undefined), this leaves only options A), B), D), and E) as viable options in regard to switching/not switching. This means that there are exactly two chances for the contestant to win the car by switching and two chances for the contestant to lose the car by switching. Thus, at this point it makes no difference what the contestant does because the odds are 50-50 that she will win the car no matter if she keeps the door she initially chose or switches.

It does make a difference if the host knows where the car is and always initially opens a "goat door."

### Re: The monty fucking hall problem.

Posted: Tue Aug 12, 2008 8:59 pm
Carleas wrote:You don't need cards at all. Check it:
From the initial choice, there are three possible outcomes: Goat1, Goat2, or Car.

If you pick Goat1, Monty reveals Goat2. There are now two outcomes, switch to Car, or stay with Goat1.

If you pick Goat2, Monty reveals Goat1. There are now two outcomes, switch to Car, or stay with Goat2.

If you pick Car, Monty reveals either Goat1 or Goat2. There are now two outcomes, switch to Goat2 or Goat1, or stay with Car.

If we assume that Monty always reveals a goat, actually there are four possible outcomes:

If you pick Goat1, Monty reveals Goat2. You switch, you win.

If you pick Goat2, Monty reveals Goat1. You switch, you win.

If you pick Car, Monty reveals Goat1. You switch, you lose.

If you pick Car, Monty reveals Goat2. You switch, you lose.

Try this with three cards and you'll demonstrate the truth of it for yourself. All you have to do is to end the game if "the car" is revealed when the first card is turned over and start over because we know that in the actual game Monty will not reveal the car when he opens the first door.

### Re: The monty fucking hall problem.

Posted: Tue Aug 12, 2008 9:43 pm
Wow ... I didn't know you could switch after the first door is opened.

Therefore, if the host actually knows which door has the car behind it, then the contestant can visually pick up on the clues intuited by him / her between the host and the contestant to directly improve the chances for picking the final 50-50 choice. If the host only knows what he's told (to open door X and reveal a goat) and actually does not know which final door has the car, then this knowledge cannot be intuited between the host and contestant.

This observation manipulates the final odds of being 50-50 or something else ... 0-100.

### Re: The monty fucking hall problem.

Posted: Tue Aug 12, 2008 11:40 pm
realunoriginal wrote:Wow ... I didn't know you could switch after the first door is opened.

Therefore, if the host actually knows which door has the car behind it, then the contestant can visually pick up on the clues intuited by him / her between the host and the contestant to directly improve the chances for picking the final 50-50 choice. If the host only knows what he's told (to open door X and reveal a goat) and actually does not know which final door has the car, then this knowledge cannot be intuited between the host and contestant.

This observation manipulates the final odds of being 50-50 or something else ... 0-100.

No, there're no clues that need to be "visually picked up on." According to the way the problem is worded, you already know that there will be a goat standing behind the door that the host opens.

This is strictly a problem of probability.

If the host opens a door with the car behind it, then it's a different problem.

Of course, anyone who doubts any of this can run through it himself with three playing cards.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 12:06 am
Reality Check wrote:This is strictly a problem of probability.

No, it's a problem of certainty...

Let's assume the host knows what is behind every door: then winning the "game" is contingent on both probability and whether the host "approves of" the contestant (for known / unknown factors) for reasons I shall now explain.

The Probability

From the beginning, there are three doors to choose from. A contest may pick the prize at a 33% chance odd compared to the 67% chance of the goats. However, this is rather inconsequential to the "game" at large.

The Certainty

From the beginning, the certainty exists in the host knowing which door has the prize and which two have the goats. Now, factor in this context: the contestant either severely dislikes, dislikes, likes, or severely likes the particular contestant (for any personal reasons). In all likelihood, the game show host is white, because this is a common trend in America. Therefore, if the contestant is non-white, then the host may or may not dislike that person more than if that person were white-skinned. This, among many other factors, contribute to the later suggestion of the host to the contestant...

If the host severely dislikes the contestant, then he (a male) will likely not even give the contestant a chance if that (black) contestant picked the first choice in error (a goat). My conclusion is this: if the (black) contestant picks a goat from the beginning, then the host may expose the goat right then & there, because he (being white) severely dislikes the contestant. Thus, the game is prematurely ended (for seemingly no reason).

If the host severely likes the contestant, then he will likely give the contestant a chance if that (white) contestant picked the first choice in error (a goat). My conclusion is this: if the (white) contestant picks a goat from the beginning, then the host may expose the other goat right then & there, because he (being white) severely likes the contestant. Thus, the game is continued (for seemingly no reason).

Furthermore, if both the host and contestant are white, then the host may or may not go so far as to actually suggest which door the prize is in directly to the contestant on the spot if you know how to look for these suggestive clues, a subtle nod, a wink, a pat on the shoulder, etc.

Thus, the game is rigged by the host knowing beforehand which doors the goats were in and which door the prize was in.

Welcome to the "game" ... I wonder if it's as fun when the host doesn't know which things are behind which doors???

...I bet not!

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 12:56 am
RC, your second post reveals the flaw in your exposition. You're double-weighting the case in which you have picked the car. There are only three doors to pick, so that there are two possible goats for Monty to pick from does not change the odds. In light of that, it shows that in two cases switching gets you the car, and only in one does it not. Observe the following fancy diagram:
untitled.gif (6.97 KiB) Viewed 1411 times

You'll see that initially there are three possible paths. What your explanation does is make it seem as though Monty's choice of one goat or the other (the path on the right) actually defines a new path. But no matter which goat Monty picks, there is only one door hiding a car, and one path that follows from it.

Now the case where Monty doesn't know (which will turn out to be the same case):
There are initially six outcomes. Say we have doors A, B, and C, and the car is behind C. In the format "your choice - Monty's choice", we have,
A - B
A - C
B - A
B - C
C - A
C - B

Off the bat, we eliminate the two where monty picks the car. We are left with the same 'four' possibilities as you listed above. But again, you only actually have three possible initial choices, A, B, or C, and of those choices, two will reward switching doors after Monty's door has been opened, and one will not (Note that you're only given the option of switching if the game is still continuing, i.e. Monty picked a goat-hiding door).
The reason is that even if he doesn't know what's behind the doors, Monty's choice is constrained by yours. After you pick your door, the other two doors have ~66% chance of hiding the car, and that fact doesn't change when you discover that some specific one of those doors conceals a goat.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 1:27 am
Carleas, take three playing cards, two deuces and an ace. Turn them face down. Choose one for yourself, keeping it face down. Turn over one of the other two cards. If it is an ace, reshuffle the cards and start over.

If it is a deuce, either switch to the other card and turn it over or turn over the card you originally selected.

If you do this ten times, it's very likely that you'll win about 4-6 times and lose about 4-6 times. If you play this game 20 times, it's even more likely that you'll win 50% of the time and lose 50% of the time. If you play it 100 times you will be even more likely to split 50-50.

But don't take my word for it, try it.

Think of it this way. Shuffle three cards and don't choose a card before turning one over. Simply turn over any of the three cards. If it is the ace then reshuffle and start again. If it is a deuce, then the odds that either of the two remaining cards is the ace has to be 50-50.

Nothing magical happens simply because you select a card before one of the cards is turned over.

Don't know if that helps or not, but again, just play the game yourself with cards or dominoes or whatever until you are satisfied that switching doors when the host doesn't know which door (or card or domino) is the winner, wins only 50% of the time. It goes very quickly if you either always keep the original card you select or if you always switch. It doesn't matter because either way you are going to win roughly 50% of the time.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 1:34 am
RC,

First of all, probability doesn't work like that.

Second of all, you have to choose a card before you turn any card over, because the act of turning a card implies that you chose it.

The 50-50 divide implies that the card you set aside is a "second set of card(s)", which is the conceptual mistake. You are essentially saying that with three cards to start: choose one and this splits the remaining cards into "two sets". Regardless of that context, the probability remains the same: you have a 33% chance of choosing an ace or a deuce when it is unknown from the beginning. It does not change until you change the rules for success of choosing "set 1" (of the card you chose) versus "set 2" (of the two cards you did not choose).

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 2:18 am
Carleas wrote:Now the case where Monty doesn't know (which will turn out to be the same case):
There are initially six outcomes. Say we have doors A, B, and C, and the car is behind C. In the format "your choice - Monty's choice", we have,
A - B
A - C
B - A
B - C
C - A
C - B

Off the bat, we eliminate the two where monty picks the car. We are left with the same 'four' possibilities as you listed above. But again, you only actually have three possible initial choices, A, B, or C, and of those choices, two will reward switching doors after Monty's door has been opened, and one will not (Note that you're only given the option of switching if the game is still continuing, i.e. Monty picked a goat-hiding door).
The reason is that even if he doesn't know what's behind the doors, Monty's choice is constrained by yours. After you pick your door, the other two doors have ~66% chance of hiding the car, and that fact doesn't change when you discover that some specific one of those doors conceals a goat.

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.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 4:19 pm
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.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 8:06 pm
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.

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.

### Re: The monty fucking hall problem.

Posted: Wed Aug 13, 2008 11:47 pm
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.

### Re: The monty fucking hall problem.

Posted: Thu Aug 14, 2008 3:16 am
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?

### Re: The monty fucking hall problem.

Posted: Thu Aug 14, 2008 4:26 am
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.

### Re: The monty fucking hall problem.

Posted: Thu Aug 14, 2008 4:32 am
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.