The monty fucking hall problem.

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.

again i suggest a deck of cards for those people too incompetent to do the mental math if you don’t ‘get’ it grab some cards if the people in question thinks its faith 200 rounds with a deck will prove them wrong, the correct card *will be in the dealer’s hand consistantly enough to say the prediction is correct.

seriously not all my threads need to be dragged down by philosophy based on the standards of a seven year old. real world ideas have testable predictions, the worth of claims depends on the accuracy of those predictions, not any mental midget’s ideas about thier worth…

What about this Cyrene?

en.wikipedia.org/wiki/Newcomb’s_paradox

Is this philosophically related to your point?

I see it as meaning that there is alot more difficulty to determining probability than it might seem when we just think of dividing fractions.

Another guy I’ve been reading thinks that it’s just a prisoner’s dilemma cut in half.

Game theory rules.

Here’s what I consider to be the most easily grasped explanation for this problem:

Assume that there are ten doors to choose from to begin with, with goats behind nine of them and a car behind only one.

You pick one of the ten doors. Your chance of selecting the door with the car behind it is 10% and the chance that the car is behind one of the other nine doors is 90%.

Then the game show host, who knows which door the car is behind and will not open that door, opens eight of the remaining nine doors to reveal goats behind them. In no way does the fact that eight of the other nine doors have now been opened to reveal goats behind them affect the 10% chance that you selected the correct door to being with. It doesn’t affect it, because YOU ALREADY KNEW THAT AT LEAST EIGHT OF THOSE NINE DOORS HAD GOATS BEHIND THEM!

You are then given the option of keeping the door you picked or of switching to the only other unopened door.

Since it is obviously 90% likely that the car is behind the sole remaining unopened door that you didn’t pick and only 10% likely that it is behind the door that you did pick, only an idiot would keep the door that he originally picked, right?

Well, the same principle applies whether there are three doors to choose from to begin with or ten or a million. The only difference is, is that the odds that you selected the door with the car behind it to begin with obviously get much worse the more doors that there are to choose from. This also means of course that the odds that the car is behind one of the doors that you didn’t select get much better.

I believe the point that Cyrene is making is how cognitive probabilities are processed in more basic and “stupid” people. On a game show, for the host to pick all the incorrect doors and leave the final “chance” at 50-50, then people misinterpret that the odds of picking the correct door ends up as a 50% chance. In reality, those who are more intelligent see through the “game” of the “show” … that they detect that the host is misleading the contestant and the viewers (conceptually) by a “sleight of mind”. In the end, it appears as the pick comes down to a 50% chance, when in reality, the chance was very low to begin with (depending on the amount of doors w/goats behind them).

The “problem” here is how the probability changes in a real-time context. The human mind works cognitively in this manner (compared against an intelligent or ignorant person) by anticipating possible changes and adhering expectations toward them. Most people cannot account for this, because they are literally stupid. However, a more intelligent person will pick up on the clues and eventually understand that the door situation stayed consistent in context from the beginning to the end and the “game show” is a mere illusion to those who are fooled by it.

What would really throw such an “intellect” off is if the game show host randomly picks the correct door from time-to-time rather than leaving it to the last. Then, the odds become randomized again in expectation only … and the “game” loses its appeal (of anticipation of coming down to the final two doors rather than seeing the result straight-out).

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.

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.

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. :slight_smile:

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.

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.

Relative certainty. Exactly.

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.

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.”

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.

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.

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! :evilfun: :laughing: =D>

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
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.

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.

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).

](*,)

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.