Monty Hall Problem

[Mobyseven]

Whatever.

You call it function, he calls it motives, you're talking about the same thing.

No, we're bloody well not. Do me a favour - go to a dictionary. Check the meaning of 'function'. Now check the meaning of 'motive'. Are they the same?

Frickin' no.

Monty's function in the scenario is known and important - he is there to open a door with a goat behind it (not randomly, mind you) and offer you a choice to switch.

Monty's motive in the scenario is unknown and irrelevant - why he does what he does has nothing to do with the solution to the problem, unless you propose a motive that changes his function (as people have been doing), in which case you aren't talking about the Monty Hall problem anymore.

The proper Monty Hall puzzle specifies Monty's behavior as always being forced to show a non-prize door, and offer the switch. However, the puzzle as described in the OP of this thread does not specify that. At best, it's ambiguous. As I mentioned before, I have never seen the puzzle presented in the wild (as opposed to its Wikipedia page), where this constraint was explicit. And without this constraint, you have to make some assumptions about how Monty behaves (his function/motive) in order to come to a solution.

 

[Herzblut]

(Again, I acknowledge that if the host knows where the car is and will show every contestant a goat every time, then he can win 2/3 of the time by switching. But both scenarios are consistent with the OP, and without making an assumption about how Monty operates, you can't say that the 2/3 answer is always correct.)

Yes, as I stated before:

if you make the result of the opened door conditional, then you change winning chances to be based on a conditional probability. The switching-wins probability under the condition that Monty has opened a goat-door is 1/2, if the probability of a goat-door opened by Monty is 2/3. This is the case for instance if Monty has randomly opened the door and

- might as well have been opening a car-door with 1/3 probability,
- in which case the player would have lost the game
- and not permitted to switch to the opened door.

This is crystal clear. But, again, the latter case and its speculations are simply excluded by construction, according to my understanding of the problem.

 

[jsfisher]

The minimum problem solvable is the one I presented before, I think:

Monty opens a door and offers the player to stay or switch to another door of his choice.

Well, I think we have all now circled to some common ground where we all agree things hinge on how each of us might interpret the wording of the problem. So we are discussing semantics now and not probability.

This may be as much agreement as we can get, and I am willing to leave it there.

That aside, though, the Monty Hall problem is still a beautiful problem (preferably worded with less ambiguity to our mutual satisfaction) because the result is so counter-intuitive.

 

[Herzblut]

Well, I think we have all now circled to some common ground where we all agree things hinge on how each of us might interpret the wording of the problem. So we are discussing semantics now and not probability.

I'm not a native speaker, so what is unclear about

The player picks a door, then Monty opens one of the two other doors and reveals what is behind and offers the player to switch to another door of his choice.

 

[jsfisher]

I'm not a native speaker, so what is unclear about

The player picks a door, then Monty opens one of the two other doors and reveals what is behind and offers the player to switch to another door of his choice.

I am a native English speaker, so you have me at a disadvantage.

As written, it makes uses the present tense in a clumsy way. Is it present tense with future meaning (these are things that must be) or present tense with past meaning (these are things that happen to be)?

I favor the latter interpretation, but I can understand why others might prefer the former.

At the risk of confusing things further, consider the following:

The player will pick a door, then Monty will open one of the two other doors that will reveal a goat, and then Monty will offer the player the option to switch to another door of his choice.​

Here, I think with everything in future tense, the ambiguity is gone. Now, let's try it all in past tense:

The player picked a door, then Monty opened one of the two other doors that revealed a goat, and then Monty offered the player the option to switch to another door of his choice.​

With everything in the past tense, is it clearer that Monty's motives may have a role in the analysis? I am hoping so.


Unfortunately, the original problem statement wasn't at all clear. The use of the present tense gives in characteristics of both the future and past tense versions. And, as I said, I favor the past tense interpretation, but your actual mileage may vary.

 

[Michael C]

Herzblut, you just wrote this:

Yes, as I stated before:

if you make the result of the opened door conditional, then you change winning chances to be based on a conditional probability. The switching-wins probability under the condition that Monty has opened a goat-door is 1/2, if the probability of a goat-door opened by Monty is 2/3. This is the case for instance if Monty has randomly opened the door and

- might as well have been opening a car-door with 1/3 probability,
- in which case the player would have lost the game
- and not permitted to switch to the opened door.

This is crystal clear. But, again, the latter case and its speculations are simply excluded by construction, according to my understanding of the problem.

But yesterday you wrote this:

OK, now if you change the rules and assume, Monty is randomly choosing a door (but not yours) with two possible outcomes, then changing to another door (of your choice) still gives you 2/3 winning chance.

So which answer do you consider to be correct?

 

[Herzblut]

So which answer do you consider to be correct?

Both, the two versions describe different scenarios. They differ in what happens when Monty reveals a car (lost/won).

 

[Herzblut]

As written, it makes uses the present tense in a clumsy way. Is it present tense with future meaning (these are things that must be) or present tense with past meaning (these are things that happen to be)?

It is present tense of a technical description. The description of a random experiment. http://cnx.org/content/m13470/latest/

And I didn't say that Monty revealed a goat.

 

[Michael C]

Both, the two versions describe different scenarios. They differ in what happens when Monty reveals a car (lost/won).

Fine, let's look at this version

OK, now if you change the rules and assume, Monty is randomly choosing a door (but not yours) with two possible outcomes, then changing to another door (of your choice) still gives you 2/3 winning chance.

In this version, you don't say what happens if Monty reveals a car. What does happen if he reveals a car in this version? Does it mean win or lose?

 

[Herzblut]

In this version, you don't say what happens if Monty reveals a car.

Of course I do. The player switches to another door of his choice and wins.

 

[Michael C]

Of course I do. The player switches to another door of his choice and wins.

Ah: a new version. The player may also choose the open door. A strange game, but never mind:

If the car is revealed, the player can either:
- switch to the door with a car (duuh), so his chances of winning are 1
- switch to the other closed door, so his chances of winning are 0
- keep his door, so his chances of winning are 0
If a goat is revealed, the player may either
- switch to the door with a goat, so his chances of winning are 0
- switch to the other closed door, so his chances of winning are 1/2
- keep his door, so his chances of winning are 1/2

In no case does switching increase his chances to 2/3.

 

[articulett]

actually it does-- at the outset his odds were one in 3-- once he picked the goat door-- it still is one in 3... that means the remaining door still has 2/3 chance of being the car. He doesn't change his original odds that his own door has the car-- ever. He just gets information by opening the other door-- the information he gets is either: He definitely has a goat (as would be the case if the car was revealed) or he his original 1/3 chance of having the other goat if a goat was revealed. Really.

You are incorrect MichaelC. Humans equate 2 choices with 50/50 odds... but that is a logical fallacy. Surely you'd understand this if there were 100 doors, right? and one car-- We'd expect him to reveal a goat... it is more likely than not that a goat will be revealed. In your scenario, If he has the car behind his door--there is 100% chance he'll be opening a door with a goat... if he has a goat behind the door-- he still has a 50/50 chance of turning over a goat. But he doesn't know whether he has a goat or a car. Getting a goat doesn't make it more likely that he has a car. His odds don't suddenly get better; he just gets information. Getting a car makes it 100% likely that he has a goat. The same is true whether there are 3 choices or 10 choices or a 100 choices. The odds are always in favor of the one who switches.

This is counter intuitive but very worth understanding.

His original odds of his door having a goat can never get better; --it just shows itself to be worse if he reveals a car behind another door.

 

[Herzblut]

Ah: a new version.

Can't you read? What's your comprehension problem with OK, now if you change the rules and assume..? You also forget, that it was your idea to randomize Monty's output, not mine. You changed the rules.

The player may also choose the open door.

What's your comprehension problem with switching to a door of his choice?

If a goat is revealed, the player may either
- switch to the door with a goat, so his chances of winning are 0
- switch to the other closed door, so his chances of winning are 1/2
- keep his door, so his chances of winning are 1/2

In no case does switching increase his chances to 2/3.

Yeah, right. Because

If a goat is revealed, and the player switches to the other closed door there may be
- a goat behind that door, so his chances of winning are 0
- a car behind the door, so his chances of winning are 1.

In no case does switching give any chance other than 0 or 1. Zero if you lose and one if you win. Same as in any other game you play.

I see you have fully incorporated the concept of probabilities.

 

[articulett]

A funny thing about humans is that they tend to go with their first choice-- as soon as they place a bet they tend to see their odds as increasing-- they feel more confident with their choice... but your odds don't suddenly change when you place your bet or pick your door... your feeling that you are more likely to win, does not make you more likely to win. This is true even with lottery winners. Before playing, they may correctly assess their odds-- but once they've committed, they "feel" as if their odds go up.

My whole state (Nevada) is built on gambling odds and the bizarre psychology and irrationalities people have in regards to the subject-- their belief in "winning streaks" and so forth.

 

[CFLarsen]

but your odds don't suddenly change when you place your bet or pick your door...

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

 

[Herzblut]

actually it does-- at the outset his odds were one in 3-- once he picked the goat door-- it still is one in 3... that means the remaining door still has 2/3 chance of being the car. He doesn't change his original odds that his own door has the car-- ever. He just gets information by opening the other door-- the information he gets is either: He definitely has a goat (as would be the case if the car was revealed) or he his original 1/3 chance of having the other goat if a goat was revealed. Really.

Not really. If Monty does not consistently (necessarily), but conditionally, reveal a goat the player's winning-by-switching chances in case a goat has been revealed become dependent on how probable it was the goat showed up.

For instance, assume Monty consistently opens the car-door whenever possible. Monty now opens a goat-door and that means that the player has picked the winning door with 100% probability and will always lose by switching.

But he doesn't know whether he has a goat or a car. Getting a goat doesn't make it more likely that he has a car. His odds don't suddenly get better;

Yes, they might do so. In the above scenario, it's even certain that he has the car.

However, the strategy of consistently switching gives the player an average winning chance of 2/3, no matter what. Actually, that's my point. #sigh#

 

[lionking]

Not really. If Monty does not consistently (necessarily), but conditionally, reveal a goat the player's winning-by-switching chances in case a goat has been revealed become dependent on how probable it was the goat showed up.

For instance, assume Monty consistently opens the car-door whenever possible. Monty now opens a goat-door and that means that the player has picked the winning door with 100% probability and will always lose by switching.


Yes, they might do so. In the above scenario, it's even certain that he has the car.

However, the strategy of consistently switching gives the player an average winning chance of 2/3, no matter what. Actually, that's my point. #sigh#

Jeez how many times does it have to be repeated?

 

[articulett]

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

 

[CFLarsen]

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

Educate yourself.

 

[Herzblut]

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

No. Let's define:

A = car is behind one of the other doors
B = Monty reveals a goat

The (conditional) probability of A under the condition that B has taken place is given by

P(A|B) = P(AnB)/P(B)

where P(AnB) shall be the probability of "A and B". In case P(B)=1 this reduces to

P(A|B) = P(A) = 2/3

which is the Monty Hall case.

However, if P(B) is not 100% this 2/3 is no longer necessarily correct. For instance, if Monty selects a door without knowing where the car is:

P(B) = 2/3
P(AnB) = 2/3 * 1/2 = 1/3 ==> P(A|B) = 1/2

I presented this to avoid any useless debate about a given mathematical truth.