• Quick note - the problem with Youtube videos not embedding on the forum appears to have been fixed, thanks to ZiprHead. If you do still see problems let me know.

Monty Hall Problem

Dr Adequate said:
The shortest argument goes like this. When you picked, there was a 1/3 chance you were right first time. After the unveiling, there's still a 1/3 chance you were right first time --- nothing can change that. This means that there's a 2/3 chance that the car's behind the unselected drawn curtain. [/B]

Wrong, further information can alter the probability; indeed this monty problem is highly unusual in that it doesn't. If for instance Monty's choosing of the door was random, but yet still revealed a goat, this would mean your original choice has increased in probability from 1/3rd to 1/2 of being the car.
 
Cabbage said:
Say you randomly pick a positive integer. What is the probability the number you pick is even?

[/B]

Randomly pick a positive number?? How pray would you do that?? How indeed could you do that even with the help of a genuine random process in the world??
 
Paul C. Anagnostopoulos said:
Haven't we assumed the probability of picking a given integer = 0 in various other conversations we've had here? I thought that was because it was infinitessimally small, so we called it zero. But all the probabilities would still add up to 1. No?

~~ Paul

Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

Almost everyone said I was stupid and that only an infinite string of numbers could guarantee it :) People never did realise that they were wrong and said that I don't understand because I packed in Maths at the age of 16 LOL

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D
 
Interesting Ian said:
Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

If the finite string has length n, you can in principle calculate the probability that a given string of length m never occurs. That probability is nonzero.

I don't want to open up what was obviously a huge can of worms, but look: Pick any length-n string (n 0's for instance) that doesn't contain your length-m string. The probability of that particular length-n string occurring is 1/10^n. That's small, but it isn't zero. And there are a great many strings that don't contain your target string, each of them having a probability of 1/10^n of occurring.

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D

I think I see where you're going. This gets into the kind of arguments I have with people who can't understand how you can say that there are infinitely many integers, but all of them are finite. I think you're coming down on the side of the angels with that one. There is indeed a mathematical distinction between infinite (the cardinality of the set of integers for instance) and finite but unbounded (the magnitude of any particular integer).

So if what you're saying is "if you wait long enough, the probability is 1 that the length-m will occur in finite time", then you are correct. That's different from picking one particular length-n string.

I'm going to guess that you argued that "probability = 1" means "certain". That's another eternal argument in the math forums. It doesn't. I'll bet the term "almost certain" came up in that discussion.

Sorry if I've given flashbacks to the other readers.
 
rppa said:
If the finite string has length n, you can in principle calculate the probability that a given string of length m never occurs. That probability is nonzero.

I don't want to open up what was obviously a huge can of worms, but look: Pick any length-n string (n 0's for instance) that doesn't contain your length-m string. The probability of that particular length-n string occurring is 1/10^n. That's small, but it isn't zero. And there are a great many strings that don't contain your target string, each of them having a probability of 1/10^n of occurring.



I think I see where you're going. This gets into the kind of arguments I have with people who can't understand how you can say that there are infinitely many integers, but all of them are finite. I think you're coming down on the side of the angels with that one. There is indeed a mathematical distinction between infinite (the cardinality of the set of integers for instance) and finite but unbounded (the magnitude of any particular integer).

So if what you're saying is "if you wait long enough, the probability is 1 that the length-m will occur in finite time", then you are correct. That's different from picking one particular length-n string.

I'm going to guess that you argued that "probability = 1" means "certain". That's another eternal argument in the math forums. It doesn't. I'll bet the term "almost certain" came up in that discussion.

Sorry if I've given flashbacks to the other readers.

I was never talking about an upfront specified finite number. So maybe we agree there.

Yes I think probability = 1 is "certainty". In order for it not to be certain there has to be a possibility of it not being certain. Logical possibility is not relevant. A logical possibility can not occur ie it cannot be a possible future event. It can only have occurred in retrospect.
 
hgc said:
I know most of you are familiar with this one already, but we must settle it once and for all...

Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?
I wrote a program to test this problem, and keeping your original door limits your odds to about 1/3 and switching bumps you up to 1/2.

Edit to add: Scratch that. I reread the problem again, and I saw I made a mistake. I assumed that one of the doors which can be eliminated was the door which was chosen. I've edited my code below (edits appear in green), and I'm getting the correct answer of 2/3.

Code:
Private Sub CommandButton1_Click()
Dim SwitchYes As Long
Dim SwitchNo As Long
Dim StayYes As Long
Dim StayNo As Long

Dim Doors(3) As Boolean
Dim Choice(3) As Boolean

Dim MyDoors As Integer
Dim MyDoors2 As Integer
Dim MyChoice As Integer
Dim MyChoice2 As Integer

Dim SwitchTrials As Long
Dim StayTrials As Long

Dim I As Long
Dim J As Long

    Randomize
        
    For J = 0 To 100000
        For I = 0 To 3
            Doors(I) = False
            Choice(I) = False
        Next I
        
        MyDoors = Int(Rnd * 3) + 1  'This is the grandprize door
        MyChoice = Int(Rnd * 3) + 1 'This is the original choice
        
        Doors(MyDoors) = True
        Choice(MyChoice) = True
        
        If Int(Rnd * 2) + 1 = 1 Then    '<=== Changing some of these values can affect
            'switch                     'whether you want the program to switch everytime,
            For I = 0 To 3              'stay everytime, or decide at random whether to
                Choice(I) = False       'switch door or stay.
            Next I
            
            Do
            'This eliminates one of the goat doors. [color=green]Correction: This eliminates a goat door of the two remaining unchosen doors.[/color]
                MyDoors2 = Int(Rnd * 3) + 1
                If MyDoors2 <> MyDoors [color=green]And MyDoors2 <> MyChoice[/color] Then
                    Exit Do
                End If
            Loop
            
            Do
            'This chooses one of the remaining doors
                MyChoice2 = Int(Rnd * 3) + 1
                If MyChoice2 <> MyChoice And MyChoice2 <> MyDoors2 Then
                    Choice(MyChoice2) = True
                    Exit Do
                End If
            Loop
            
            If Choice(MyDoors) = True Then
                SwitchYes = SwitchYes + 1
            Else
                SwitchNo = SwitchNo + 1
            End If
            
            SwitchTrials = SwitchTrials + 1
        Else
            'stay
            If Choice(MyDoors) = True Then
                StayYes = StayYes + 1
            Else
                StayNo = StayNo + 1
            End If
            
            StayTrials = StayTrials + 1
        End If
    Next J
    
    MsgBox "Switch Trials (" & SwitchTrials & "):" & vbNewLine & _
           "Switch Doors And Won Car: " & SwitchYes & " / " & Left(SwitchYes / (SwitchTrials + 1), 6) & vbNewLine & _
           "Switch Doors And Won Goat: " & SwitchNo & " / " & Left(SwitchNo / (SwitchTrials + 1), 6) & vbNewLine & _
           vbNewLine & _
           "Stay Trials (" & StayTrials & "):" & vbNewLine & _
           "Stayed And Won Car: " & StayYes & " / " & Left(StayYes / (StayTrials + 1), 6) & vbNewLine & _
           "Stayed And Won Goat: " & StayNo & " / " & Left(StayNo / (StayTrials + 1), 6)
End Sub

Here are the graphical results:
 
TeaBag420 said:
How long does it take you to convinince them that you have a bigger penis, um, I mean, that they have failed to solve the problem.

About 20 minutes.

It's been days here and statistically I would bet that someone has a bigger penis than you.

Maybe. I'm a good sport, and I'll pay up if I lose. This, however, hasn't happened yet.

Being right and getting paid are two different things, Swinging Dick. I smell a liar.

You should get out more often, then. And it's Bwana Dik, not Swinging Dick.
 
Paul C. Anagnostopoulos said:
Had a known that the fate of the rational world depended upon rigorously specifying the Monty Hall problem in Parade magazine, I would have been much less cavalier about it, and certainly would have used more bold words.

You people are over the top. :bowl:

I love you guys.

~~ Paul

Paul, I realize this is just an attempt at a cheap put-down on your part, but I think this is fair dinkum considering how much press this conundrum got in the Skeptical Inquirer, pushed as it was by that guy (I'm terrible at names) small brunette, physician in the Tampa Bay area, had two white Trans-Ams in the 1990s, Gary Posner maybe? Anyway, whether the name is right or not, he had a column in SI and talked about it, and since then, getting the 2/3 answer has become a shibboleth of having a "Skeptic" tattoo on your forehead, which is mostly if not completely unrelated to being skeptical.

I wasn't the one who elevated it to Shibboleth Status, but I do note that when I subscribed to SI, I saw only one letter to the editor about it that displayed real skeptical thinking.
 
garys_2k said:
:D

Yeah, there are several. I was thinking of the leafy garden plant, though. the one with "a short stem and a dense globular head of usually green leaves that is used as a vegetable" (or so says Merriam-Webster).
(Looks around...:confused: ) I don't see anyone around here fitting that description. My leaves are usually red.






;)

Seriously, though, it looks like I've managed to partially hijack this thread from Monty Hall into a more general probability discussion; that wasn't my intention, but maybe that's a good thing...

(reads some of the recent posts)

...or maybe not.

Anyway, I think I've said all I can possibly say about Monty; anymore and I would simply be repeating myself (though in reality I probably reached that stage three pages ago). If you still don't agree with me, I don't guess I'm gonna change that.

Regarding picking an integer at random vs. a real number at random, yes, of course there's no mechanism by which this experiment can be carried out; it's a thought experiment, but certainly a worthwhile one, with all sorts of applications, I'm sure.

To have a uniform probability distribution on a set, the set does need to be either finite or uncountable; a countable set is where such a notion breaks down.

In the uncountable case, there may still be some "limitations". For example, take a uniform probability distribution on the real interval [0,1], Lebesgue measure, say.

So you can say things like (as has been mentioned before), the probability of your number being less than 1/3 is 1/3, or the probability of it being between 3/7 and 5/7 is 2/7, or that it has a probability of 0 of being in the Cantor set.

The probability of a particular number being picked is zero. Probability zero does not mean "impossible", which is demonstrated by the simple observation that while every particular number has zero probability of being picked, some number has to be picked.

Similarly, probability one is not synonymous with "certain".

Anyway, the "limitation" I was referring to is that there are some subsets of the real interval [0,1] for which it doesn't make sense to ask what the probability of our number being in that set is. Such sets are not in the domain of our probability function. I believe the axiom of choice is required to guarantee the existence of such sets.
 
Interesting Ian said:
Wrong, further information can alter the probability; indeed this monty problem is highly unusual in that it doesn't. If for instance Monty's choosing of the door was random, but yet still revealed a goat, this would mean your original choice has increased in probability from 1/3rd to 1/2 of being the car.
And if he reveals that the door that you originally chose has a car behind it, you'd be a fool to switch.
 
epepke said:
Paul, I realize this is just an attempt at a cheap put-down on your part,
Indeed, I was just yanking various chains. Also, it was an opportunity to use the bowling smilie, for reasons that now escape me.

but I think this is fair dinkum considering how much press this conundrum got in the Skeptical Inquirer, pushed as it was by that guy (I'm terrible at names) small brunette, physician in the Tampa Bay area, had two white Trans-Ams in the 1990s, Gary Posner maybe?
You can remember that he had two white Trans Ams, but not his name? You sound like me! [I wonder if the plural should be Transes Am?]

Anyway, whether the name is right or not, he had a column in SI and talked about it, and since then, getting the 2/3 answer has become a shibboleth of having a "Skeptic" tattoo on your forehead, which is mostly if not completely unrelated to being skeptical.
Ah, now this is interesting. I had not picked up on this becoming some sort of skeptical right of passage. Certainly, in that context, blithely assuming the conditions of the problem without explicit clarification would be a mistake.

In the interest of being a "true skeptic" and not a "skeptical buffoon," :D I will retract my statement that all this concern over the precise wording of the Monty Hall problem is pedantic. All future statements of the problem should be worded precisely. However, I stick with my claim that that will not help most people see the light, because it is not any possible ambiguity that is the barrier to understanding.

~~ Paul
 
Interesting Ian said:
Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

Almost everyone said I was stupid and that only an infinite string of numbers could guarantee it :) People never did realise that they were wrong and said that I don't understand because I packed in Maths at the age of 16 LOL

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D

Hey, except for me that said it was not guaranteed to appear even IN an infinite string. But I'm not getting into it again, for my comments read the thread.
 
Paul C. Anagnostopoulos said:
Good point, dodge. We can generate random integers between 1 and n, but then each has a probability of 1/n and half are even. But we cannot generate random integers between 1 and infinity, so it's difficult to run the experiment.

Actually, it's easy to pick a random integer between 1 and infinity.

Pick a random decimal number n between 0 and 1 (most random number generators do this) and your random integer is 1/n, rounded off.
 
TeaBag420 said:
So you offer two possibilities, both of which say it's a math problem, but previously you said it's not a math problem. Care to jack, um, I mean explain?

Perhaps you would care to "firetrucking" read what I said.

There are two possibilities both of which are math problems. But if you don't know which it is, then it is not a math problem.

It is a case, as I have said numerous times, of clarifying the question.

In fact 99% of the debate in this forum is generated by not clarifying the question.

(edited for spelling)
 
rppa said:
I'm happy to see a non-probabilistic analysis.

The answer of course is yes, switch.

Gee, I guess I'm a learning disabled retart, but somehow that doesn't look like an analysis.

Could you amplify just a little? Starting by defining "to your advantage"?
If you want a car, you maximize your chances of getting a car. If you prefer a goat (not that there's anything wrong with that) think about how many goats you could fu...uh, get in exchange for a car. Either way, you win. That's "to your advantage" Johnny Shortbus.
 
TeaBag420 said:
Originally posted by rppa
I'm happy to see a non-probabilistic analysis.

The answer of course is yes, switch.

Gee, I guess I'm a learning disabled retart, but somehow that doesn't look like an analysis.

Could you amplify just a little? Starting by defining "to your advantage"?
If you want a car, you maximize your chances of getting a car. If you prefer a goat (not that there's anything wrong with that)

No, no, that's fine. OK, so "to my advantage" means "maximizing the probability that I will get a car."

Now that's a well-defined mathematical problem. It's simple to solve if we have a model for what the probability of getting a car is under various scenarios.

But you claim that the answer to this maximization problem does not require that knowledge. All it requires is the knowledge that in the game you just played, where you don't know what's behind your door, Monty opened a door containing a goat.

So please enlighten me. Without making any more assumptions about Monty's behavior, tell me what the probability of getting a car is if I switch, the probability if I don't switch, and why switching is better. Because what you've written, basically "obviously the answer is to switch" still doesn't look like an analysis to me.
 

Back
Top Bottom