Strange Famous

A friend mentioned this to me, and we have had a discussion on it. To give the background, here is a link to an extremely patronising wikipedia article about it.

Monty Hall problem - Wikipedia, the free encyclopedia

The basic setup is:

In my opinion this is completely wrong.

If you are in the same scenario but you dont make any guess and they open box 3 and there is no prize there, the possibility that it is 1 or 2 is equal by any acount

Given that what you guess has no physical effect on whether the prizes is in 1 or 2, how can the fact you say "1" alter the probability of where the prize actually is? The phyical reality is not affected in any way by what you guess.

If this was acted out in actual life 99 times, then the car would be in box 1 33 times, box 2 33 times, and box 3 33 times (on average). Whether you change your choice or not, the box has the same possibility to be in either of the remaining boxes. Whatever the person chooses and whatever the show's host does, it makes no difference.

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"Do not tell lies, and do not do what you hate,
for all things are plain in the sight of Heaven. For nothing
hidden will not become manifest, and nothing covered will remain
without being uncovered."

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ratbastid

You're denying that changes in the situation result in changes to the odds. I mean look: since he eliminated one door, at LEAST you'd have to concede that your odds improved from 1 in 3 to 1 in 2, right? Since the door with the goat is now out of play, there are really two doors to pick between?

The timing of the scenario makes that not an accurate description of the odds, though, because the switch gets done AFTER your selection.

When you picked your door, your odds of that door winning are 1/3 and the odds of EITHER of the other two doors being the winner is 2/3. When the goat door is eliminated, that doesn't change--either the closed door you didn't pick, or the door with the goat is the winner, 2/3 of the time. Since we know the goat door isn't the winner, that means the closed door you're being offered carries that whole 2/3.

This isn't a matter for opinion. If you play it out, you'll see that switching, counter-intuitive though it seems, is the winner 2/3 of the time.

Lasereth

This problem causes a lot of Internet arguments.

This poll is based on whether the reader assumes the problem is based on conditional statistics or unconditional statistics. If read in an unconditional manner (it appears that most read it this way), the answer IS 1/2 or that it doesn't matter if you switch or not. Conditional statistics problems are based upon a completely different set of mathematics and prove that the answer is that you should switch because of the 2/3 chance of winning.

So basically the answer is founded on which way you look at the problem. Saying that the answer is 2/3 for switching is true, but the answer is also "doesn't matter" or 1/2 if you consider it unconditional.

Wikipedia: "According to Morgan et al. (1991) "The distinction between the conditional and unconditional situations here seems to confound many." That is, they, and some others, interpret the usual wording of the problem statement as asking about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open when the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, nor make clear whether a conditional or an unconditional answer is required, making the answer that switching wins the car with probability 2/3 equally vague. Many commonly presented solutions address the unconditional probability, ignoring which door was chosen by the player and which door opened by the host; Morgan et al. call these "false solutions" (1991). Others, such as Behrends (2008), conclude that "One must consider the matter with care to see that both analyses are correct.""

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ratbastid

That's not what that Wikipedia quote is saying, though. That quote says there are lots of ways to SET UP the problem.

The fact is, a computer program running this thing thousands of times a second can demonstrate that, given the classic setup of the problem, switching is a 2 to 1 winner.

Here's a simulator where you can try it: The Monty Hall problem

Martian

Mathematics are not up for debate. The formula this whole thing is based on doesn't care about your opinion.

It's an interesting little quirk of probability that can be proven both mathematically and experimentally. You can have an opinion counter to that I suppose, but it's a demonstrably false one and that's the end of it.

/thread

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telekinetic

Monty Hall is basic statistics, albeit counter-intuitive.

I've always had the best luck explaining it like this:

I have a deck of cards. If you picked the Ace of Spades, I'd give you $100. So you pick one. Then, I flip over all of the cards except for one (so you can see every card other than, say, the 10 of Hearts and the Ace of Spades), and tell you if you want, you can switch from the card you originally picked to the only other remaining card face down. Would you do it?

Clearly you would, as your original card only had a 1/52 chance of being the Ace, whereas (because I sorted) my card has a 51/52 chance.

Now think about a smaller group, say 5 cards. Again, you pick one, I flip over three showing you they aren't the Ace. Yours still has the same 1/5 chance of being right, but mine has 4/5 (since either yours is right or mine is, and they have to equal 1)...do you switch?

Now we're down to our 3 cards. You pick one--it has a 1/3rd chance of being right. I flip up one card showing you it isn't the ace. Since your card still has a 1/3rd chance of being the ace (it's probability isn't retroactively affected!), mine has to have a 2/3rd, since the ace is SOMEWHERE.

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twisted no more

filtherton

I don't see how the traditional setup of this problem could be interpreted in an unconditional way. The whole point of the problem is that the hosts' choosing of a door gives you more information than you had at the outset.

Redlemon

SF, try this out to make it a bit more understandable.

Imagine that there were 100 doors. You select door 1. There is a 1% chance that your door is the correct door.

Monty opens doors 2 through 99, which are all empty.

Which looks like a better bet now, door 1 or door 100?
----
edit: I didn't see TK's post before I posted, sorry!

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GreyWolf

What many people don't get about this is that the host will ONLY open a door that has a goat behind it and offer to let you switch for the unopened door. It would be useless to open the door with the car (which you've missed 2 times out of 3), because you would then know you had a goat, and would be pissed off.

So, the opening of the door with the goat isn't really random, and so the probabilities are affected because of that condition set on the opening.

IF THE HOST WERE SIMPLY TO OPEN A DOOR AT RANDOM (revealing the car 1 time in 3), THEN THE PROBABILITY OF GAINING BY SWITCHING IS 50%. BUT HE'S NOT, SO THE ODDS SWING TOWARDS THE SWITCH, BECAUSE HE'S ELIMINATING HALF THE BAD CHOICES (a conditional probability).

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warrrreagl

This.

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Strange Famous

I tell you again wikipedia and other people in this thread are all wrong

If there are 100 doors, or 1000 doors.... the chance that it is behind any door is equal in this scenario

If I choose 1 and he opens 2 through 999 - there is a 1/1000 chance it is behind door 1, and a 1/1000 chance it is behind door 1000

Whatever I choose, whatever Monty Hall does - it makes no difference! The human choice does not affect which door the prize is behind. They put the car before the horse... this isnt game theory. The actual and material reality is there is a 1/3 chance of it being behind any of the doors. If it is revealed that it is not door 3, the die has already been cast.

Whether I change to 2, or stick with 1 - this has no material effect on the actual position of the prize. Everyone accepts this?

Given that this is accepted, and certain, and intrinsic in the puzzle - it MUST follow that my choice does not affect the probability of where it is, it is just as likely to be 1 as it is 2!

__________________
"Do not tell lies, and do not do what you hate,
for all things are plain in the sight of Heaven. For nothing
hidden will not become manifest, and nothing covered will remain
without being uncovered."

The Gospel of Thomas

Baraka_Guru

I think you are confusing the effects of human agency (choosing a door will have no impact on where the prize is) with probability (the measurable likelihood of an outcome based on risk-based choice or random chance).

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Humankind cannot bear very much reality.
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dreadxxx

Just watched this excellent documentary from an excellent site the other day:

topdocumentaryfilms.com/alan-marcus-go-forth-multiply/

Practical demonstration in part 2

Strange Famous

To look at this is real terms, lets say we played the game three times, and the results happened to go by the numbers

Each time I stick

Game 1 - I guess 1, the prize is in 1. Monty opens 3, offers me a swap, I say no - I win

Game 2 - I guess 1, the prize is in 2. Monty opens 3, offers me a swap, I say no - I lose

Game 3 - I guess 1, the prize is in 3. Monty opens 2, offers me a swap, I say no - I lose

If I stick each time, I win 1/3 of the time

If I twist each time

Game 1 - I guess 1, the prize is in 1. Monty opens 3, offers me a swap, I say no - I lose

Game 2 - I guess 1, the prize is in 2. Monty opens 3, offers me a swap, I say yes - I win

Game 3 - I guess 1, the prize is in 3. Monty opens 2, offers me a swap, I say yes - I win

...

I just typed that and it makes absolutely no fucking sense.

There is a 1/3 chance it is in any of the boxes! How can it possibly be so that whether I stick of twist can effect in anyway what is in the box???

Will the sky fall in now???

__________________
"Do not tell lies, and do not do what you hate,
for all things are plain in the sight of Heaven. For nothing
hidden will not become manifest, and nothing covered will remain
without being uncovered."

The Gospel of Thomas

Baraka_Guru

The probability in question is applied to the second choice, not the initial choice. You're maintaining the initial probability throughout, but it's changed by Monty giving you a second choice after the initial condition has changed.

__________________
Knowing that death is certain and that the time of death is uncertain, what's the most important thing?
—Bhikkhuni Pema Chödrön

Humankind cannot bear very much reality.
—From "Burnt Norton," Four Quartets (1936), T. S. Eliot

telekinetic

The fact that switching gives you 66% chance of getting a prize, compared to not-switching giving you a 33% chance isn't up for discussion. It does. This is not up for debate any more than 1 + 1 = 2.

If you would like to see this for yourself, go here, play with the simulator until you you have a statistically significant amount of samples (30+) or just have it run a couple hundred automatically for you:

The Monty Hall problem

We can have a discussion about WHY you get a better chance if you switch. We cannot have a discussion about IF you get a better chance if you switch.

---------- Post added at 11:05 AM ---------- Previous post was at 11:04 AM ----------

Look at your own examples. You have a 1/3rd chance of winning (win/lose/lose) if you don't switch and a 2/3rd chance of winning (lose/win/win) if you do. Each of your three options are statistically equally likely! This is probably one of the best demonstrations I've seen as to why you get a better chance if you switch--I'm not sure how you could type such a perfect explanation and yet still not understand it!

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twisted no more

Baraka_Guru

Oh, the irony.

Good job, Strange Famous! Seriously.

__________________
Knowing that death is certain and that the time of death is uncertain, what's the most important thing?
—Bhikkhuni Pema Chödrön

Humankind cannot bear very much reality.
—From "Burnt Norton," Four Quartets (1936), T. S. Eliot

Strange Famous

I understand what I typed, but I dont understand how it can be possible!

__________________
"Do not tell lies, and do not do what you hate,
for all things are plain in the sight of Heaven. For nothing
hidden will not become manifest, and nothing covered will remain
without being uncovered."

The Gospel of Thomas

Baraka_Guru

Therein lies the magic of mathematics!

__________________
Knowing that death is certain and that the time of death is uncertain, what's the most important thing?
—Bhikkhuni Pema Chödrön

Humankind cannot bear very much reality.
—From "Burnt Norton," Four Quartets (1936), T. S. Eliot

Ourcrazymodern?

Just give me the fucking goat.

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BE JUST AND FEAR NOT

ratbastid

You're so sure your intuitive solution is right that you've thrown logic out the window. Pray tell, good sir, what the hell happens the other 998 times, then?

You're telling me 1 time in 1000 the door you chose before was right, and 1 time in 1000 the door the host left closed was right, and 998 times, a door with a goat actually has a car???

Hektore

When you pick your first door out of 1000 your chances of choosing correctly are .1%. The probability that the car is not behind the door is 99.9%. The probability that it is not behind the door you chose initially will always be 99.9%, no matter how many doors they choose to open, until the car's location is actually revealed. Which means if they chose to reveal every door you did not pick except 1, and they all contain goats, the probability that it will be behind the last closed door you did not pick is 99.9%.

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cremaster

Hi Strange Famous,

I do love this problem. The way I try to explain it is to ask you to think of Monty as a gardener or a goat weeder to be more precise.

The original garden of possibilities has the following configuration;

goat – goat – car
goat – car – goat
car – goat – goat

or 6 goats to 2 cars.

If you fence away box 1 possibilities from Monty by picking it first he can't weed out any goats from it. He is left with boxes 2 and 3.

In them he has the following possibilities;

Goat – Car
Car – Goat
Goat – Goat

or 4 goats to 2 cars, but then he gets to work. Because he has to weed out a goat each game (by opening a goat door), in the above grid he has to remove 3 goats leaving car – car – goat or 2 to 1 chances of a car.

So the question becomes do you stay in your unweeded patch or jump the fence into Monty's weeded one? Answer is simple, jump every time.

Don't get too stressed if it doesn't come to you right away as I have read an autobiography of the great mathematician Paul Eros and he too got it wrong initially.

Could be a little too much weed in the above explanation but hey.

GreyWolf

If you look at your example, and read my previous post, you'll see your answer.

Monty is NOT opening a door at random. If he did that, sometimes (1 out of 3 times) he'd reveal the car. But he never does. If you work your examples so that he opens a door at random, you'd have more choices, and you would end up only winning half the time, as you expect. But because he ONLY opens a door with a goat behind it, he changes the probabilities of the outcome from the original conditions, favouring the switch.

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Jinn

Strange, are you a deliberate troll?

This is half a dozen threads of yours that I've read where you posit the obvious wrong solution and then proceed to defend it even against immeasurable evidence (and I believe you know you're wrong), presumably just to generate ire. It's getting old.

Check out the simulator link that rat provided:

http://www.grand-illusions.com/simulator/montysim.htm

Do "Run" 100 times as many times as you want, and select keep vs. change. If you can't see it, then this thread is over because you're simply not contributing.

Here are my results. Does that not look exactly like 2/3 and 1/3 to you?

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Ourcrazymodern?

Well, that wasn't very loving, but this goat smells wonderful.

...in perfect agreement, they wandered off pretending they had argued.

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Redlemon

Strange, once you understand the Monty Hall problem, try the Girl Named Florida problem.
That wasn't so bad, right? Just wait...
I still can't wrap my head around this one.

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filtherton

This is why conditional probabilities are fun.

The first question is interesting because it isn't what it seems. It refers to something that has ostensibly already happened and so it requires one to map out all the different ways that such a thing could have happened. If it were posed differently, say, "If you know a family already has one daughter, what are the odds that their next child will be a girl?" you'd get the expected 50%.

There is a subtle distinction between what has happened compared with how it could have happened and what could happen compared with how it could happen.
So if you decide that birth order isn't important, then you do get 50% for both examples.

Strange Famous

Did you note the post where I admitted I was wrong when I went through it choice by choice?

To be honest, I get a bit fed up of people calling me a troll, especially in a situation where I admitted I was wrong.

__________________
"Do not tell lies, and do not do what you hate,
for all things are plain in the sight of Heaven. For nothing
hidden will not become manifest, and nothing covered will remain
without being uncovered."

The Gospel of Thomas

Lasereth

Strange Famous, I've sat thinking about this problem for hours and I thought it was based on semantics alone. I was wrong, but I never understood why. Your post above with the 6 possibilities is the first time I've ever understood why the answer is 66% with switching. I love how you didn't understand it yourself when you typed it out but you helped me get it. Sometimes it helps to type these things out I guess ha ha ha ha ha.

It also helped me understand it like this: Monty will not only always open a door with a goat, but he'll also always NOT open YOUR door, and even furthermore, he'll never open a door with the car behind it. So I guess it does help if you go further with the explanation. For some reason this made a lot more sense to me and it clicked all of a sudden.

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"A Darwinian attacks his theory, seeking to find flaws. An ID believer defends his theory, seeking to conceal flaws." -Roger Ebert