Why is the Monty Hall Problem so controversial?

[Schrödinger's hat]

Controversial*

 

I've been terrible at spelling lately, not sure why. :/

[Xittenn]

I'm not 100% sure on what you asked me. Perhaps you could re-state the question, maybe adding a little redundancy just in case I still don't follow. I would be happy to attempt a formal answer if I knew exactly what was being asked.

 

I think part of the issue here, is that the simplest way to proove this is by simple exhaustion. There are numerous tables and tree diagrams on the wiki page that constitute a proof from a frequentist standpoint.

 

If you prefer symbol pushing, then you can throw Bayes theorem at it repeatedly, and there are a number of ways to approach the answer.

 

 

I was observing it as the number of doors increases indefinitely . . . I actually believe the probability boost gained from this trick decreases as you increase the number of doors renormalizing on 1/2. I could be wrong, but I have found a few papers that sort of suggested this.

 

The reason we're concerned with your standpoint is that this community is here to help people learn and understand things. If we perceive you do not understand something we will endeavour to help you (even if we sometimes do it rather inefficiently).

This is also an excellent example of how intelligent and mathematically experienced people can have difficulty understanding the Monty Hall problem or accepting the conventional answer. Ie. a good example of 'why is it so controvercial?'

 

 

I think I'm having a nervous breakdown right now, I'm not sure because I can't feel my face and I'm crying in the library again. So with regards to my clarity and whether or not I'm sober, I might be kind of messed up in a sense . . . . I swear I tried to step back and I couldn't as always . . .

[Schrödinger's hat]

If you have n doors and the host only opens one goat-door, the probability boost from switching (to one of the n-2 doors) does decrease.

 

If you have n doors, total and the host opens n-2 goat-doors, the probability boost increases.

 

The frequentist rationale:

There are n choices initially. For n-1 of these choices, the remaining door must have a car (because you picked a goat initially).

For 1 choice, the remaining door has a goat.

So given that you are about to pick a door:

P(Car|Switch) = (n-1)/n

P(Car|Stay) = 1/n

 

Which approaches 1 as the number of doors increases.

 

I can expand an analysis for arbitrary number of doors in the style of the Zomorodian paper if you wish. Modifying the asinine cretin's program to do so should also be fairly straightforward (change some 3s to ns and modify the revealing logic slightly).

The program I wrote and posted -- not so much. I wrote it to be more interactive (and thus it's slightly more complicated) and didn't plan it at all so it's rather a mess.

 

 

A hands on experiment can also be more convincing. I suggest the following refinement of my ace of spades example.

Shuffle a deck of 52 (or n where n>=3) cards.

Pick a card at random, give the rest of the deck to a friend (or pretend to be said friend).

Ponder on the following questions (and maybe follow along on the analogous steps of one of the derivations):

What is the probability you have the ace of spades?

What is the probability that your friend has the ace of spades?

 

Get your friend to sort through their 51 (n-1) cards, and put 50 (n-2) cards that are not the ace of spades in their shoe. Get them to hold the remaining card.

What is the probability that the ace of spades is somewhere about their person?

What is the probability that the ace of spades is in their shoe?

What is the probability that the ace of spades is in their hand?

 

Finally, given this information what is the probability that the ace of spades is in your hand?

 

What is the probability that you would be holding the ace of spades if you swapped cards?

 

Also, I apologise for the sober comment. It was somewhat insensitive.

Edited March 31, 2012 by Schrödinger's hat

[Xittenn]

Yeah thanks for the code cretin, something to play with. I'll most definitely have to take a coarse on statistics in the fall!

 

And no worries Schrödinger, I get a little crazy sometimes. My thoughts have never been trained and or fully developed so I often fall apart and wonder off into unusual places--it just sucks when it happens in front of people and during exams! More so exams . . . .

[the asinine cretin]

Also, fixed my applet to reflect this, and added a x1000 and random button

Edit, whoops: http://jsfiddle.net/uLzmY/3/

Cool. And hey, that's a nifty site. I wasn't aware of it.

 

Yeah thanks for the code cretin, something to play with. I'll most definitely have to take a coarse on statistics in the fall!

No problem. I wouldn't mind taking a course in statistics some time too. It's interesting.

[Schrödinger's hat]

Cool. And hey, that's a nifty site. I wasn't aware of it.

 

You can update and fork someone else's fiddles, too. The number in the url just increases when you save it.

It may do something fancy like ip checking and force you to fork, not sure

Edited March 31, 2012 by Schrödinger's hat

[Xittenn]

A recent comment made by the Hat as a reply to a blog post reminded me of this thread and my dissatisfaction with what most are considering good replies to a math question. So I Googled again, but this time with the more explicit search parameter of 'bayes's theorem monty hall problem.' I found a very nice little blog entry that very simply states the problem with, what is in my opinion, a proper mathematical solution. It is a very simple read and clearly delineates the problem as having a simple solution that most individuals with a grade twelve maths credit, or even less, will understand--mathematically. It very nicely does away with any hodge podge of information, which includes the idea that the state of the emotional relationship between show host and contestant need be considered--because honestly this isn't what most people are asking when they ask the question!

 

Bayes Theorem and the Monty Hall Problem

 

and also the mistake that I was making was in this step

 

We now calculate the various components:

 

P(Open(B) | Choice(A), Choice(C)) = 1

 

Because the host never selects the door you choose or the one with the prize.

 

P(Open(B) | Choice(A), Choice(C)) =/= 1/2

 

Statistics is not something I've spent a lot of time on . .. .

Edited April 19, 2012 by Xittenn

[dgb]

This is a trivialization . .. if I am in one of three groups there is a 2/3 chance that I have chosen a group that has both a car and a goat. If the second goat is destroyed either I beat the odds and am in the group that had a 2/3 chance or I am in the group that has a 1/3 chance not the other way round. If I then make a choice between two options I have a 50/50 . . . . and will remain so in my mind until I see a mathematical proof!

 

And not a 2/3 chance of winning the car, but a 2/3 chance of either having picked the car or of being in the group that the car is in . . . where the one third you picked the double goat!

 

 

The trick to understanding this problem is that it is a bit of a "cognitive illusion". The situation is engineered so that at the end you are faced with either a goat or a car; the host knows where everything is and ensures this is the case. The SEQUENCE of events is critical. Imagine you are blindfolded and sit at a table with two blue balls and one red ball. You pick one up in your right hand and then another is placed in your left hand such that you end up with one of each (non-random engineered situation). Think about how likely it is that the red ball is now in your right hand. It is in fact twice as likely that you picked up a blue ball at the start and therefore twice as likely that at the engineered endpoint the red ball has ended up in your left hand.

[anotherfilthyape]

You are thinking the problem must be solved mathematically but you are ommiting the fact that their is a higher chance that the game show host will not reduce the chances unless you got the car on your firts choice... This is the black swan fallacy, thinking that it is a matter of total chance...

[MonDie]

I haven't studied probability yet, but I think I thought of a good way to explain the problem. I haven't read the whole thread yet, but it doesn't look like anyone has tried this.

 

---

In the image below, the boxes decrease in size to show decreasing probability. For example, there is an equal chance that the contestant will choose any door. But if they choose the right door, there are two doors that could open. If they choose the wrong door, there is only one door that could open. So, the two possible outcomes after choosing the right door each have only half the probability of choosing the right door in the first place. On the other hand, the one possible outcome that can occur after choosing a wrong door is as likely as choosing that wrong door. For example, if door 1 is correct, there is a 1/6 chance that the contestant will choose door 1 and door 2 will open, and there is also a 1/6 chance that they will choose door 1 and door 3 will open. This is because the 1/3 chance of choosing door 1 (the correct door) is split into 1/6 portions when we are considering the two different events that might follow. However, there is a 1/3 chance that the contestant will choose door 2 and door 3 will open, and there is a 1/3 chance that the contestant will choose door 3 and door 2 will open. This is because there is only one door that can open after you choose a wrong door.

 

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In both of the following examples, door 2 was opened to reveal a goat.

 

The first map represents the viewpoint given in the problem. The contestant knows which door they chose, so they can use that information to deduce the possibilities. They chose door one, and door two was revealed to be a goat door. This leaves them with two lit up squares on the bottom (two possibilities). One of those possibilities has half the chance of having occurred than the other, which is why it is represented by a smaller box. This is the reasoning behind the 1/3, 2/3 split.

 

The second one represents a situation in which the first contestant chooses a door, but then that contestant is replaced by another contestant who knows the rules, but does not know which door the first contestant chose. The second contestant only knows that door two has been opened to reveal a goat. Based on the information this second contestant has, each door is equally likely ( 50/50 ). This is represented by the four lit up boxes at the bottom of their possibilities map.

 

These two viewpoints don't contradict one another. Each person calculates the best they can with the information they have, but the ultimate reality does not lie within their calculations. A door cannot be two-thirds right or half right; the right door is the right door irrespective of these imagined concepts.

 

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I thought of a better way I could have drawn it. I could have made the possible number(s) for each row lit. e.g. For the second map, I'd light up all 1&2's in row 1, all 12&3's in row 2, and all 2's in row 3. Then I could have changed the color of any paths that were completely lit up to show that nothing invalidated them as possibilities.

 

Edited June 27, 2012 by Mondays Assignment: Die

[GammaTheGreat]

This diagram was helpful to people in my freshman level critical reasoning class...

 

Door 1 Door 2 - Door 3 result if staying at door #1 -- result if switched

Car Goat Goat -- Car Goat

Goat Car Goat -- Goat Car

Goat Goat Car -- Goat Car

Note the outcomes, it explains the concept to people who aren't particularly probabilities-savvy.

Edited June 27, 2012 by GammaTheGreat

[MonDie]

That diagram is basically the same except that it takes into account less variables.

I made the prior diagram because I was trying to think about how the probabilities could be different from different perspectives. I was also trying to prove that my method of diagramming works by using it to get accurate probabilities.

Edited June 27, 2012 by Mondays Assignment: Die

[GammaTheGreat]

That diagram is basically the same except that it takes into account less variables.

I made the prior diagram because I was trying to think about how the probabilities could be different from different perspectives.

 

It seems that I skipped a page of the thread on accident. Thanks for letting me know. Yeah you're right, but I just put it there as a tool to make the original problem ridiculously simple (hopefully unnecessarily so). I would (really) hope everyone here can read your chart, but I did have some trouble explaining it to some of the students using a similar chart for my purposes, as some people are chart-tarded.

 

My mistake, and thanks.

Edited June 27, 2012 by GammaTheGreat

[MonDie]

This diagram was helpful to people in my freshman level critical reasoning class...

 

Door 1 Door 2 - Door 3 result if staying at door #1 -- result if switched

Car Goat Goat -- Car Goat

Goat Car Goat -- Goat Car

Goat Goat Car -- Goat Car

Note the outcomes, it explains the concept to people who aren't particularly probabilities-savvy.

I know that diagram was meant to be a simple explanation for simple people, but, now that I think about it, that diagram doesn't cover the concept fully. It glosses over how to evaluate the probabilities for each door by considering each variable (known or unknown), and it does so by reducing the choices from 1,2,3 to stay/switch. A critical viewer might think, "Well, I see that when the situation is repeated, it's better to repeatedly switch rather than repeatedly stay. However, what does it have to say about each particular instance? That is, the diagram shows that it's generally best to switch, but is there always more justification for switching? How would my choosing of a door impact the probabilities for each door in each instance?"

Edited June 27, 2012 by Mondays Assignment: Die

[Joatmon]

I know that diagram was meant to be a simple explanation for simple people, but, now that I think about it, that diagram doesn't cover the concept fully. It glosses over how to evaluate the probabilities for each door by considering each variable (known or unknown), and it does so by reducing the choices from 1,2,3 to stay/switch. A critical viewer might think, "Well, I see that when the situation is repeated, it's better to repeatedly switch rather than repeatedly stay. However, what does it have to say about each particular instance? That is, the diagram shows that it's generally best to switch, but is it always best to switch? How would my choosing of a door impact the probabilities for each door in each instance?"

 

Nobody is saying it's always best to switch. In fact you can expect to win a goat one time out of three if you do switch. So one time in three you can expect it to be a bad thing to switch. But two times out of three you can expect it to be a good thing!

 

 

[MonDie]

Nobody is saying it's always best to switch. In fact you can expect to win a goat one time out of three if you do switch. So one time in three you can expect it to be a bad thing to switch. But two times out of three you can expect it to be a good thing!

I was suggesting that the simpler explanation doesn't do enough to instill a secure understanding. I'll edit that.

Maybe I should read my probability book before talking more.

Edited June 27, 2012 by Mondays Assignment: Die

[ecoli]

I know that diagram was meant to be a simple explanation for simple people, but, now that I think about it, that diagram doesn't cover the concept fully. It glosses over how to evaluate the probabilities for each door by considering each variable (known or unknown), and it does so by reducing the choices from 1,2,3 to stay/switch. A critical viewer might think, "Well, I see that when the situation is repeated, it's better to repeatedly switch rather than repeatedly stay. However, what does it have to say about each particular instance? That is, the diagram shows that it's generally best to switch, but is there always more justification for switching? How would my choosing of a door impact the probabilities for each door in each instance?"

 

Choosing a door changes the probability, because the host opens a door with a goat. However, choosing a specific door doesn't change the prior probability since you don't have additional information.

 

It's always better to switch if you play the odds (like it's always better to not hit on 20 in blackjack), it doesn't mean you'll always be right,

 

think of it this way, if you play the game 100 times in a row, if you always switch you'll win 67 times. If you decide not to switch for any number out of those hundreds, the total number of trials that you win will almost certainly be lower (though you could calculate the exact probability of that if you were so inclined)

[GammaTheGreat]

As I explained, it is the diagram I used for a freshman level critical reasoning class. It elicits a sufficient understanding for their purposes. However, since you brought it up, its a relatively simple concept and its practical applications are limited. That being said, for my purposes, it's just a tool concept to help someone understand the larger concept for later learning.

 

Also, since part of the Problem is that you can't know if you should switch before choosing to do so/not do so, regardless of what lies behind each door in any situation, it's probably in your best interest to switch doors. The Problem doesn't contend that it will always turn out in your favor, only that, on average, its probably in your best interest to switch doors.

 

I might be misunderstanding what you're claiming, or maybe even the Problem on the whole, so please do elaborate. I'm here to learn! =]

 

EDIT: sorry to restate you, ecoli. I hadn't loaded your post yet.

Edited June 27, 2012 by GammaTheGreat

[MonDie]

I wasn't actually asking those questions, they were examples of questions. After all, this is a thread about why the problem is controversial.

Some of those concerns were my concerns as I was trying to solve the problem, especially the part about redefining the choices (1,2,3 vs. stay/switch). The concern was because even though I knew it was generally better to switch, it really bothered me when I was unable to define the probabilities in terms of the individual doors rather than stay vs. switch. I solved that with my chart.

There were also times when I wondered why the probabilities that generally apply over the course of many trials necessarily applied in each individual trial. I solved that with my chart too. There could be other situations in which the proportions of probability will change depending on the values of the variables (e.g. if this, 1:2 ratio of probability, if that, 2:3 ratio probability), but this is not one of those situations. This was the lesser concern of mine, although both concerns might have really been the same concern.

 

I would (really) hope everyone here can read your chart, but I did have some trouble explaining it to some of the students using a similar chart for my purposes, as some people are chart-tarded.

If your chart was like mine in the ways I am assuming, did you try writing fractions below the bottom boxes? For example, in the bottom row, the small boxes would have "1/18" below them and the medium boxes would have "1/9" below them because that is the probability of them being the final outcome when no variables are known. This would emphasize the half value of the small boxes compared to the medium boxes.

You could also put "1/9" by the middle row boxes and "1/3" by the top row boxes to further drive the point.

Edited June 28, 2012 by Mondays Assignment: Die

[GammaTheGreat]

In that case, I would understand if you wanted to solve for probabilities given that the assumptions change. I guess my revised position would be that I agree with those that don't see a legitimate controversy regarding the Problem - if you stay within the original assumptions (the rules Monty follows). You're correct (as you know) that the probability changes if the assumptions (rules) of the Problem changes.

 

Your point is now understood! I should have taken it upon myself to understand it more clearly from the beginning. Thanks for the added clarification though.

 

And as far as the chart goes, I should have said I utilized a similar format of chart - a similar structure. It was just a case of people needing more (and different) exposure and examples, rather than needing "better" examples. Sorry, I have been pretty unclear. Try not to hold it against me, I just started posting with you all a few days ago.

[MonDie]

Don't worry about it. I could have more explicitly stated that those were only example questions.

 

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I derived another rule from my diagram that seems counter-intuitive.

Assume you chose door one, and door two opened to reveal a goat. This is displayed in the top diagram of my previous image. One way we could explain the 1/3 2/3 split is by saying this, "If door one was right, the host could have chosen to open either door two or door three. However, if door three was right, the host could only open door two. Therefore, door three is more likely the right door."

 

Consider another example that applies this logic in an exaggerated way. People are sending two-digit numbers to a program. The program randomly chooses a person and displays their number. You type in "62," and the program displays "62."

We don't know the chances of it choosing you randomly. However, if it chose you, it would have to display "62." If it chose someone else, the chances of it displaying "62" would be 1/100. This makes it more probable that the program chose you and not someone else. This is the same logic, but it's exaggerated. The number displayed by the program is analogous to the door opened by the host.

Edited June 29, 2012 by Mondays Assignment: Die

[MonDie]

Interestingly enough, it all hinges on the fact that the game's host knows where the prize is. I ran a simulation on this years ago, and it turns out to be a 50/50 chance if the host didn't know the location of the prize. If he didn't know and opened the door with the prize, there would be that "waah waah waah" sound, and the host would voice condolences for the contestant picking the wrong door.

 

Indeed, assume you picked door one, the host opened door two, AND door two opening didn't trigger the "waah waah waaaah."

 

Edited June 29, 2012 by Mondays Assignment: Die