Talk:Monty Hall problem

From Wikipedia, the free encyclopedia
Jump to: navigation, search
Former featured article Monty Hall problem is a former featured article. Please see the links under Article milestones below for its original nomination page (for older articles, check the nomination archive) and why it was removed.
Main Page trophy This article appeared on Wikipedia's Main Page as Today's featured article on July 23, 2005.
          This article is of interest to the following WikiProjects:
WikiProject Statistics (Rated B-class, Top-importance)
WikiProject icon

This article is within the scope of the WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page or join the discussion.

B-Class article B  This article has been rated as B-Class on the quality scale.
 Top  This article has been rated as Top-importance on the importance scale.
 
WikiProject Mathematics (Rated B-class, Mid-importance)
WikiProject Mathematics
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
B Class
Mid Importance
 Field: Probability and statistics
One of the 500 most frequently viewed mathematics articles.
A selected article on the Mathematics Portal.
WikiProject Game theory (Rated B-class, Mid-importance)
WikiProject icon This article is part of WikiProject Game theory, an attempt to improve, grow, and standardize Wikipedia's articles related to Game theory. We need your help!
Join in | Fix a red link | Add content | Weigh in
B-Class article B  This article has been rated as B-Class on the quality scale.
 Mid  This article has been rated as Mid-importance on the importance scale.
 
WikiProject Television Game Shows (Rated B-class, Low-importance)
WikiProject icon This article is within the scope of WikiProject Television Game Shows, a collaborative effort to improve the coverage of game shows on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
B-Class article B  This article has been rated as B-Class on the project's quality scale.
 Low  This article has been rated as Low-importance on the project's importance scale.
 

XKCD[edit]

I don't know if you folks have seen this before, but I just ran across it for the first time. http://www.xkcd.com/1282/  :) --Guy Macon (talk) 09:05, 27 February 2015 (UTC)

He was lucky in that he won the right goat ;) Martin Hogbin (talk) 10:13, 27 February 2015 (UTC)
Love will find a way... :) --Guy Macon (talk) 10:44, 27 February 2015 (UTC)

The host opening a door is irrelevant[edit]

I wonder if this might be considered original research, but I've come to the realisation that the step where the host opens a door is irrelevant to the probability of getting the car:

Say, I have 3 boxes, only one containing the car. I pick one at random. There's 1/3 chance that I picked the box with the car (in which case, I should stick) and 2/3 chance that I didn't (in which case, I should swap).

Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities.

Is this something worth noting in the article? Thanks, cmɢʟeeτaʟκ 19:40, 24 March 2015 (UTC)

Alas, your conclusion that "Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities" is incorrect. It actually makes switching a better choice (2/3 chance of winning the car) and not switching a worse choice (1/3 chance of winning the car). BTW, you are in good company. Nearly 1,000 people with PhDs (including at least one world-renowned mathematician) came to the same wrong conclusion that you did. --Guy Macon (talk) 20:35, 24 March 2015 (UTC)
No I can see his point, it's the one that's usually referred to as combining the doors. The opening of the door is the mechanism by which the 2/3 option is collapsed into one choice (the remaining door). However the two 1/3 probabilities of the two unchosen doors are collapsed it's the same. SPACKlick (talk) 20:38, 24 March 2015 (UTC)
Perhaps his description of the game before any doors are opened ("There's 1/3 chance that I picked the box with the car (in which case, I should stick) and 2/3 chance that I didn't (in which case, I should swap") misled him. Yes, at that point there is a 2/3 chance he choose wrongly, but swapping at that point (which is not what happens in the real game) would only let him choose one of the remaining two doors, and thus there would still be a 1/3 chance of winning. --Guy Macon (talk) 20:47, 24 March 2015 (UTC)
Yeah it depends on what he's thinking there. If he's thinking a combined doors style version where "Switch" at that point means "get any cars behind those two doors" then he's right but it's been done. If he hasn't thought about the fact there are two doors left then maybe he's been misled. SPACKlick (talk) 21:19, 24 March 2015 (UTC)

I think that the point that cmglee is making is that the hosts opening of door that is known to hide a goat does not change the probability that the originally chosen door hides the car. This has been a much discussed point in the past. Is that fact obvious? Martin Hogbin (talk) 08:36, 25 March 2015 (UTC)

"Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities" looks like an imprecise misrepresentation of the "intended clean paradox": cmɢʟee, did you mean "before my ultimate decision, DELIBERATELY showing a goat, and not the car" (as per the intended clean paradox), or did you intend to say "before my ultimate decision, COINCIDENTALLY showing a goat, and not the car"? (see University of California San Diego, Monty KNOWS Version and Monty DOES NOT KNOW Version).
The intended clean paradox says "Before my ultimate decision, DELIBERATELY showing a goat, but NEVER the car". It follows that the chance of the first chosen door still remains 1/3, and the chance of both combined unchosen doors still remains 2/3, yes. But this chance of both combined unchosen doors of 2/3 has collapsed to the still closed unchosen door that the host offers now as an alternative for the ultimate final decision. Gerhardvalentin (talk) 11:32, 25 March 2015 (UTC)

A couple ways to clarify the meaning of the odds[edit]

To look at it from a different angle, the contestant has a 2 in 3 chance of choosing a door with a goat. That means the remaining pair of doors have only a 1 in 3 chance of containing two goats. When Monty reveals a goat from the remaining two doors, then the other door will have a goat one-third of the time. Therefore, the contestant halves the odds of getting a goat when switching doors, the same of course of doubling the odds of getting a car.

Note: this also hinders the question as to the importance of Monty Hall revealing a goat.

People get hung up on the idea that the contestant is asked to make a second choice among the remaining two options, leading them to think their odds at that point are 1 in 2. There are two doors. One has a goat, and the other has a car, so a random choice should get the car half the time, they think. It's important to point out that the problem revolves around an initial selection and actions based on that selection. There are two doors, but a car needn't appear behind both of them with the same frequency.

The choices appear to be equal, but they are not because Monty Hall acts on the contestant's choice. His behavior relies on the contestant's initial choice. This is not the same as being offered a choice between two equivalent choices. What happens in a seemingly similar case when there are six contestants in a singing contest and one will be selected to go home? If all things are equal, each contestant puts his/her odds of going home at 1 in 6. But as the MC declares one contestant after another to be safe, the remaining contestants no longer rely on those 1 in 6 odds. Had I placed an initial bet that one of the final two contestants remaining would go home, I would no longer view the odds of my winning that bet as 1 in 6, but I'd think I had a fifty-fifty chance at that point. I wouldn't imagine that the other contestant had a 5 in 6 chance of going home that day just because I made an initial random choice of a contestant. The difference in these scenarios is that the judges and the MC have no knowledge of my bet and take no actions based on it; whereas, Monty preserves the game show contestant's choice when he acts. — Preceding unsigned comment added by Summers999 (talkcontribs) 13:21, 20 April 2015 (UTC)

It is always good to approach the problem from different angles. Different ways of looking at the problem work for different people. It does seem though that once you understand why the answer is 2/3 it is very hard to put yourself back into the mind of someone who does not understand the correct solution.
However, we cannot put everybody's preferred solution or explanation into the article because WP is based on what is said in reliable sources. If you can find a good source that gives your explanation then it could be added to the article. Martin Hogbin (talk) 08:19, 21 April 2015 (UTC)

Are the odds ever 50-50 ?[edit]

It's been established that the contestant and Monty have produced a pair of doors containing a car and a goat that do not have the same odds of producing a goat. Yet, on Marilyn vos Savant's website where she discusses this teaser, she perhaps yields too much ground to the academic furor that confronted her. She made this concession:

"Suppose we pause at that point [when two doors remain], and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she’ll randomly choose the one with the prize are 1/2, all right. But that’s because she lacks the advantage the original contestant had—the help of the host." [1]

But are the chances 50%? Probabilities don't care much for who does the choosing, and neither the contestant nor the green woman has insider information about the whereabouts of the car behind the doors. The contestant can know that the odds of finding a car behind the switched door are 2/3, but the odds don't change just because someone doesn't know the odds. If I know that four sides of a die are red and two are blue before I paint over the sides, the odds of someone who has never seen the die rolling red are not 1/2 due to her unawareness.

If the odds of randomly selecting a car among the two blind choices are 50-50 for anyone other than the contestant, then they must be 50-50 for everyone. We know that they are not, so the green woman approaches a stacked deck, so to speak. Otherwise, we end up saying that all odds between situations with two outcomes are 50-50 so long as we know nothing about the underlying conditions. That's not correct.

Vos Savant refers to the help of the host as being a missing ingredient, but that help actually settled the odds prior to his final selection, and it must've done so for anyone choosing between the two doors. We are now convinced that this problem is a matter of frequency of appearance of the car behind particular doors after particular actions have settled the odds, so vos Savant needn't have made any concessions to the contrary.

Thoughts? — Preceding unsigned comment added by Summers999 (talkcontribs) 14:24, 25 April 2015 (UTC)

The question is what odds are you talking about? If you pick randomly between two choices the odds of your choice being the correct one are 50-50. Note how I worded this - the odds we're talking about are the odds of "your choice". If we roll your red/blue die the odds are 4/6 that it ends up red and 2/6 that it ends up blue. If you roll it 100 times and I randomly guess red or blue, about 50 times I'll guess red and 50 times I'll guess blue. Of the 50 times I guess red I'll be right about 4/6 of the time, so about 33 times. Of the 50 times I guess blue I'll be right about 2/6 of the time, so about 17 times. Altogether I'm right about 50 times out of my 100 guesses, i.e. 50-50. If I don't know 4 sides are red and 2 are blue, my odds of correctly guessing are 50-50 whatever the "actual" odds may be because I have no choice but to randomly guess. If I always guess red (because it's my favorite color), I'll be right about 4/6 of the time, but now we're not talking about the odds of a random guess but the odds that the die ends up red (which is a different question).
If you want to talk about this further let's move the discussion to the Talk:Monty Hall problem/Arguments page. -- Rick Block (talk) 16:40, 25 April 2015 (UTC)
I'm only referring to a single selection of unequal odds. Namely, if the contestant or a green alien chose one of the remaining doors, it would still be better to choose a particular door. And this is true whether or not someone asked to select knows of the contestant's original choice. Even two of three random selections will yield a car from the alternative door, just as blind rolls of the die I described will yield red twice as often as blue. Summers999 (talk) 21:32, 25 April 2015 (UTC) user:Summers999
Like I said - if you want to talk about this further let's move it to Talk:Monty Hall problem/Arguments. -- Rick Block (talk) 04:09, 26 April 2015 (UTC)