Talk:Monty Hall problem

The trick

lies in that the host will always open a door with a goat, If the host opens one of the remaining doors at random and reveals the car (by accident) the whole paradox falls to pieces 184.88.37.235 (talk) 13:38, 9 December 2011 (UTC)

Is that not already clear enough in the article? ~ Ningauble (talk) 16:37, 9 December 2011 (UTC)

the article seems pretty adamant that the host will always pick a goat door and not that he picks a random door that may or may not contain the car 184.88.37.235 (talk) 20:45, 9 December 2011 (UTC)

No, this doesn't only "seem" so, but it is explicitely said so with intent, as this is the basic of the question "Is it to your advantage to switch your choice?"  Gerhardvalentin (talk) 21:15, 9 December 2011 (UTC)

That's not "the trick," as you call it, that's the entire point of the problem, and is made clear in every solution presented in the article — Preceding unsigned comment added by 108.45.70.208 (talk) 06:26, 19 January 2012

R code for simulating the problem

Hello All.

I code faster than I read, so decided to get my head around the problem by simulating it in R. This is 50 lines of messy code (apologies, I'm sure it could be improved!). I'm not sure if this is the kind of thing moderators would feel it would be useful to include, but I thought I may was well offer it up! Hope I've posted correctly.

Best,

--BmrE2 (talk) 18:27, 11 December 2011 (UTC)

#Painfully verbose simulation code moved 2011-12-13
#An opaque simulated replication of vos Savant's answer in one line for RockAndStones:
source("http://monty-hall-r-simulation.googlecode.com/files/Monty%20Hall%20Simulation%20in%20R.r");win.props

@BmrE2, thanks, the code is fantastic. When you optimise it to three lines, please just post a weblink. Users of this page are mainly involved in the discussion of logical aspects of the paradox and its variations, rather than checking if a particular program works.RocksAndStones (talk) 22:15, 11 December 2011 (UTC)

It could be done as a one-liner in APL, but this is not germane to the purpose of this talk page, which is for discussing changes to the Monty Hall problem article itself. ~ Ningauble (talk) 18:21, 12 December 2011 (UTC)

Thanks for the feedback, both. I agree that it wouldn't have a useful place in the article, apologies for filling talk pages with more junk! My reaction to this article seems pretty common... --BmrE2 (talk) 01:46, 13 December 2011 (UTC)

Intuition suggestion

The article seems to focus a lot on pitting math and experimentation against a failure of intuition, but there's another tactic to try too: elicit a stronger (and correct) intuition and pit it against the false one. In the example of a million doors, I'm thinking we can ask the reader to, for example, "Just imagine the host marching down the line of a million doors, opening them one by one...and then he skips one! That door seems very suspicious - why did Monty skip that door? It is very likely that Monty is skipping that door because the car is behind it. So obviously, the player should switch to that door. The only way the player would lose is if he'd picked the car already, and the host was skipping any old door - and that's very unlikely."

What do people think of this idea? —AySz88\^-^ 22:34, 28 December 2011 (UTC)

To clarify, this sort of idea is referenced several times (in "million doors" and "N doors"), and the mathematics is described, but it's never presented in a way that appeals to intuition. —AySz88\^-^ 23:26, 28 December 2011 (UTC)

i like it, as so much of this problem is the result people ignoring math and going with their faulty intuition. what you're presenting is similar to expanding the "missing dollar" riddle, so that people would be less susceptible to that faulty intuition — Preceding unsigned comment added by 108.45.70.208 (talk) 06:29, 19 January 2012 (UTC)

I am in favour of more way to help readers to understand this problem, however, they should be based on reliable sources. That does not mean that we must copy exact wording from sources but that we should give explanations like those in sources. The proposed explanation is essentially that given by vos Savant. Martin Hogbin (talk) 09:52, 19 January 2012 (UTC)

I'm a bit confused about how your comment applies. Are you saying that my explanation is too similar to something that von Savant wrote? I actually wrote it myself, from the information currently in the article; I didn't refer to any of von Savant's articles (though I wouldn't be surprised if she took a similar approach at some point). Could you point me to exactly what you say it's similar to? —AySz88\^-^ 05:31, 26 February 2012 (UTC)

The "million door" analogy is from vos Savant's very first column about this problem, see http://www.marilynvossavant.com/articles/gameshow.html (her first column is the initial question plus the response in red text). On the other hand, note Ruma Falk's article which (at the very end) says (of vos Savant's million door analogy) "There can hardly be a more smashing argument. Only that its truth depends on the understanding that if the prize is behind door no. 1 [the door the player initially picks], the host decides by a fair draw which of the remaining 999 999 doors to leave closed. If, however, you know that for some reason or other that the host is determined to leave door 777 777 closed [the door the host skips as he opens all the rest], whenever possible, observing that situation will render that door as likely to hide the prize as door no. 1. You would then have no reason to hurry to switch". The point here is that how the host decides what door to open, in the event that the player has initially (however improbably) chosen the door hiding the car is critically important. You're never worse off switching, and if you decide before seeing which door the host opens that you're going to switch you'll win on average (n-1)/n times (where n is the number of doors) - but given the knowledge of which door the host opens, your chance of winning by switching depends on how committed the host is to leaving the door that he doesn't open closed. If we call this p (he'll leave this door closed with probability p in the case where you happen to initially pick the car), your chances of winning by switching are 1/(1+p) - regardless of how many doors there are. In the 3-door version p is generally assumed to be 50% meaning if you initially pick the door hiding the car the host opens either remaining door randomly - and your chance of winning by switching is therefore 1/(1+1/2) = 2/3. In the million door version, assuming the host randomly picks which door to leave closed if you've initially selected the car means p is 1 / 999 999 - so your chances of winning by switching switching are 1/(1+1/999999) = 999999/1000000. If you don't know what p is, your chances of winning by switching are somewhere between 1/2 and 1 - on average, (n-1)/n where n is the number of doors - but if you don't know what p is the bottom line is you don't know your probability of winning by switching. It's at least 1/2, might be as high as 1, and averages (n-1)/n where n is the number of doors - but it's actually 1/(1+p). -- Rick Block (talk) 07:47, 26 February 2012 (UTC)

Although our explanations should be supported by reliable sources there is nothing is WP policy that says that we must quote sources verbatim, in fact this is not the preferred approach.

I therefore suggest that AySz88 should propose some form of wording that he considers the most intuitive for discussion. We already have a reliable source that supports that solution and we are free to try to improve the wording. Bear in mind that what might seem very intuitive to one person might not seem so to another. Martin Hogbin (talk) 10:24, 26 February 2012 (UTC)

@Rick Block: Although your reasoning:

• "depending on p, the chance to win by switching must be within the range of at least 1/2 to even 1/1, without exception, and on average it is 2/3"

is mathematically sound and completely correct, the literature on MHP shows very clear that this cannot be of any relevance whatsoever for the actual question the MHP asks for, the actual decision to be made,

• as long as p is unknown. And, for the MHP, p is unknown.

And as long as p actually is unknown, p can be of effect and of relevance only in mathematics lessons on probability theory, but remains without any effect for the MHP, and can never have any influence on the right decision to be taken: definitely to switch in any case and without exception, based on and tightened to the only definitely "known" solid base, on the chance to win by switching of 2/3. To actually switch to the door offered.

The literature on MHP has shown that abundantly, meanwhile. Gerhardvalentin (talk) 10:49, 26 February 2012 (UTC)

Other simple solution - John Tierney

John Tierney in the New York times has an explanation - not sure if it's fancy enough to qualfy as a 'solution'.

When you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average.

If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.

Works for me ! Anyone else find that helpful ? --195.137.93.171 (talk) 17:31, 6 January 2012 (UTC)

OK Tierney is mentioned throughout the article and that NYT page is linked, but I feel it would be worth adding the above wording somewhere.

I feel it would also be worth adding a section on 'Consequences and applications'. I am particularly interested to find out how I can make money out of it ! --195.137.93.171 (talk) 17:37, 6 January 2012 (UTC)

Is this substantially different from the wording that starts the "Other simple solutions" section:

“

An even simpler solution is to reason that switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3 (Carlton 2005).

”

Just curious how you think Tierney's wording differs. -- Rick Block (talk) 19:46, 6 January 2012 (UTC)

This is a typical case of overcomplicating a problem. One of the doors is always removed. It is always a goat. It becomes a non-factor. This means that there is NO DIFFERENCE if the door is removed BEFORE OR AFTER THE CHOICE! It leaves the contestant with a 50/50 chance of having picked the car on the first try. — Preceding unsigned comment added by MrShemp (talk • contribs) 00:26, 12 January 2012 (UTC)

Well - every single reliable source says the chances are 1/3:2/3. I'd be happy to try to help you see why, but this is not the place. If you're interested, please start a thread on the talk:Monty Hall problem/Arguments subpage. -- Rick Block (talk) 01:38, 12 January 2012 (UTC)

MrShemp, quite simply your logic and math are wrong. if door 2 contains a goat, and this is revealed to you before you make your choice, you definitely would not choose door 2. if door 2 contains a goat but this is not revealed to you, you may still pick door 2.— Preceding unsigned comment added by Bpjoyce10 (talk • contribs) 06:37, 19 January 2012

Totally overthought this problem

There's no need for all these mathematical proofs. If you stick, you have a 1/3 chance of picking the prize. If you switch, however, you are trying NOT to pick the prize. By switching when you do not win the prize, you will win every time. This means a 2/3 chance of winning. — Preceding unsigned comment added by 94.7.196.157 (talk) 01:23, 21 February 2012 (UTC)

You are quite right, of course, and you are not the first person to point this out. Martin Hogbin (talk) 09:38, 21 February 2012 (UTC)

Yep, it really is very simple. Yet the majority of people get it wrong when they first encounter it. That is what makes a good riddle: the answer is only obvious when you see it. And this is what makes the literature, and this article, so overwrought: explaining the obvious is no simple matter. ~ Ningauble (talk) 20:14, 21 February 2012 (UTC)