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Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks! |
Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks! |
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:Is the earlier explanation, in the section titled "Why the probability is 2/3", more clear to you? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 01:17, 18 July 2007 (UTC) |
:Is the earlier explanation, in the section titled "Why the probability is 2/3", more clear to you? -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 01:17, 18 July 2007 (UTC) |
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Okay, I've spent entirely too much brainjuice on this. It probably has one or more glaring errors. But it's the algebra-ized version of the way I view the problem in my head (which has more to do with colors than with symbols; go figure). That said, I'm not so good at this whole color→algebra thing, so please fix it up: |
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{| border=1 |
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|+Probabilities in an ''n''-door problem with ''d'' non-contestant-selected doors still closed |
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| || You select a winning door || Your door is now a winner || Remaining Door(s) contain a Winner || You've already lost |
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|- |
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| Host picking randomly || <math>\frac{1}{n}</math> || <math>\frac{1}{n}</math> || <math>\frac{d}{n}</math> || <math>1 - \cfrac{n - d + 2}{n}</math> |
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|- |
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| Host picking only losing doors || <math>\frac{1}{n}</math> || <math>\frac{1}{n}</math> || <math>\cfrac{n - d}{n}</math> || <math>\varnothing</math> |
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|} |
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<span style="font-family: monospace">[[User:Jouster|Jouster]]</span> (<span style="font-size: smaller; background: black;">[[User Talk:Jouster|<span style="color:white">whisper</span>]]</span>) 23:48, 18 July 2007 (UTC) |
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== Choice? or Strategy? == |
== Choice? or Strategy? == |
Revision as of 23:50, 18 July 2007
Please note: The conclusions of this article have been confirmed by experiment |
There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. If you find the article's arguments unconvincing, then please feel free to use the space below to discuss improvements. |
Monty Hall problem is a featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Wikipedia community. Even so, if you can update or improve it, please do so. | ||||||||||||||||
This article appeared on Wikipedia's Main Page as Today's featured article on July 23, 2005. | ||||||||||||||||
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Current status: Featured article |
Mathematics FA‑class Low‑priority | ||||||||||
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Archives |
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Archiving notes
I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)
- I've done similarly to produce Talk:Monty Hall problem/Archive3. In keeping with Wile E. Heresiarch I moved the page so the edit history is with the archive page, and copied back the current (March 2006) discussions. └ UkPaolo/talk┐ 13:04, 10 March 2006 (UTC)
- Likewise for Talk:Monty Hall problem/Archive4. Gzkn 06:15, 27 December 2006 (UTC)
- Talk:Monty Hall problem/Archive5 archived via pagemove.--Father Goose 05:49, 22 May 2007 (UTC)
- Likewise for Talk:Monty Hall problem/Archive4. Gzkn 06:15, 27 December 2006 (UTC)
The new images
I'm all for racial harmony and all, but why do the images include a bald black man as the player? Could we be using a more abstract, symbolic face? (I know I know, I'm a racist because I wouldn't have noticed if it were a white guy. Granted, but I think the question is still valid.) --P3d0 02:00, 3 May 2007 (UTC)
- Cause the player is Curly Neal. Perhaps I should've gone with Mr. T?--Father Goose 05:11, 3 May 2007 (UTC)
Deal or No Deal
The two mentions of Deal or No Deal contradict one another: under 'Sequential Doors' is an explanation of the critical difference between the NBC Prime Time SMASH HIT and the Monty Hall problem as stated, but under the 'History of the problem' heading, Deal or No Deal is essentially considered to be a minor variant with the same sorts of conclusions. The numbers directly contradict one another as well. Though I'm fairly positive the first description is accurate, I'll refrain from editing for now so people can argue and yell and scream and then somebody smarter than myself can fix it. I also kinda like the fact that both are in this article, as it seems the Monty Hall problem's best contribution to society is watching people flailing about entirely confused yet certain that they're right in the face of contradicting evidence. 66.188.124.133 17:32, 16 May 2007 (UTC)
- I think there might be a valid place in the article for a section comparing and contrasting Deal or No Deal with the Monty Hall problem. They differ (and DOND was incorrectly analyzed in the section that was recently deleted) because the cases removed from play are chosen by the contestant, not the host of the show, and are chosen without knowledge of which case holds the big prize; thus, it is possible (and common) for the highest prize to be eliminated through the case-opening during the game. Thus, if it gets down to the last two cases, and one of them still holds the million dollars, then the odds are 50-50 for the contestant whether he/she keeps the original case or swaps it. *Dan T.* 19:49, 17 May 2007 (UTC)
DEBUNKED!!
The problem with this is that the 3 scenarios are actually 4.
For the first scenario where the person picks the car it is listed as host showing "either Goat A or B". Actually these are two different scenarios:
Scenario 1: Contestant picks car, Host shows Goat A, Contestant switches, Contestant LOSES
Scenario 2: Contestant picks car, Host shows Goat B, Contestant switches, Contestant LOSES
Scenario 3: Contestant picks Goat A, Host must show Goat B, Contestant switches, Contestant WINS
Scenario 4: Contestant picks Goat B, Host must show Goat A, contestant switches, Contestant WINS
You can see that switching yields the expected 50% success.
It is rather alarming to me that this is missed by experts. I believe Quantum computing falls into this same "smoke and mirror" science. —The preceding unsigned comment was added by 143.182.124.2 (talk • contribs).
- Mindful of the note at the top of the page, the key is that scenarios 1 or 2 occur 1/3 of the time, while scenario 3 and 4 occur 2/3 of the time. The decision tree from the article lays out all four options. As an aside, however, the talk page of an article is not really the place to discuss the topic of the article, only to discuss improvements. A better place to ask questions is reference desk. Thanks, --TeaDrinker 19:33, 5 June 2007 (UTC)
- This objection has been brought up twice recently, so I think it's worthwhile to specifically note and address it in the article. I've made an attempt to do so.--Father Goose 20:18, 5 June 2007 (UTC)
- I've reverted this, tweaking a single word in the original that may help clarify. The detailed decision tree analysis is already presented not too far away. -- Rick Block (talk) 02:45, 6 June 2007 (UTC)
- Now that I think about it, a different, possibly better way to address it is to change my diagram so that there's two panels in the upper right corner; switching to Goat A or switching to Goat B. I'll see if I can pull this off without making it ugly.--Father Goose 05:54, 6 June 2007 (UTC)
Re: new B+ rating
Might I ask for more specificity as to what this article needs in order to become an FA? It's not clear how to act upon the suggestions left in the Mathematics rating box.--Father Goose 02:31, 23 June 2007 (UTC)
- Well, the article is a Featured article, so I think pretty much by definition it has to have a FA rating. -- Rick Block (talk) 02:41, 23 June 2007 (UTC)
- The comments are at Talk:Monty Hall problem/Comments. -- Rick Block (talk) 03:33, 23 June 2007 (UTC)
- Yes, that's what I was referring to. It's not clear how to act upon them. "Might be too long" is extremely vague, and the other comment seems to be demanding redundancy.--Father Goose 06:19, 23 June 2007 (UTC)
- Many thanks to Rick for his constructive response to my concerns, and for signing the current rating with a good comment. Geometry guy 21:14, 23 June 2007 (UTC)
- Yes, that's what I was referring to. It's not clear how to act upon them. "Might be too long" is extremely vague, and the other comment seems to be demanding redundancy.--Father Goose 06:19, 23 June 2007 (UTC)
- The comments are at Talk:Monty Hall problem/Comments. -- Rick Block (talk) 03:33, 23 June 2007 (UTC)
From my talk page, but it makes more sense to comment here:
- Hello and thanks for the curiosity and comments. Yes, I did see the star, but on the grading scheme, FA-Class is for articles which have "received featured article status after peer review, and meet the current criteria for featured articles". I came to this article because the maths rating was not signed and dated, so I looked at the article and read the recent FAR. In my view the latter got side-tracked by irrelevant inline citation arguments and I don't think the article should have passed: in fact I doubt it would survive a good article review at the moment. Now I can't sign a rating I don't agree with, so I changed it, and added my comments to the rating. I do actually quite like the article (it is great for the portal, for example), and the most obvious flaw is pretty easy to fix: see WP:LEAD.
- However, this is just my opinion. If someone else believes that the article does currently meet the FA criteria, they are of course free to uprate the maths rating and replace my comments and signature by their own. Geometry guy 10:47, 23 June 2007 (UTC)
The current lead is a teaser, which is great for a magazine, but not for an encyclopedia. I'm sorry that my other concerns are vague, but it seems that other editors believe that this article meets FA standards, in which case please will someone replace my comment and signature by their own. Thank you! Geometry guy 10:56, 23 June 2007 (UTC)
- The lead used to include the solution, but there was an objection to this (to the extent that for a while there was a "spoiler" warning). I've been viewing the current version as a exception to WP:LEAD based on the desire of some readers to not have the solution displayed in the lead. The recent FAR was apparently OK with this. We could of course change it back. Any other opinions on this? -- Rick Block (talk) 17:00, 23 June 2007 (UTC)
- Yes I noticed that in the FAR the view was expressed that information should be taken out of the lead: certainly detail should not go in the lead, but the lead is supposed to be able to stand alone; indeed I believe some fixed editions of WP will contain only the lead for some articles. As regards including the solution, I guess one has to seek guidance from WP:SPOILER. My view is that at least the problem should be summarized briefly. Geometry guy 17:41, 23 June 2007 (UTC)
- Anyone have any objections restoring the lead to approximately this version? -- Rick Block (talk) 20:33, 23 June 2007 (UTC)
- My feeling is that this is the kind of precise formulation of the problem that SandyGeorgia wanted removed from the lead at the last FAR, and that it would be better to summarize (approximately) the problem rather than state it in full. Geometry guy 21:21, 23 June 2007 (UTC)
- At one point there was a summary statement of the problem, but it was constantly wordsmithed to include this or that constraint on the host's behavior or to avoid this or that ambiguity. Including the problem statement as a quote was a device to end the wordsmithing. The other problem with summarizing the problem statement is that the "correct" solution depends on the detailed constraints assumed or placed on the host's behavior. To conclude switching results in a 2/3 chance requires the host to always offer the choice to switch, to have knowledge of what's behind each door, and to always open a "goat door". Without including these constraints, any solution offered in the lead can be argued to be incorrect (and people do argue this, believe me). If anyone has a suggestion for an unambiguous, but summarized, statement of the problem please speak up. Short of that, quoting the Parade version seems to me like a reasonable approach (it's only 72 words). -- Rick Block (talk) 22:29, 23 June 2007 (UTC)
- I think moving the Parade version into the lead is a good approach. It doesn't look like the article would need to be rewritten much at all to move it in there and still present the information in an almost identical way.--Father Goose 02:08, 24 June 2007 (UTC)
- Here's my suggestion. -- Rick Block (talk) 02:55, 24 June 2007 (UTC)
The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. A widely known statement of the Monty Hall problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Since there is no way for the player to know which of the two unopened doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3.
The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, when Marilyn vos Savant offered the problem and the correct solution in her Ask Marilyn column in Parade, approximately 10,000 readers, including several hundred mathematics professors, wrote to tell her she was wrong. Some of the controversy was because the Parade statement of the problem fails to fully specify the host's behavior and is thus technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.
- That looks very good, and it's got an excellent concise explanation of the solution. That might run into the "spoiler" objection again, but I'm not sure this article should ever be considered a suitable place to first hear (and try to solve) the riddle anyway.--Father Goose 05:03, 24 June 2007 (UTC)
- I agree. I think if anyone complains, they can be pointed towards WP:SPOILER, which states "Concerns about spoilers should play no role in decisions about the structure or content of an article, including the article's lead section", and WP:LEAD, which states "The lead should not "tease" the reader by hinting at but not explaining important facts that will appear later in the article." Taken together, I would say that is pretty decisive! Geometry guy 12:02, 24 June 2007 (UTC)
- Not to mention: "Spoilers and spoiler warnings should not be used in articles on non-fictional subjects" from WP:SPOILER. --P3d0 19:36, 11 July 2007 (UTC)
About the length of the article
I found another exchange in the FAR which resonates a little bit with my reaction to the article and may help clarify the vaguer part of my comment. I quote:
- Comment: I am of the opinion that the article could be cut in half, and its quality would dramatically improve. I am aware that not everyone shares this opinion. But in its current state, I find it overwhelming, directionless, and confusing. - Abscissa 02:49, 10 January 2007 (UTC)
- Abscissa has a point. The article contains several explanations of the same result, and many of them could be struck. I suspect, however, that one will make sense to one reader, and another to another reader, so I'm not sure which ones, if any, to remove. This should be taken to the article talk page. Septentrionalis PMAnderson 03:07, 10 January 2007 (UTC)
I wouldn't go as far as Abscissa, but it is not dissimilar from my reaction. Anyway, this exchange seems to have got lost in a sandwich between the arguments about inline citation that opened the review, and the impressive copyediting drive that ended it. I couldn't find it in the talk archive. Did anything come of it? Geometry guy 17:41, 23 June 2007 (UTC)
- Would it be completely unreasonable to determine the pedagogy at work in each of the explanations, and make subpages for each one? Or, better yet, to select the consensus-best explanation of the problem, and relegate the others to an "alternate explanations"-type page? Jouster (whisper) 18:27, 23 June 2007 (UTC)
- I think the biggest problem is not article length but lack of structure. The "Aids to understanding" section lumps together too many things: fundamental explanations of the problem (Bayes, "Why the problem is 2/3", and the decision tree); explanations which aren't so much fundamental as alternative ways of viewing the problem; and an orphaned "Sources of confusion" section. Can we split it into "Fundamental explanations" and "Alternative ways of viewing the problem"? (Something like that.) "Sources of confusion" repeats a few things found elsewhere in the article; most of it can probably be merged with that stuff.
- At an absolute minimum, "Combined doors" and "Venn diagram" can be merged together: they say nearly the same thing.
- --Father Goose 02:40, 24 June 2007 (UTC)
- I wrote that comment that was quoted above. I still hold that opinion. My brother and I were laughing at the "quantum version" of the problem. Who is the target audience for that? At least people have given up trying to convince us that this is a problem of game theory... we (editors of this article) hold a place amongst the lamest edit wars on Wikipedia. - Abscissa 19:30, 4 July 2007 (UTC)
A flaw in the variations?
I am just wondering if it is a flaw in one of the Variations Other host behaviors “The host does not know what lies behind the doors, and the player loses if the host reveals the car.”
The answer is allegedly “The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.” Shouldn’t switching here also result in a 2/3 probability of victory (since it is purportedly the same result (so far) as in the original Monty Hall problem)? --85.164.95.143 14:15, 9 July 2007 (UTC)
- We may safely discount what the host does here, since he's choosing a door at random. (Though I think that description might need some work, or be outright misleading. But I just woke up, so I could be out of my mind.) The host choosing a door at random is the key to the whole thing, really—if the host is running off of the same set of information we have (namely, none), then we have a straight 50/50 shot at picking the right door amongst the two that remain. We haven't "narrowed the probability field", if you will, at all through our (and the host's) earlier actions. Jouster (whisper) 18:28, 9 July 2007 (UTC)
I believe as well there is a flaw here. Let's extend the problem to 100 doors the same way we reason the standard problem. The host luckily opens 98 goats in a row. But once this has happened, certainly you would switch, because even without the knowledge of the host, you only had a 1/100 chance of landing on the car in the first place, which has not changed. Your odds of WINNING this sort of game are of course lower because of the probability of the host opening the door to the car causing a loss. However, after he opens a goat, even if it is random, it does not affect the probability of you having picked the car in the first place, and thus the advantage of switching. Again, the 100 doors intuition here seems to support this argument. — Preceding unsigned comment added by 151.190.254.108 (talk • contribs)
- The difference is whether the host acs knowingly or by random chance. In the 100 door version if the host knows then any player is still in the game after 98 doors have been opened and still has a 1/100 chance that the originally picked door hides the car. If the host doesn't know, 98/100 times the player loses (because the host randomly opens the door with the car). The player picked a random door. The host opened 98 random doors. Whether the player picked the car door is now a 50/50 chance, made so by the host's random choices. In the "host knows" case the only random event affecting whether the player wins the car is the player's initial choice. In the "random host" case, the host's actions affect the player's chance of winning as well. -- Rick Block (talk) 13:54, 17 July 2007 (UTC)
- No, because there are only two base circumstances involved in the 100-door variation: you have a goat, or you have a car. All other chances are randomly distributed.
- Let's try another way--suppose we ran this test as, "Contestant picks door #1, hosts opens doors #2-#99, switch or not?", 100 times, one time with a car in each of the 100 positions. 98 times out of 100, the game would fail because the host would reveal a car. One of the remaining two times, switching would lose, as the car is in position #1. The remaining time, switching would win, as the car is in the last position.
- If the host has no knowledge of the doors' contents, it's all random. Jouster (whisper) 14:00, 17 July 2007 (UTC)
- Perhaps I'm confused as to what probability we are talking about. The probability of winning the game when the host doesn't know anything, with the rules making you lose if he reveals a car, is in fact 0.5 for the 3-door experiment. I was talking about the conditional probability of winning by switching given that you already have not lost from the host opening doors.
- Let's say we run your example. What I am talking about is only that situation when the 98 doors the host opened revealed goats. It is in this case that I feel it is advantageous to switch doors, because your original probability of picking a goat was 0.99. The method by which the other possible goats were eliminated does not change the experiment at this point. It is only the overall probability of winning that is affected because you can lose before this situation even arises.
- I started doing a little bit of testing with a random number generator. The numbers agree with what I felt, which leads me to think we are in fact talking about different points. You win the game about half the time, like you said. However, if we discount the games in which you lose because the host opens the door with the car, of the remaining times you win twice as much as you lose by switching. Think of this: if you believe that the overall probability of winning the game is 0.5, then of those 50% losses, some (turns out, half) must be as a result of the host picking the car, and the other reason is because you got to the point in the game where you switch doors, and you lost as a result of that switch. Overall, you win the game by switching 50% of the time, lose by host 25% of the time, and lose by switching 25% of the time. Once we are past the host selecting, by random, those conditional numbers support what I was saying. I'm guessing though, that we were arguing two different points. I do agree that the overall chance of winning the whole game is 0.5 — Preceding unsigned comment added by 151.190.254.108 (talk • contribs)
- I'm somewhat-confused, but let me try this:
- With n doors, no matter how many goat doors we reveal, or by what method, it has no effect on the probability that you originally picked a car.
- This is not true. The method does matter. If the method is an omniscient host is opening doors known (beforehand) to have goats behind them then the probability you originally picked a car does not change. If the host is randomly picking, then the chance that your door has a car (assuming the host doesn't reveal the car) is 1/m where m is the number of doors left.
- (as an aside, to relate this to the original problem) If the host chooses doors containing only goats, #1 still applies, but the probability that the remaining door is a car increases for each goat he reveals.
- This is the only case in which #1 applies.
- If the host chooses doors randomly, without foreknowledge, the apparently increased chance to have chosen a car door, in violation of #1, is an illusion due to the increased chance of the host having opened the car door, and thus you having lost.
- No, it's no illusion. You're left with a random choice of a randomly selected set of doors. At this point, all doors are equal (including the one you originally picked).
- With n doors, no matter how many goat doors we reveal, or by what method, it has no effect on the probability that you originally picked a car.
- Try it with n=3, where you can exhaustively list the possibilities. Or, more simply, since you've correctly stipulated that the doors eliminated randomly do not affect the experiment in any way, try it with n=2. q.e.d. Jouster (whisper) 19:07, 17 July 2007 (UTC)
- So, with n=3, your original chance is 1/3. If the host randomly opens a door 1/3 of the time you now lose (because the car is revealed). In the other 2/3 of the time you now have a 1/2 chance that the door you originally picked hides the car. Switch or not, you'll win the car 1/2 the time. The total chance of winning is 0 for the 1/3 of the time you immediately lose, plus 1/2 * 2/3 from the other case, i.e. a net total of 1/3 - whether you switch or not. -- Rick Block (talk) 01:17, 18 July 2007 (UTC)
- I'm somewhat-confused, but let me try this:
I'm thinking now I am mistaken because I read another type of explanation that I could not seem to refute, but your point #3 is still somewhat confusing because it seems to contradict #1, not show how #1 doesn't work.
Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks!
- Is the earlier explanation, in the section titled "Why the probability is 2/3", more clear to you? -- Rick Block (talk) 01:17, 18 July 2007 (UTC)
Okay, I've spent entirely too much brainjuice on this. It probably has one or more glaring errors. But it's the algebra-ized version of the way I view the problem in my head (which has more to do with colors than with symbols; go figure). That said, I'm not so good at this whole color→algebra thing, so please fix it up:
You select a winning door | Your door is now a winner | Remaining Door(s) contain a Winner | You've already lost | |
Host picking randomly | ||||
Host picking only losing doors |
Jouster (whisper) 23:48, 18 July 2007 (UTC)
Choice? or Strategy?
I believe that some who are confused by this problem are focused on the difference between a Strategy and a Choice. The Host presents the player with a Choice, and if it were truly acted on as a Choice then the probability would not be 2/3 winning. Only when the player examines the rules and develops a Strategy which is adhered to without fail, does the probability turn to hir favor. Those who have watched The Price is Right remember the indecision on the faces of players as they ponder what to do. In my (admittedly basic) understanding of probablilty, everything hinges upon a priori decisions. Thus, I posit that if a Player goes into the game with a Strategy of sticking to their original choice, they'll have 1/3 chance of winning. The Strategy of always switching when offered the inevitable Choice yields 2/3 Chance of winning. This has been established. But I think that it is important to note that if the Player has no Strategy, and truly decides to randomly Choose between switching and staying, then there is a 1/2 chance of success. I believe I am correct in this assessment, and I think it would make a good additional to the article to explain this, since I believe many of those who are confused are thinking in these terms. i.e. Explain the difference in probability between a true Strategy (Choice decided in advance) vs. random Choice. Of course, if I'm completely wrong on this (which is why I have placed this in discussion), I still think it might clarify the article to point out that this is a Strategy, and that the player is not really making a Choice - their outcome is predetermined when Monty opens a door. P.S. I note that one of the external links explains the exact case I present of 1/2 odds with a random Choice. Shouldn't that be mentioned somewhere in the article for completeness? BrianWilloughby 18:29, 11 July 2007 (UTC)
- I think this is a very complicated way to look at the problem and is almost certainly not how most folks are approaching it. If a player chooses randomly to stay or switch, then (over numerous iterations) the aggregate probability will indeed be 1/2. The complication is that on any given iteration, the odds are 2/3 on switching vs. 1/3 on staying - regardless of when the player decides whether to switch. A player who randomly decides who only plays the game once does not have a 50/50 chance, but a 1/3 chance (if the random decision was stay) or a 2/3 chance (if the random decision was switch). I suspect this distinction between the outcome of a strategy and the outcome of an individual choice is quite frankly beyond most people's grasp of probability. An example might help clarify this. If there are 100 players who choose randomly, likely 50 will switch and 50 will stay. Of the 50 who switch, we'd expect about 2/3, i.e. roughly 33, to win the car. Of the 50 who stay only 1/3 will win, i.e. only 17. 17+33 equals 50, so on average these 100 players have a 50% chance of winning although no one of them individually has a 50/50 chance. Contrast this with a pool of 100 players who all switch. Each one of them has a 2/3 chance of winning and we'd expect out of 100 about 67 will win the car. The aggregate probability of this group is the same as the individual probability of each player in the group (unlike the previous case). -- Rick Block (talk) 18:59, 11 July 2007 (UTC)