Talk:Monty Hall problem/Arguments

From Wikipedia, the free encyclopedia
Jump to navigation Jump to search


Reductio ad Absurdum of the "vos Savant" Solution[edit]

Your first choice had a 1/3 probability of being a car, as did the other two doors. If you continue to say that it has a 1/3 probability, after the host's intervention, then you must say that the remaining unopened door still has a 1/3 probability of being a car, which is clearly absurd.

What has happened is that the host has introduced new information by opening a door, so the original logic tree must be modified or restarted from scratch. Nigelrg (talk) 21:14, 15 May 2017 (UTC)

Indeed, once the host has opened another door to reveal a goat, then the only chance to win the car is a 50-50 choice to stay or switch to other door, as probability 0.5 either way. Conversely, some people claim Monty Hall would show the car if behind the first door chosen; otherwise the car would be behind the other door, as 100% chance of winning car by switching door (not merely ​23). Of course, such a game would be absurd; hence, the only sensible game would include the chance of the car behind first door chosen, as again 50% chance of win, whether stay or switch. There is no other sensible conclusion. -Wikid77 (talk) 03:27, 16 May 2017 (UTC)
@Nigelrg: You are exactly correct. In fact, numerous mathematicians have described vos Savant's reasoning as somewhat less than complete (see https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions). Before the host opens one of the doors, the probability the car is behind each door is 1/3. After the host opens one, the probabilities must now be reevaluated. We're certain the probability the car is behind the door the host opens is 0, but what about the other two?
@Wikid77: Wikid77 - please pay attention to this.
If the player picks door 1, and the car is behind door 2, then the host MUST open door 3, so the composite probability the car is behind door 2 AND the host opens door 3 is 1/3 * 1, i.e. 1/3.
If the player picks door 1, and the car is behind door 1, then the host can open either door 2 or door 3. Let's say the host doesn't care which door he opens in this case (e.g. he flips a coin to decide). This makes the composite probability the car is behind door 1 AND the host opens door 3 equal to 1/3 * 1/2, i.e. 1/6.
If the host opens door 3, these are the only two possibilities. To express these as conditional probabilities, we divide each by their sum. 1/3 + 1/6 = 1/2, so the conditional probability (given that the host has opened door 3) that the car is behind door 1 is 1/6 / 1/2 = 1/3. And the conditional probability (given that the host has opened door 3) that the car is behind door 2 is 1/3 / 1/2 = 2/3.
As it turns out, the probability the car is behind door 1 doesn't change. But we really can't just assume that. We should compute the conditional probabilities. As one source puts it "The host can always open a door revealing a goat and the probability that the car is behind the initially chosen door does not change, but it is not because of the former that the latter is true." -- Rick Block (talk) 04:09, 16 May 2017 (UTC)

@ Rick Block. I believe that your analysis only holds true if the player has decided whether to stay or switch before the host opens the door. The first post of the first discussion thread makes this point, and the main article addresses it in some depth, as stated in the unidentified post above: https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_of_the_simple_solutions. No such stipulation is made in the description of the game, so we should assume that the player makes his decision after the host opens the door. I prefer to use Occam's Razor, and cut out unnecessary detail. Because the player has no information about the location of the car, other than the fact that it's behind one of two closed doors, he/she has a simple 50:50 choice.Nigelrg (talk) 02:43, 18 May 2017 (UTC)

@Nigelrg: First, the analysis I presented above is not "mine", but it is what the most reliable sources in the field of probability use. It most definitely does NOT hold true only if the player decides whether to stay or switch before the host opens the door. It evaluates the probabilities explicitly AFTER the host opens a door. Yes - you don't know for sure which of the two remaining doors the car is behind, but you have "partial" information. This partial information (the fact that the host MUST open door 3 if the car is behind door 2, but has a 50/50 choice which door to open if the car is behind door 1) is what lets you conclude that the chance of winning the car by switching (AFTER the host opens a door and there are only two doors remaining) is 2/3.
Lets take this a bit further. Say we have a deck of 52 cards, and the ace of spades "wins". If I (the "host") shuffle the cards, give you one (face down), and now (looking at the rest) discard all but one making sure I'm not discarding the ace of spades is it now 50/50 that you have the ace of spades? Or is it 1/52 that you have it and 51/52 that I have it? There are only two choices - you have it or I have it. You don't know for sure which one it is. In case it's not obvious the analogy to the MHP is this - instead of putting a car behind one of 3 doors we're putting the ace of spades among 52 cards. Instead of the player picking a door, we're dealing a random card to you. Instead of the host opening one "losing" door resulting in only two doors being left - the host host is discarding 50 losing cards resulting in only two cards being left. Please try this (really, no kidding) - say 20 times and let us know how it turns out (how many times do you end up with the ace of spades vs. how many times does the host end up with it). The notion that "you don't know for sure" and "there are only two choices" necessarily leads to the chances being 50/50 is simply incorrect. This is the entire point of the Monty Hall problem. It goes against a deeply held belief. The bottom line is that humans suck at conditional probability (or, perhaps, schools suck at teaching elementary probability concepts). -- Rick Block (talk) 06:49, 18 May 2017 (UTC)

I'll think over your interesting post later. I had some further thoughts myself, but your post may have negated them. Re: your last sentence, it's not just schools in general. Your views are opposed by post-graduates in math from some of the world's better universities. They're not infallible, of course, but it's a bit like climate change. When a large body of experts say one thing, there's a high probability that they're right.Nigelrg (talk) 18:05, 18 May 2017 (UTC)

I'm dealing with your post piecemeal :-) The deck of cards example is only relevant to one option in the Monty Hall problem - the case in which the player has pre-decided to stick (once you've picked a card, you can't change it). Therefore I don't think it's relevant to the problem as a whole≈Nigelrg (talk) 20:05, 18 May 2017 (UTC)

That's not right. In the formulation of the card game aboive, it does not preclude that the player could still switch cards with the dealer, turning around the probabilities of having the ace of spades. This would make the game equivalent to a Monty Hall game with 52 doors, where the host opens 50 doors which don't have the car, and the player can decide to switch or not after that. Diego (talk) 23:06, 18 May 2017 (UTC)

Re: "partial" information. I think you've neglected the situation when the car is behind door 3, so the host must open door 2. Therefore the host has 2 possible actions when the player's door hides the car, and only 1 possibility when it doesn't. Either way, the player gets new information. I came back to this post months later and corrected it. I made a typo/brain fade when I originally wrote it.I hope my correction didn't change any responses.≈Nigelrg (talk) 21:00, 18 May 2017 (UTC)Nigelrg (talk) 06:07, 11 November 2017 (UTC)

My views are NOT oppposed by post graduates in math (well, not any in probability or statistics) anywhere in the world. This is a classic problem - it appears in many elementary probability textbooks (with the exact answers I'm giving you). So, yes, it's a little like climate change in that the science is settled. But this is math, so not only is the science settled it's actually proven.
@Nigelrg:I thought you wanted to talk about a case where the host has already opened a door and there are only two possibilities for where the car is - for example, the player picked door 1 and then the host opened door 3. In this case the car is manifestly not behind door 3. I'm not 'neglecting" the situation where the car is behind door 3 - we're explicitly talking about only a subset of cases where the host has opened door 3, which means the car is not there.
Here's yet another way to think about it. Imagine 300 shows where the player has initially picked door 1. We'd expect the car to be behind each door about 100 times (right?). So, now the host opens door 2 or door 3. If we want to think about only the shows where the host has opened door 3 we're not talking about 300 shows anymore - but only some subset of the entire 300. Can you answer the following questions (thinking about 300 shows where the player has picked door 1 and the car is behind each door 100 times)? -- Rick Block (talk) 14:58, 19 May 2017 (UTC)
In how many of the shows where the car is behind door 1 does the host open door 2? ______
In how many of the shows where the car is behind door 1 does the host open door 3? ______
In how many of the shows where the car is behind door 2 does the host open door 2? ______
In how many of the shows where the car is behind door 2 does the host open door 3? ______
In how many of the shows where the car is behind door 3 does the host open door 2? ______
In how many of the shows where the car is behind door 3 does the host open door 3? ______
In how many shows overall does the host open door 2? ______
In how many shows overall does the host open door 3? ______
In how many shows where the host opens door 3 is the car behind door 1? ______
In how many shows where the host opens door 3 is the car behind door 2? ______
If you pick door 1 and the host then opens door 3, are you more or less likely to win the car if you switch?
This is a good example, Rick. Answer: It is completely impossible that you ever can be less likely to win the car if you switch.
This is valid not only in that (given?) case if you pick door 1 and the host then opens door 3 (in order to show a goat), but this is valid in any case, regardless which door you may pick, and regardless which other door the host may be opening (in order to show a goat).

Once more: You never are (nor can be) "less likely" by switching. This is valid in any situation given.

Your chance to have picked the car by luck is 1/3, so on average the risk to lose the car by switching is still 1/3, but
your chance to have picked one of the two goats (=wrong guess scenario) is 2/3, so on average the chance to win the car by switching doors is 2/3. And the host's opening of a door (to show you the SECOND goat) does not alter the scenario you're actually fixed in.
The chance to have picked the car by luck (lucky guess scenario) is only 1/3, and the chance that you are fixed in the wrong guess scenario, having picked a goat (thereafter the SECOND goat has already been shown to you !) consequently is even 2/3.
So in 2/3 of all cases you will win the car by swithing doors.

As to the host's behavior, he will never give you any hint on the scenario you're actually (unchangingly) fixed in (Henze, Mladinow et al). --Gerhardvalentin (talk) 12:20, 20 May 2017 (UTC)

Nigelrg seems to think that the analysis of the situation AFTER the host opens a door is that there are two doors and we don't know where the car is, so therefore (??) the chances are 50/50. The point is that the structure of the imaginary show (the real show wasn't actually run like this) gives us some information. -- Rick Block (talk) 15:50, 20 May 2017 (UTC)
Some readers believe in miracles, may be. Or they forget about the significant progress of the advancement made in course of the imaginary "show". But reality never will. --Gerhardvalentin (talk) 17:02, 20 May 2017 (UTC)
This post is very useful to settle the topic.[1] — Preceding unsigned comment added by 190.199.242.101 (talk) 23:15, 16 August 2017 (UTC)
Thank you. IP190.199.242.101, for your compliment. --Gerhardvalentin (talk) 16:08, 25 April 2018 (UTC)
Bottom line is that the idealised Monty's behaviour is precisely equivalent to his telling you, "If you have picked the wrong door, then the prize is HERE". And you had 2 chances in 3 of being wrong. What more needs saying? Fredd169 (talk) 13:12, 23 December 2017 (UTC)

In probability theory no single case has any meaning; 'control groups' need to be used[edit]

There is a way to explain that several answers -and most logic behind it- can be right, using the 'different control groups' idea. Which is based on the simple fact that in any single case, probability has no meaning. So in probability 'thinking', we always have to use bigger amounts that we also use in statistical evidence, to make it clear that if we follow certain rules (consistent behavior) and repeat it enough times to rule out 'coincidence', the calculated probability is actually exactly that of repeating rules. For that, we first have to decide what control group of consistent behavior the players are in.

Because of the way of questioning: 'if you are the contestant, what should you do?', the contestant has to try to determine what control group he is in, to understand the (behavior) rules of the group, and then decide what to do himself in this single case. Which is of course nonsense, because 1.) his deviation in a single situation has nothing to do with probability calculation, which applies to (endless) repeating, and 2.) his decision about his own behavior automatically creates the same control group, in which all cases follow exactly the same rules. Let me explain by examples, starting with the unexpected.

Assumptions in all cases are the obvious, which is not the problem. The actual problem is in the two questions being asked:
1.) "Do you want to pick door No. 2?"
2.) Is it to your advantage to switch your choice?

Example winning probability = 1/2

Most people think that switching does not change the probability of winning. Let's suppose the contestant is one of those. Let's also suppose that the contestant chooses randomly to switch or not. Now it becomes clear that his control group will answer question 1 randomly with yes or no. Just like the host randomly chooses between two doors that have goats. Rule = random. Now we also know that half of them switches and uses the 2/3 probability, while the other half uses the 1/3, equaling to 1/2. Because we cannot apply probability calculation to a single case, we have to apply these common rules of the whole group (both halves) to this one contestant, because in repetition he is sometimes switching and sometimes not. The correct answer to question 2 is that there is no advantage of switching, because it happens randomly, which is the rule. (And which is very counterintuitive.) The contestants will give the right answer, but not with the right reason.

Please notice that these two questions can be seen as really different questions, or actually the same. When seen as different questions, the first one is about the personal preference of the contestant, while the second one necessarily is a logical one, which can be answered right or wrong. It seems obvious (because of the missing quotes in the second question) that these questions are meant to be one: 'please estimate your advantage and then make your choice to switch or not'. The reason leading to the first answer is a given fact that must apply to the entire control group, while the correct answer to the second question is the result of that reason and thus behavior, which has opposite dependency between both questions as is meant to be.

1/3 < Example winning probability < 1/2

In this situation the contestant still thinks that switching does not matter, but he tends to stick to his initial choice, let's say he does this in 70% of the cases. Chances of winning are (7/10 * 1/3) + (3/10 * 2/3) = 0.4333.. He will answer question 2 with 'no', which is incorrect, because he still switches in 30% of the cases. Given his behavior, the correct answer is 'yes, my winning probability increases from 0.3333 to 0.4333 switching as often as I do', but he is not giving that answer of course. This situation is probably most realistic and also makes it very well clear that a single case or choice does not mean anything in probability; we need to know the consistent processes behind it. In one case he will switch doors, in another he will not, which is not even distributed equally.

Note: the counterintuitive thing about this whole 'one case has no probability' approach, is that it seems that it doesn't matter whatever you choose, having only one chance, which seems to happen a lot in everyday life. Vos Savant helps us to understand: suppose that in 999,999 of 1,000,000 cases you are stupid enough to randomly switch, and only in one case you have a short moment of seeing the light, convincing you to switch. This will not significantly change the odds. However, if this would be a consistent reality (it happens to you once in a million), the only way to calculate it right, is to use the 0.500001 probability of switching. But what if you, when you have this bright moment, simply are aware of that brightness? How could it not be significant? There is only one way to be (probably :)) sure about that, which is your ability to recognize such bright moments, proven by experience. This completely changes the control group of situations, because now you have to add a second group of cases that overall have a high rate of brightness. Which shows us again that only the extent of consistent behavior matters, but also that the odds are really relative and depending on the control group(s) used!

Example winning probability = 2/3

First of all I'd like to emphasize that the key is not in the knowledge, but in the behavior (by knowledge). (Then again it's the knowledge of behavior that enables us to create a realistic control group.) If we check the standard assumptions from the chapter in the article with the same name, we see that the three doors are distinguished as 'originally chosen', 'opened' and 'remaining closed'. Not as numbers. We can safely make the assumption that host nor contestant is consistently paying attention to these numbers. Because they have no reason for that. We only know that the host, when offering the switch, is explicitly using the name of the number of that remaining closed door. It may well be that he is simply reading the number displayed on it, at that moment. It does not mean that any player is aware of the number of the originally chosen door at any time. Moreover, since the possible knowledge of the number of the chosen door does not change the behavior of any player, it does not make sense to use this information in any scenario. Also the information from the puzzle: "You pick a door, say No. 1" may be interpreted as: "this may also be No. 2; we only use the numbers as relative distinction to each other, not as absolute information to recognize one certain door." Because of this, conditional solutions using the specific numbers as absolute information are not only unnecessarily complex; they are even wrong when this interpretation is right. Only if (additional) knowledge changes behavior, it should be used in calculation. We have no clue that this is the case, so the simple solution is most appropriate.

Most important is again the question: is there consistent behavior and what is it? One contestant in one case never shows consistent behavior, whatever he thinks, answers or does. Suppose he knows all about probability theory. He will answer: "Yes, I want to pick door No. 2. If all cases in my control group will always and consistently show the behavior to switch when offered, it will be in the average and relative advantage of this group compared to those groups that show consistent behavior to not switch respectively to randomly switch. There is however no way to know anything about my own chances in this single case, if even such a probability exists." — Preceding unsigned comment added by Heptalogos (talkcontribs) 15:27, 8 January 2018 (UTC)

--Heptalogos (talk) 23:59, 7 January 2018 (UTC)

@Heptalogos: The sentence ' He then says to you, "Do you want to pick door No. 2?" ' is a statement, not a question. The Monty Hall Problem does not require us to answer it. We are asked 'Is it to your advantage to switch your choice?' to which the solution is 'Yes' and the value is 2/3. To illustrate the meaning probability has in any single case, such as this, consider a single fair dice to be thrown once only. The probability it will land a six is 1/6. Freddie Orrell 22:51, 9 January 2018 (UTC)

@Freddie, that's exactly the misconception I mean: there is no way to know the probability for one single throw. We can only make correct statements about probabilities for greater amounts of repetitions of the exact same behavior under the exact same conditions. Statisticians have learned that, and most of them will not deny it, but they simply 'forget' about it sometimes. Because usually it's not an issue. The reason for that is that statisticians always create their own reality in theory as well as in empirical testing. They may however fail to relate a single case in real life to the experiment correctly; nobody will ever know. Look at my first example: the contestant randomly chooses to switch. This will lead to a probability of 1/2 of winning. Because if we test this behavior, the contestant will switch in half of the cases and will not switch in half of the cases. Any single case (switching or not) makes no sense, because that's not his consistent behavior. Stupid statisticians will argue that it is: if he switches, they will use that knowledge and use only 'switching behavior' in their calculations and tests. If he doesn't switch, they will use only 'non switching behavior'. Then the tests of course will prove the theory. :) The whole horrible thing about this, when you think it around one thousand times, is that there is no objective probability at all in any single case. We only obtain meaning or significance from 'law', which is consistency, recognition by repetition. What is the consistent behavior of the contestant? Is there even consistent behavior? Nobody knows, not even the contestant himself: will he react exactly the same in many and all cases? So I read about the assumption that the contestant has to decide what his consistent behavior is, which is one step in the right direction. But still he can decide or 'promise' whatever he wants, that doesn't make it full proof consistent behavior of course. Wrap up: in one single case the contestant can switch, and if he consistently always switches, his chance of winning is 2/3, but if he consistently randomly switches, his chance of winning is 1/2. Actually we can make as many groups and probabilities as we want. To convert a single case to a group, we always have to know the exact rules that apply. Of course we only know that if we turn it around: creating cases by performing rules. Which is what we do in empirical testing. --Heptalogos (talk) 22:47, 10 January 2018 (UTC)

@Heptalogos: You say the probability that a single fair dice thrown once will land a six is unknown. And if I take a fresh, fair coin and toss it once, is the probability it will land heads also unknown? Freddie Orrell 23:34, 10 January 2018 (UTC)

@Freddie, yes it is unknown. Probability in a single case is not even existing as such. There is simply average probability, of control groups. Which is the only thing we can empirically test and therefore prove. The idea that any single case in such a group has it's own probability is just that: an idea. Besides that it cannot be proven, it doesn't even make sense (as ideas sometimes do), because any unique case has the freedom to be influenced by significant forces that do not apply consistently to the whole group. Even if we create cases by performing rules, we should use computing (a program that follows only predefined if-then statements resulting finally in series of 0 and 1) and even then we should rule out any uncommon electric currency that can switch the bit. In empirical testing we usually rule out such deviations, at least afterwards. (If any inconsistency happens, it's a formal excuse to deny the outcome and run the program again.) Then we have probability as an outcome. Which we apply to any case we can reason as being part of our control group. Which is first of all wrong anyway, because it's a single case. And which secondly has a greater possibility to be wrong when the 'natural' single case (control group) has more complexity/diversity than a computing program, like the human individual certainly has, therewith not being in a controlled lab but in a vivid place with lots of stimuli. But most of all, this reasoning is not necessarily the expertise -to put it mildly- of scientists anywhere in the field, publishing articles that dominate Wikipedia. Which becomes very clear by disagreements 'internally'. Wrap up: the outcome of empirical testing is actually still theoretical related to any real life situation. Creating this relation rationally is still theoretical and very tricky. Besides that, it's actually a second probability issue itself. :) --Heptalogos (talk) 10:07, 11 January 2018 (UTC)

@Heptalogos: The unknowns or uncertainties you describe are accounted for in probability, for example by the principles of indifference and propensity. While it is indeed impossible to predict the outcome of a single case, such as a coin or dice, this allows its probability to be determined in the form 1/n. Freddie Orrell 19:25, 11 January 2018 (UTC)

@Freddie, the issue is not about those uncertainties. There are two issues:
1. It is in fact an incorrect statement that one has a probability of 1/2 of throwing head in a single case. We can only state that when repeating enough times, the average probability is 1/2. The advice to 'always switch doors' doesn't make sense to a contestant who is once in a lifetime in a show and has only one opportunity to switch. There is absolutely no way to predict his chances.
2. The question "Is it to your advantage to switch your choice?" can have two correct answers, both yes and no. It can only be answered correctly if you can repeat enough times, fully depending on your consistent behavior. So if you for example decide to consistently switch randomly, it's a very strange question, because it doesn't seem to realize issue 1: in a single case of consistently randomly switching, it is not in your advantage whatever you choose. But it is in your advantage to consistently randomly switch, compared with consistently not switching. --Heptalogos (talk) 21:47, 11 January 2018 (UTC)

@Heptalogos: You are simply repeating the same fallacies over and over - I give up. Best wishes, Freddie Orrell 22:56, 11 January 2018 (UTC)

@Freddie, you are right that I am trying to describe the same issue (it can be reduced to one issue) in many ways. It's because I think it's too hard to understand for most people. I believe it's really not a fallacy, but (indeed) an issue that covers the entire area of probability theory. However, the Monty Hall problem raises this issue by me, for two reasons:
1.) there is not at all any consistent behavior even possible;
2.) the whole discussion has gone to a level that is so exactly and well defined, that it seems ridiculous that the article does not even mention (i.e. that anybody in the professional field did not even publish) the fact that no single case has a probability. I think I will start searching for it, or initiating it. Any help is welcome. --Heptalogos (talk) 13:13, 13 January 2018 (UTC)

Stop over analyzing this[edit]

Let's examine the facts:

The original choice can only act against a 2/3 for proportion (of goat/doors) and 1/3 for odds (of finding the car)
The removal of the door changes the proportion of goat/doors from 2/3 to 1/2
If we do a sheer guess, our calculation would tell us that there's 2 doors, 1 guess left, so we'd think it's 1/2
But the trick here is that the (as of yet) not chosen remaining door is not a random door, and therefore a sheer guess isn't possible
Via our initial pick, we disallow our chosen door from being eliminated by the host; our 1/3 choice door stays in the sample space
And based on the parameters of the question, the car cannot be removed by the host; thus, since the car stays available, the total of all odds must equal 1
If the host removes 1/3 and our 1/3 choice stays in the game, then to equal 1 there's 2/3 remaining which must attributed to the as of yet not chosen other door
The question isn't asking us "what is the proportion of remaining goat to doors after the host removes one?" (and FYI, that proportion is indeed 1/2)
Rather, the question is asking us "Is it better to switch?"
Thus, in the final analysis, there are two numbers which apply: The final goat/doors ratio; which is 1/2 and odds attained if one switches; which are 2/3
98.118.62.140 (talk) 14:39, 3 February 2018 (UTC)

Another way to think about this[edit]

If the aim were to find 1 of the 2 goats, instead of the car, but all other rules were the same, then what? The host must remove a goat - he cannot remove the car. Thus, the other door which remains must contain what? Either a goat or a car. If it contains a goat, you are already on a car. If it contains a car, you are already on a goat. When you first picked a door, you picked one door from a 2/3 chance of getting a goat. If you stay, staying with your first choice will win you a goat 2 times for every 3 times you play. But, since the host must remove a goat, if you choose again, you will only get a goat 1 time for every three switches. Why? Because 2 out of 3 times you are already on a goat after your first choice. Thus, you only can possibly switch to a goat 1 out of 3 times. Thus, finding the car is the opposite of this: 1 out of 3 first choices get you the car. And if you stay with that choice, you can't then put yourself into the remaining 2 out of 3 pool; because you are stuck at 1 out of 3. When you switch, you leave the 1 of 3 pool of choices and you switch to the 2 of 3 pool of choices. And the 2nd choice is indeed two of three; 3 doors, one chosen by you, one removed by the host (no car) and one which you leave behind when you switch. If you switch, in only 1 time out of 3 plays will you accidentally leave the car behind, because in only 1 out of 3 plays, will you have actually picked the car to begin with. The odds of switching improve over the original odds because the removal of the door by the host enriches the prevalence ratio of cars to the available remaining choices. There initially is 1 car in a pool of 3 doors, then there is 1 car in a pool of 2 doors, which means it looks like this: 1/3 (car prevalence of 1st choice) ÷ 1/2 (car prevalence of 2nd choice) = 2/3 if switch. Of course, this last sentence looks like Ma & Pa Kettle math [1] Xerton (talk) 18:26, 4 February 2018 (UTC)


There are ONLY 2 choices left: "Impossible"/Confusing to have a present x/3 statistic.[edit]

The fact is, when there are only 2 doors left to choose from, there are then ONLY 2 choices left; and it is therefore impossible to have an x/3 statistic among those at that point in time because you no longer have a choice of 3 but only of 2. Thus, the way some of the article is written is incredibly confusing for the 9/10 people who notice this clear and indisputable fact: There are only 2 choices left.

Seeing that even a famed mathematician thought similarly makes it clear that if you are going to use x/3 when you have left only 2 choices, then you'd better be crystal clear about the logic of application, which from reading much of the article isn't that convincing or clear; because using an x/3 stat with there are only 2 choices left makes little sense to most of us.

Also, placing that remaining x/2 likelihood back in time when the INITIAL chance was 1/3, and continuing to use x/3 makes little sense to most of us.

Saying anything new about x/3 when there are only 2 choices left makes little sense to most; so Why be so confusing with the math? we think. The fact is, the REMAINING chance has only 2 choices possible, and so therefore the remaining statistic MUST be in reference to that: It must be stated as x/2.

Hardly a clear justification was made in my reading of the most of the article for using x/3 at all, at the point where there were only 2 choices left. It is that justification that needs to be made, and clearly applied, which would help the hapless reader understand at least some of the gobbledygook; because otherwise, it continues to seem nonsensical. And after all, isn't this conundrum in particular one that needs to be crystal clear to the general reader who was mystified in the first place when reading the "Ask Marylin" article? Misty MH (talk) 23:55, 9 March 2018 (UTC)

I hand you a die with 2 white sides and 4 black. You only have 2 choices. Roll it, what do you think your odds are if you choose white? — Preceding unsigned comment added by Nijdam (talkcontribs) 11:17, 10 March 2018 (UTC)
The formula ---> (#favorable events / #possible events) is applicable only if you have the same information about the options. Remember the division distribute equal amounts. For example, if I have a cake and two persons and I want to give each one the same amount, then I should give 1/2 to each. But certainly that's not the only way to distribute it. I could give 4/5 to one and 1/5 to the other, or give all the cake only to one, etc.

The probability is a measure about how much information we have. The confusion is to think that the options must be equally likely in any case. In this game, yes, we have two options, but do we have reasons to distribute the probability equitably? No. The contestant's door was chosen randomly from three, meaning that it has 1/3 probability to be the correct. On the other hand, the host knows the positions and must leave the car hidden (because he must reveal a goat door), so always the contestant failed at first, the other door the host leaves closed is the correct one, and we know this is more likely to have happened than the opposite.

In the second selection, we don't have a random door vs another random door. Basically, you have an option selected by someone who hits the correct 1/3 of the time and another chosen by someone who chooses the correct in the other 2/3 cases. Which one do you prefer? The important thing here is nor how many options do you have, but that they were chosen by two different people, one with more knowledge than the other. The host's closed door tends to be the correct with more frequency than the contestant's selection. — Preceding unsigned comment added by 190.36.105.224 (talk) 05:28, 13 March 2018 (UTC)

@Misty MH: I guess Leonard Mlodinow, in his book The Drunkard's Walk, shows it best, in underlining the neglected, but obviously constricted role of the host in the MHP. His role leads to conditional probabilities of surprising 1/3 : 2/3. About Mlodinow, Just have a look there.
It's hard for anyone to grasp that the host can act "randomly" only in 1/3 of all cases. He can act randomly only in 1/3, only in case that the guest (by luck) selected the door with the car (i.e. only if the guest actually is in the "lucky guess scenario").
Only in this special case (1/3) the guest should stay and never switch.

But in 2/3 of cases the guest will be in the "wrong guess scenario" (2/3), having selected one of the two wrong doors (two out of three). Because in that "guest's wrong guess scenario", the host's two doors having the car and a goat resp. a goat and the car, the host nevermore can act randomly, because he never shows the car but the second goat only. So his selection which door to open, in openening always the door with the goat, leads to conditional probabilities. He keeps the secret car undisclosed behind his still closed second door. In that 2/3 switching wins the car for sure.

So in 2/3 of all cases, as a "conditional probability", switching doors helps the guest to get the car.

The contestant, in opening a door, does not know the scenario he actually is in. He only knows that - by the host's constraint to never showing the car - by switching he will win the car with double (conditional) probability. Admittedly, on the long run only.

I agree with you, this fact that in 2/3 the host will never act randomly leads to a conditional probability of 2/3 to win the car by switching is hard to tell, using a few words only. --Gerhardvalentin (talk) 15:15, 22 April 2018 (UTC)

I read this as a criticism of the article, not the math. What Misty MH is saying is that the article is not clear, because it immediately delves into the "simple" solutions which more or less ignore the conundrum most people encounter which is that at the point of the decision there are only 2 doors involved, not 3. In particular, we know vos Savant's explanation from her original column is unconvincing based on the reader reaction she got. -- Rick Block (talk) 14:53, 23 April 2018 (UTC)
Yes, you are right. As to my understanding, Misty MH says that the article should clearly show that any assemblage of both unselected (host's) two doors is more likely to contain the prize. Exactly they have probability of 2/3, compared to any originally selected door of only 1/3.

And it's a fact that the host AVOIDS to show the car, but in any case he shows a goat. Solely in the contestant's lucky guess scenario, the host actually having two goats, he can act randomly. But if the CONTESTANT SELECTED ONE OF THE TWO GOATS (with twice chance) the host avoids to show his car, the "strict condition" is that he will show the SECOND GOAT only. So the 2/3 chance to win by switching doors is a typically "conditional probability". It's that easy.

And the famous mathematician Henze says that strictly speaking "math" is unnecessary to solve the MHP. And also Misty MH says: Why be so confusing with the math?

I say once more: math is rather unnecessary to fully understand vos Savant's MHP. That means that the whole ado about the host's biased behaviour belongs to a completely different article, you can name it "Index of lessons on conditional probability theory, based on the famous MHP"

And Marilyn vos Savant says that only in case that you already do have additional information of a factual existing preference of the host, and its extension, it might be sensual to do maths. But this never will be the case in the MHP, so any of similar considerations do never address the MHP of vos Savant, that definitely excludes any "additional (hidden?) information". No hidden hints are given. Such assumptions have to live their own life completely outside of vos Savant's famous MHP paradox.
(Though mathematically fully correct, M.et al. have disqualified themselves, in inventing ungiven additional assumptions, far outside of vos Savant's MHP, similar to "if wishes were horses, beggars would ride" and similar to "if you know that the biased host is paralyzed, then ...").

Any of such unbased considerations, only basing on ungiven and unproven assumptions (suitable for lessons in probability theory only), are obfuscating for the reader and prevent understanding the core of the famous paradox:

"Two still closed doors, why does the door offered have double chance, compared to the door originally selected?"
Answer: because there is'nt a new simple probability of 50:50, but because there meanwhile emerged a conditional probability of 1/3 : 2/3.

See theoretical physicist Leonard Mlodinow, who worked together with Steven Hawking, his diagram (1/3 : 2/3 on the long run) here:
The article, for years and years, has been confusing. It is time to detoxify it from disconcerting, misleading and truthless "conditional bias" math garbage that clearly belongs to a different article.--Gerhardvalentin (talk) 09:55, 25 April 2018 (UTC)

@Rick Block EXACTLY! You said it much better than I did! :) — Preceding unsigned comment added by Misty MH (talkcontribs)