Talk:Monty Hall problem/Simple solution

Why is the simple solution wrong? And equivalently the combined doors solution. Let me start with the last. `

But before it may be sensible to state what I (and many others, of course) consider to be the MHP.

• A car is placed random (uniformly) behind one of 3 doors, goats behind the other doors.
• The player may choose one of the 3 doors. His choice is independent of the placement of the car.
• Before the chosen door is opened, the host opens one of the 2 other doors, and exhibits a goat. (Strategy of the host: always opens one of the remaining doors, always showing a goat, acting randomly.)
• The player is offered to switch to the other closed door.
• What wiil be the players decision? And why?Nijdam (talk) 22:15, 10 February 2010 (UTC)

Combined doors

The reasoning is as follows: The chosen door has probability 1/3 to hide the car, hence the two other doors have probability 2/3 to hide the car. When one of these doors is opened, this door has probability 0 to hide the car. Hence the other one must have probability 2/3 to hide the car.

The error lies in confusing the probabilities before opening a door and there after. The fact that for the opened door the "probability" of hiding the car is 0, means the considered "probability" is the conditional probability given the opened door. To decide that the remaining door has "probability" 2/3, we need to calculate the conditional probability. Unconditional the two not chosen doors have together probability 2/3. But it is not immediately clear that conditional this sum is also 2/3. This has to be proven.

A possible defense of this solution is to say that the "condition" is that the host opens "a door" rather than "an identified door", and that this then means P(car behind the player's door|player picks a door, host opens "a door") is the same as P(car behind the player's door|player picks a door). Using "opens a door" as the condition, then the prior and posterior probabilities of the chosen door are the same, and the other two doors have combined prior and posterior probabilities 2/3. However, in this case, the other two doors each have individual prior and posterior probabilities of 1/3 as well - if the condition does not affect the probability of the player's chosen door it does not affect the probability of either of the other doors! This is a logically consistent, but highly counterintuitive way to view the situation. The point remains that if the "probability" of the opened door is said to be 0, this probability must be a conditional probability given the condition that the host has opened this specific door and, if this is the case, then determining whether the prior probability of the player's door is the same as the posterior probability requires evaluating the conditional probabilities. -- Rick Block (talk) 15:29, 9 February 2010 (UTC)

Even if the host opens an identified door, it is possible to show that the posterior probability must be exactly equal to the prior probability in the case that the host chooses randomly. Thus the 'condition' may be safely ignored. Martin Hogbin (talk) 00:31, 10 February 2010 (UTC)

@Marti: Do you see any difference between the prior and the posterior probability?Nijdam (talk) 00:41, 10 February 2010 (UTC)

Not in the case that the host chooses a legal door randomly. They are provably (without the use of conditional probability) equal. Martin Hogbin (talk) 22:58, 10 February 2010 (UTC)

Do you not agree that the posterior probability that the players original door hides a goat is exactly and provably equal to the prior probability if the host chooses which legal door to open randomly? How can they possibly be different? No new information is revealed. Martin Hogbin (talk) 22:50, 11 February 2010 (UTC)

@Martin: I asked you if you see the difference between the prior and the posterior probability. Implicitly I deduce you see indeed the difference. But you still argue about how the posterior probability is calculated. And I repeatedly explained it's unimportant. Rick and I also tried to make clear to you that two probabilities may have the same value and yet be different! It is exactly about what you indicate with "provable". It has indeed to be proven that the prior and the posterior probability have the same value. This is in fact our only point.Nijdam (talk) 11:54, 12 February 2010 (UTC)

What do you mean by 'two probabilities may have the same value and yet be different'? Do you just mean that they apply to different circumstances (before and after the host has opened a door) or something else? Martin Hogbin (talk) 22:19, 12 February 2010 (UTC)

Right! Nijdam (talk) 00:30, 13 February 2010 (UTC)

I take it you mean that they apply to different circumstances. We are back to 'the host says the word "door" '. The circumstances after he does this are not the same as those before this but we know that this does not affect the result. We know that the posterior probability is identical to the prior probability. We know they are exactly the same thing. There is no posterior probability because there has been no event which might possibly alter the probability that the car is behind the originally chosen door. It is what I called a null event. The event that the host opens a specific door is also a null event; it can be ignored; there is no prior and posterior. Martin Hogbin (talk) 15:28, 13 February 2010 (UTC)

Sorry Martin, we're indeed back where we have to be. The prior and posterior probabilities have the same value, but are not identical. Rick and I have shown you this in different ways, in words, formulas, tables. It's a pity, but you seemingly do not want to understand this. It has nothing to do with your thoughts about conditioning events. Nijdam (talk) 21:39, 13 February 2010 (UTC)

You make that assertion without proof. I have shown, by considering several counterexamples that you refuse to address, that the only possible measure of which events constitute a condition is that of is whether it is considered possible that a given event might affect the probability of interest. In other words whether the probability of interest is independent of the event, or to put it yet another way, whether the probability posterior to the event can possibly be different from the probability prior to the event. If this condition is not fulfilled then any given event cannot be a considered condition of the problem. If you think that you know of a better test of which events should be considered conditions of a given problem please tell me now, bearing in mind that I have proved all your attempts so far to be incorrect. Martin Hogbin (talk) 23:54, 13 February 2010 (UTC)

editing

Martin, you should know better. We have explained to you that conditioning means an essential reduction of the sample space. One simple example: Population of UK is our sample space, Scotland is the condition. Yet the probability (rel. freq.) of women has the same value with and without the condition. But they differ in nature. Nijdam (talk) 00:44, 14 February 2010 (UTC)

I have proved this argument wrong. The sample space is originally set up taking into account all events on which the probability to be calculated might be dependent. It should not include events of which the probability of interest is independent.

The example you give is a very good one. The probability of interest (frequency of women) cannot be proven to be independent of whether sample is taken in Scotland or not. In fact this probability may not be independent of location. Thus if we are interested the probability that a person is a woman on the condition that they are in Scotland we must indeed condition our sample space. The posterior probability might be different from the prior probability.

Really Martin, what are you talking about? Let me specify the events: UW = {women in the UK}, SW = {scottish women}. I assume half the population in the UK as well in Scotland consists of women. Now it is your turn: give me the relevant probabilities if we pick one inhabitant at random. Nijdam (talk) 21:33, 14 February 2010 (UTC)

If you tell me that half the population in Scotland are women then the probability of a random person in Scotland being a woman is obviously 1/2. There is no need for or any calculation, sample space, or conditioning. Martin Hogbin (talk) 21:47, 14 February 2010 (UTC)

You seem to have difficulty to stick to the subject. Who said the calculation of the probabilities is the issue here? Nijdam (talk) 10:45, 16 February 2010 (UTC)

P(UW)=1/2 P(SW)=1/2 simply because you specified both probabilities. This has nothing to do with conditioning. Martin Hogbin (talk) 15:29, 16 February 2010 (UTC)

Carefull: P(UW)=1/2, but P(SW)!=1/2. Think why. Nijdam (talk) 16:29, 16 February 2010 (UTC)

Can you explain your notation then please. What is P(SW)? Martin Hogbin (talk) 16:38, 16 February 2010 (UTC)

Why? But okay, P(SW) is the probability to pick a scottish woman in the experiment (which is picking at random a UK inhabitant). Nijdam (talk) 21:21, 16 February 2010 (UTC)

Because I did not understand it. You want the probability that a random person in the UK will be a Scottish woman? As you have defined sex to be independent of location this is P(S)P(W) = P(S)/2, that is the probability that a random person is Scottish times the probability that a random person is a woman. Martin Hogbin (talk) 21:36, 16 February 2010 (UTC)

(outindented)No, that's not wat I wanted, but that is the meaning of P(SW). The calculation goes (S is scottish): P(SW)=P(UW and S)=P(UW|S)P(S)=P(S)/2. I defined: P(UW)=P(UW|S)=1/2, and to be precise UW and S are idependent. You got the idea, but were confused by the notation (could be interpreted in your way). But now to what I wanted. That's: P(UW|S)=1/2. What do you learn from this last expression? ~~

I do not quite understand what you mean by, 'What do you learn from this last expression?'. I learned nothing I did not know before. What are you getting at? Martin Hogbin (talk) 09:31, 17 February 2010 (UTC)

As you say beneath: Now suppose that we want to measure the probability of being a woman in Scotland. We have already restricted our sample space to people in Scotland. That is the condition, and it actually restricts the sample space. Yet it doesn't influence the probability. Nijdam (talk) 20:50, 17 February 2010 (UTC)

I have never claimed that restriction of the sample space must change the probability. What I do claim is that, if it can be shown that the probability after a given event must be equal to the probability before that event, then that event need not be taken as a condition of the problem. Whether conditioning on a specific event reduces the sample space depends on whether we choose to include that event in our sample space. Martin Hogbin (talk) 09:06, 19 February 2010 (UTC)

Martin, I really don't know what point you want to make here. Pease try to formulate it in "our" terminology. It seems you say: I can show that P(C=1)=P(C=1|H=3). Hence {H=3} needs not to be taken as a condition. There is no such thing as "need to be taken as a condition", the event {H=3} is a condition. Why else spoken of the probability after a given event? And {H=3} certainly has to be part of the sample space. Nijdam (talk) 21:24, 22 February 2010 (UTC)

You repeatedly say that {H=3} is a condition but you have never given me a valid reason why this must be so. You have given a number of reasons which I have all shown to be false. An event is not a condition if the probability of interest must be exactly the same after the event as before it. This is just like the host sneezes, because we assume P(C=1) = P(C=1| Host sneezes). An event which cannot possibly make any difference is not a condition. Martin Hogbin (talk) 22:10, 22 February 2010 (UTC)

It seems you do not understand what is meant by the term condition. You have your own ideas about conditioning, but they do not coincide with the probabilistic view. So it is useless to discuss this point, where your and my definition differ. Anyway, my definition is the probabilistic one. Maybe we better speak of probability before and after an event has happened. Nijdam (talk) 23:13, 22 February 2010 (UTC)

If it is of any help to you: how would you prove: P(C=1) = P(C=1| Host sneezes)?Nijdam (talk) 23:15, 22 February 2010 (UTC)

Now suppose that we want to measure the probability of being a woman in Scotland. We have already restricted our sample space to people in Scotland but we do not have the resources to poll all the people in Scotland. Instead we take a random sample and count the proportion of women in this. It is a well-accepted fact, widely used in surveys and censuses, that the posterior probability after the event of taking a random sample must be equal to the prior probability. The only issue is that of ensuring that the sample is truly random.

In the symmetrical MHP the host legal door choice is defined to be random, thus the posterior probability that the car is behind the originally chosen door must be equal to the prior probability. It is not that it just happens to be equal,we know that we are taking a sample uniformly at random from the full sample set so the posterior and prior probabilities must be equal; there is no condition to consider. Martin Hogbin (talk) 11:30, 14 February 2010 (UTC)

Simple solution

A similar error lies in the simple solution, reading: the chosen door has 1/3 chance on the car, the opened door has chance 0, hence the remaining door must have chance 2/3.

As soon as one speaks of the opened door having 0 chance on the car, this means the conditional probability.

Nijdam (talk) 11:52, 9 February 2010 (UTC)

?

What is the concern, what is this about? Problems are being called, but hiding where they are based on, hiding what they refer to. Telling "it is so and so", not calling the basics. That is not enough. Yes, you come closer to the core problem, however you are hiding the basics that these core problems refer to. Not calling them explicitly. So please tell what you are talking about. Are you talking about the simple perspective of the guest who just knows the "simple MHP rules" just the simple rules of the simple game? The guest who relies on the observance of the simple rules? Or are you considering additional perspectives ("Anything unlawful that could be, could be!")? Are you talking about the perspective of any observer who has additional information, or are you talking about the perspective of the host who, in advance, already has all information available? Or just stating "Lets assume: Anything what possibly could be, indeed could be"? Please call it. Using conditionals should not "happen for no reason". This reason should be mentioned. Please do not hesitate to call them.

Thank you, Nijdam, for adding the above rules. -- Gerhardvalentin (talk) 00:20, 11 February 2010 (UTC)

Gerhard: I add some comment in your text.

Combined doors

The chosen door has probability 1/3 to hide the car, hence the two other doors have probability 2/3 to hide the car.

>>Probability is not a property of the door, but of the experiment. So better say: initially .....

Repeat:
Probability 2/3 to hide the (only one!) car. So even before any door is opened, it is well-known (only one car) that at least ONE of those two unpicked doors MUST definitely have "probability 0".

>>Here you demonstrate you don't know how to handle probability. Look: each door has 1/3 chance to hide the car, but in your way of reasoning, as there is only one car, two of the doors must have chance 0, and the remaining chance 1. Doesn't this strike you as strange? The point is you're mixing a probabilistic view with the realisation of the situation.Nijdam (talk) 23:48, 2 March 2010 (UTC)

Unconditionally. Either the one or the other one, or vice versa. This fact is given by the rule that says "only one car". The error may be in the rule saying "only one car". At least ONE of those two unpicked doors MUST definitely have "probability 0". Unconditionally. On the first glance: It does not even matter which one. It will be shown before you can ask "which one". Before considering conditionals it would be advisable to explain the reason for conditionals, and the additional assumptions that could make "conditionals" necessary. Advisable to show explicitly the reason for conditionals. "A priori" conditional probabilities are not necessary.

"Conditional probabilities" are not necessary from the outset. There must be a reason for conditionals. This prior "reason", as a condition, has to be mentioned. Without any prior condition "conditional probabilities" are not necessary. So call this reason, call all these conditions explicitely (irregularities, additional assumptions, etc etc). Without the adoption of "apparent irregularities", e.g. submitting illegal messages and further information to the guest ("BOTH unpicked doors are nuts"), the use of conditional probabilities is not a good idea. Not a priori. Without calling those assumptions, the use of conditional probabilities is not necessary for the standard MHP. Regards, -- Gerhardvalentin (talk) 17:01, 9 February 2010 (UTC)

Gerhard - What does "probability" mean to you? For example, if I flip a coin once it is either heads or tails. Would you say the probability of it being heads is 100% or 0% and the probability of it being tails is 0% or 100% - or is the probability that it is heads 50% and the probability that it is tails is 50% because if I do this repeatedly (say 100 times) these are the expected proportion of heads and tails? The same logic applies to the 2 unpicked doors before one is opened. Before a door is opened they together have a 2/3 chance of hiding the car. But, just like the coin where heads and tails each have a 50% chance, each door has a 1/3 chance meaning if I play the game 300 times with the car randomly placed, I'd expect the car to be behind any particular door about 100 times. It is only conditionally, when a known one of the doors is opened that it has a 0% chance. Before the door is opened, or (this one is extremely counterintuitive) even after "a door" is opened if you don't know which one, both unpicked doors have an equal 1/3 chance. Think about playing the game 300 times and how many times the car will be behind any particular door - before any door is opened, after "a door is opened", and after "a particular known door is opened". This is the probability. Unless you know which door has been opened, you have no way to change your initial 1/3 estimate. -- Rick Block (talk) 01:30, 10 February 2010 (UTC)

Thank you so much, Rick. Yes, you are right, of course you are right. But I was not about the chance of winning only, but I especially am considering the RISK of the unchosen pair of two doors to hide one or even two goats, also. Each of those two doors has a risk to hide a goat of 2/3, yes. But that's not all you know. You have additional info in this respect. You also know that in any case there actually "IS at least one unavoidably given goat" behind this pair of two doors. That's a FACT. So you know that the "risk" of each door to contain a goat is 2/3, but you know at the same time that the risk of those two doors to hide a goat never can be spread 2/3 + 2/3 in reality (with an insignificant minor chance for containing two cars at the same time), but necessarily as a matter of fact must be spread exactly "3/3 + 1/3" or "1/3 + 3/3", likewise. Another option will never exist. You know this is a fact. And this known fact has to be considered and never may be neglected. As to the "risk of 3/3", no comment is necessary, positioning will be shown anyway. A goat will be shown without "I'm telling you what, I have two nuts". If the host has a choice between two goats to show, then he will NOT tell about, then he will disclose any additional information about, he just "shows a goat there". You only have to mention that then the risk of the "other unpicked and still closed partner door" is only 1/3. It will hide the "second goat", with a risk of 1/3 though, and it can hide the "only car" with a chance of 2/3. But I am just talking of the RISK of this pair of doors in the "simple game", and of the given facts you know. As to the risks, you know not only one aspect, you sure know at least two aspects: "2/3 + 2/3" on the one hand, and you also know the effective spreading must be "3/3 + 1/3" resp. "1/3 + 3/3", likewise. Ignoring this fact you will resort to conditional probability without needing such resort. Why not using all info you have on hand? You asked: What does "probability" mean to you, and I am asking: "What implies given information?". Kind regards -- Gerhardvalentin (talk) 12:17, 10 February 2010 (UTC)

Pardon the intrusion...

The Combining Doors solution relies on two rules:

The total probability must always equal 1

If I know the probability of 2 of the doors, I can calculate the probability of the 3rd.

Glkanter (talk) 19:27, 10 February 2010 (UTC)

@Glkanter: I hope you're serious this time. So I ask you to demonstrate your rules. For a start let us agree that random placement of the car means that for every door the chance of hiding the car is 1/3. How do you proceed? Nijdam (talk) 12:05, 12 February 2010 (UTC)

The contestant (you or I) selects door #1. Glkanter (talk) 12:38, 12 February 2010 (UTC)

The host opens door 3 with a goat. Finish with your calculations. Nijdam (talk) 14:27, 12 February 2010 (UTC)

1/3 +(2/3 + 0) = 1. The bold values are unchanged by the host's actions. The '0' represents door #3 which has a goat. I believe I'm using Devlin, among others as my source. Glkanter (talk) 16:03, 12 February 2010 (UTC)

Not correct, we agreed that for every door the chance to hide the car was 1/3. Nijdam (talk) 16:05, 12 February 2010 (UTC)

You'll need to discuss this with Devlin and the others. Glkanter (talk) 16:39, 12 February 2010 (UTC)

So you don't understand. Nijdam (talk) 00:35, 13 February 2010 (UTC)

Nijdam, those two unselected doors 2+3 (with a chance to hide the car of 1/3 each, together 2/3) may contain ONLY ONE car, so at least ONE of them HAD to contain an inevitably given GOAT with a chance of 0. - Isn't it so? This was known as soon as door 1 was chosen. Can you verbally explain or justify a different perspective? In case that door 2 should contain the second goat, let's suppose the host was not telling that in any way. Regards, -- Gerhardvalentin (talk) 17:24, 12 February 2010 (UTC)

I add, as above: the two unseleceted doors have together a chance of 2/3 to hide the car, but each of them separately has chance 1/3. That's what probability is about.Nijdam (talk) 23:55, 2 March 2010 (UTC)

You are right, Nijdam. The two unselected doors have together a chance of 2/3 to hide the car, but you KNOW that just ONE of them has NOT a chance of 1/3, but definitely must a priori have a chance of 0. Just one of those two doors. And one goat will later be shown anyway. So the second door a priori has a risk of only 1/3 and double chance of 2/3. You need not to calculate or to guess. You just KNOW it.

>>May be you need glasses, I didn't just say the two unselected doors have together a chance of 2/3, but each of them separately has chance 1/3!! And please do not speak about "risk", it is a term invented by yourself and without meaning here. But ... the good news is, you seem to be on the right trail as you speak now of a priori. You may see the light in the end. Nijdam (talk) 11:23, 3 March 2010 (UTC)

Glasses? Three doors. Prior chance on the car: 1/3+1/3+1/3=1. One door selected: chance 1/3. Two doors unselected: 2/3. And it is GIVEN INFORMATION (!) that at least one of those two doors has a chance of 0 because behind those two doors is one "given" goat. Read the rules. If you insist on the one door hiding a goat also having a chance of 1/3 though, it's your fun. Regards, -- Gerhardvalentin (talk) 12:29, 3 March 2010 (UTC)

Okay, no glasses, but binoculars! I do not know which door hides a goat, I do however know which door shows a goat, after it is opened. And yet, belief it or not, the (prior) chance for this door to hide the car is 1/3. But, and that's what it's all about, and what I have been explaining over and over, the posterior chance it hid the car is 0. I hope you understand the distinction. And if you do, the prior chance of the "combined" doors to hide the car is 2/3, right, but what about the posterior chance? I really wouldn't know, unless I calculate it. And right, calculation shows it is also 2/3. See? Nijdam (talk) 12:53, 3 March 2010 (UTC)

Thank you for your patience.

>>Don't mention

Thank you though, for trying to figure out my clumsy arguments

But listen, I told it over and over: You do not need to know which door hides a goat.

>>Well, actually I need to know, but alas, I do not know

No Nijdam, believe it: Actually neither you nor the guest need to know. The guest just has picked her door and she hopes she made the right decision. She is not wondering about "which one of the two unchosen doors will hide one given goat". Because the host did not show a goat there yet, nor had she been offered to switch to the second unchosen door yet. What you say is just not right. And you did not say WHY you are eager to know. In this stage this is no item of interest.

>>I'm eager to know, because then I can be sure to get the car! Got it?

There's just ONE "given" goat there, and one goat will be shown.

>>There are TWO goats, be it I do not know where.

The "risk" (yes, "the risk") that the door chosen by the guest hides a goat has been 2/3 (66 %). But the host shows ONE "given" goat behind the pair of the two unchosen doors. If he opens the door at random, not flashing any "illegal Morgan-info", this pair of two unchosen doors has (!)/had (!) a "risk" (yes, "the risk") to hide TWO goats of only 1/3 (33 %), i.e. the "risk" that the second goat will also be there is just 1/3 (33 %). And even IF the host should have strict "Morgan-preference" the risk of the second goat being there never is 2/3 (66 %) but is only in the range [from 1/2 to 0], [from 50 % to 0 %]. So it never is worse for her to switch. And yes, that's the question: "To switch or not to switch, that's the question".

>>(1) The word risk has a complete different meaning in scientific circles, so better not use it. And you seemingly just mean: the probability (although a probability is never greater than 1, and your risk sometimes is.) (2) The host shows A goat, indeed ONE, but not a GIVEN one. (3) The rest of your arguments I don't understand. BTW: to be, or not to be, that's the main question.

Three doors: door selected, door unselected, door unselected. In consequence of given information the prior chance (1/3, 1/3, 1/3)

>>Here youre right

must in effect be
A:(1/3, 2/3, 0) or B:(1/3, 0, 2/3), another variation never exists.

>>What the h... do you mean by "in effect".

You know about reality? I mean "in effect", say "as a matter of fact in the real world"

>>Maybe, this is, I think, where you go wrong. In a probability situation, not what really is effectuated is important, because we just don't know, but what may be effectuated.

And if a legal door is opened then at random, without signaling any further information, the posterior chance remains unchanged!

>>Nowhere in the problem statement is mentioned that the door is opened without signaling any further information,

Think you may need glasses. MY statement just was: "And if a legal door is opened then at random, without signaling any further information, the posterior chance remains unchanged!" Please read my lips!

>>I wear already glasses, but from my distance your lips still are difficult to read. Anyway, it's unimportant, because they are not correct. It's not what you say, but what is formulated in the problem statement. This doesn't say anything like: "without signaling any further information.

You may use whatever method you are acquainted with to find out that it remains unchanged: In any case it is given fact that it did not change.

>>No it is definitely not a given fact. That's why I have to use some method.

Yes, that's what I say: YOU can use ANY method you know and like to use. But that's not necessarily necessary and you do it on your own risk   :-)

>>Any probability remains unchanged. So I never will discover any change. What I however will discover is that a different probability has the same value as a former one. Interesting, isn't it?

Whatever method. It does not matter whether you use a butcher knife or a pitchfork, or even Bayes or another tool, and regardless what terminology you use to solve the problem. One of those two unselected doors DID hide a goat, you knew that, and opening a legal door at random shows that a goat STILL IS behind it.

>>Well, I'd say one of these doors does hide a goat, may be even both, and one of them with a goat is opened

If no illegal information has been signaled, the only information revealed

>>well, door 3 is opened

Okay, so A:(1/3, 2/3, 0) was shown to be correct

>>No, try to think harder. The prior probability for door 3 to hide the car is 1/3, you said so yourself. And now there is a number 0, how come? Because it is the posterior probability. And equivalently we have to deal now with the posterior probabilities for both doors 1 and 2. And on forehand I have no idea what they will be. See?

is whether it was A:(1/3, 2/3, 0) or B:(1/3, 0, 2/3), the probabilities beyond remain unchanged.

>>these A and B' were not correct

pardon? didn't hear anything.

>>Read my lips.

No matter how you "call" it. You may call it "conditional" (without any condition) or even in chinese or maths language.

>>Call what?

Call what? We are speaking about the chance that the car will be behind the second unselected and still closed door offered as an alternative to switch

>>Yes, offered to the player AFTER she made her choice of door 1 and the host opened door 3.

Morgan et al. and their inaccurate ridiculous "preference distortion" has confused and does still confuse: The clear fact has become almost impassable jelly.

>>I think you're hardly in a position to critisize Morgan

Not critisizing, no. Just stating they invented a completely different game, corrupting the golden "paradox". I just say it was better to stay with the famous pure paradox (even if you do not like it) and to be free to name forgery what it is.

>>As you do not understand the problem you will not understand Morgan.

A herculean task to make access passable again. Morgan et al. must be named as what they are: Not treating the famous Monty Hall at all, but treating a self-invented, new game. They distorted the shiny golden "paradox" beyond recognition. They invented a completely different "game", a bastard, their OWN game, outside the MHP. Everybody courtes it, everyone bows to it. It should be named what it is. -- Gerhardvalentin (talk) 16:35, 3 March 2010 (UTC)

Nijdam (talk) 22:47, 3 March 2010 (UTC)

-- Gerhardvalentin (talk) 19:12, 4 March 2010 (UTC)

Nijdam (talk) 16:18, 5 March 2010 (UTC)

If the host now opens a legal door randomly, the situation is clearly defined. Unchanged chance for the door selected=1/3, and chance for the second closed door 2/3. This is the standard-MHP that clearly shows the paradox.

But Morgan's idea is a completely different game. The extreme: In 1/3 of cases, if he has the choice, the host does not choose at random. With an extreme preference he always is signaling: "Chance of the selected door is not 1/3 but 50 %" (if he opens his preferred door) resp. "Chance of selected door 1 is not 33 % but 0 %" (if he exceptionally opens the other door).

If he always prefers to open door 3 and yet opens door 2, then he is signaling at the same time: Chance of selected door 1 is not 33% now but 0%, i.e. chance of the preferred door 3 is not 66% now, but 100%. If the host opens his preferred door 3, then he is signaling: Chance of selected door 1 is not 33% any more, but considerable 50%, i.e. chance of the locked door 2 is not 66% now but only 50%. The mistake is to take such signaling for the standard MHP. Although this is a completely different game and should not be confused with the standard MHP. The standard MHP shows the undistorted famous paradox.

The mistake is to mix indiscriminately the Morgan's game with the standard MHP, and not to clearly distinguish and differentiate. This is the fatal curse of the article. Can't you see? Please help to enlight this aspect. -- Gerhardvalentin (talk) 10:19, 3 March 2010 (UTC)

I don't want to dig into your reasoning. Just answer, in one line, my simple question, i.e. give the probabilities. And remember: each door hides the car with chance 1/3. Nijdam (talk) 00:35, 13 February 2010 (UTC)

Yes, and by swithing the guest will double his chance from 1/3 to 2/3 in the large number. The necessity of conditional probability may only be given when you are playing any different games, and not even then. In case the host, whenever possible and he is given two goats to select between, never opens one particular specific door within the pair of unpicked doors (say "never door X"), but always the "other" door (call it "always door Y"), then in 1/3 of the cases, when he has no choice but to open door X though, the chance for the car by switching does not rise from 1/3 to 2/3, but from 1/3 to full 3/3.
In contrast: In the remaining 2/3 of cases where the host can open his preferred door Y, the chance by swithing will rise only from 1/3 to just 1/2.
In the large number and in total the chance by swithing is definitely rising from 1/3 to 2/3 nevertheless, but with a difference in opening door Y or door X, respectively.
Any "simple table" will show this effect also and can show it sufficiently, not even challenging conditional probability. Conditional probability is not even necessary in such a version with differing conditions. But you can show later that maths and conditional probability is indeed able to show all of this, but that maths is only reflecting reality and effects of given assumptions. Yes, conditional probability can clearly "show resulting effects", but it never has "to proof" anything in this respect. The role of conditional probability is clearly defined. Regards, -- Gerhardvalentin (talk) 20:54, 10 February 2010 (UTC)

Before a door is opened, one of these must be true (as the contestant can have chosen no more than 1 of the 2 goats): 1/3 + (2/3 + 0) or 1/3 + ( 0 + 2/3). Why is it difficult to grasp that once the host opens door #3, it must be 1/3 + (2/3 + 0)? Glkanter (talk) 10:32, 3 March 2010 (UTC)

The chances of three doors

Nijdam, I said: Those two unpicked doors 2+3 (with a chance to hide the car of 1/3 each, together 2/3) may at best hide ONLY ONE lonely single car, so at least ONE of them HAD to hide the one inevitably given GOAT with a chance of 0. - Isn't it so? This was known as soon as door 1 was chosen. Can you verbally explain or justify a different perspective? - And in case that door 2 should contain the second goat, let's suppose the host was not telling that in any way, and you answered:

"I don't want to dig into your reasoning. Just answer, in one line, my simple question, i.e. give the probabilities. And remember: each door hides the car with chance 1/3."

But there is only one car in the MHP. Please give me any contradiction:
Each door hides the car with chance 1/3 (3 doors, 2 goats, only 1 car).
Three doors hide 2 inevitably given goats, a pair of two doors hide 1 inevitably given goat, 1 single door has no inevitably given goat.
The guest picked door 1. Consequently the pair of unpicked doors 2+3 definitely do contain at least one inevitably given goat.
In case he is given not only one goat there, but the second goat also, the host will open the unpicked doors 2 and 3 randomly, not obscenely flashing additional information. Any other obscene behaviour of the host is not provided for in the simple standard MHP, so any "flashing additional information" would be quite another game, let's call it "The flashing MHP".

The host will open door 3. Chances for the three doors to hide the car before the host openes door 3 and after he has opened door 3:

Chances to hide the car BEFORE the host opens a door
door 1 picked	unpicked door 2	unpicked door 3	total chance
1/3	1/3	1/3	3/3

But additional info says: Just ONE door of the unpicked pair of two doors definitely must contain one inevitably given goat there. Just one of them.
Answer: The consequence is "either - or", there is no third possibility:

Chances to hide the car BEFORE the host opens a door
door 1 picked	unpicked door 2	unpicked door 3	total chance
Either: 1/3	0	2/3	3/3
Or: 1/3	2/3	0	3/3

Host now opens door 3

Chances to hide the car AFTER the host has opened door 3
door 1 picked	unpicked door 2	unpicked door 3	total chance
did not happen	did not happen	did not happen	did not happen
1/3	2/3	0	3/3

Once more situation before the host opened a door and after the host opened door 3, collocating in parallel:

Chances to hide the car BEFORE the host opens a door				Chances to hide the car AFTER the host has opened door 3
door 1 picked	unpicked door 2	unpicked door 3	total chance	Door 1 picked	unpicked door 2	unpicked door 3	total chance
Either: 1/3	0	2/3	3/3	did not happen	did not happen	did not happen	did not happen
Or: 1/3	2/3	0	3/3	1/3	2/3	0	3/3

"between": I am NOT speaking of OTHER games here, for exymple that the host opens TWO or ALL THREE doors at the same time or that the host gives further information in any way, e.g. by not randomly opening the doors in case that he got two goats to show. I am NOT talking about such other games, here. —Preceding unsigned comment added by Gerhardvalentin (talk • contribs) 11:01, 16 February 2010 (UTC)

Please note once more: For this simple standard MHP it is expressly guaranteed that the host, if (only in 1/3 of cases) he can choose between TWO goats, he will always pick randomly, equivalently and symmetric and he never will flash any hint regarding the guest's first choice was the car.

As supposed: This is not a "flashing MHP" where the guest always wins by switching to door 3 (if the host prefers door 3 to be opened and exceptionally opened door 2) or will always win by switching to door 2 (if the host, if he has the choice, prefers to open door 2, but exceptionally opened door 3).

After the host opened door 3 the chance of the originally picked door 1 is 1/3, and the chance of door 2 is 2/3. Am I wrong? Why am I wrong? -- Gerhardvalentin (talk) 19:45, 13 February 2010 (UTC)

I'm sorry, I won't read this. Please answer my question above. And in no more than one line!Nijdam (talk) 21:30, 13 February 2010 (UTC)

I just told it exactly as Glkanter (1/3 - 2/3 - 0/3) but you answered him: "Not correct, we agreed that for every door the chance to hide the car was 1/3", without rebuttal. He told it in just one line, and my line is exactly the same: If the host wasn't obscenely flashing any additional hints regarding the positioning of the car by opening door 3:

The probability of picked door 1 to hide the car is 1/3 (unchanged), of the unpicked door 2 it is 2/3, and of the opened door 3 it is 0/3. Total: 3/3 - Why am I wrong?

-- Gerhardvalentin (talk) 22:32, 13 February 2010 (UTC)

Yes, I would like to hear the answer to this question. Martin Hogbin (talk) 23:56, 13 February 2010 (UTC)

Quite simply because you contradict yourself. We agreed that for every door the chance to hide the car was 1/3 and suddenly you come up with door 3 having chance 0/3. How come? Does this make sense to you? Nijdam (talk) 00:50, 14 February 2010 (UTC)

But why does the probability that the originally picked door hides the car change? Martin Hogbin (talk) 16:17, 14 February 2010 (UTC)

There we go again: why not answer the question, instead of coming up with a counterquestion (is this english?)?Nijdam (talk) 20:11, 14 February 2010 (UTC)

But mine is the question, yours is the counter-question (I think that is better). Martin Hogbin (talk) 20:27, 14 February 2010 (UTC)

(moved upwards to where it belongs) You should have read the above-mentioned axiom, you said you do not want to read it: In case the guest picked the car and the host was given two goats with the choice to select between the two unpicked doors, he will have chosen randomly, equivalently and symmetric, not giving any additional info. Read the above-mentioned axiom: "Any other behaviour of the host is not provided for in the simple standard MHP". But nobody says there were two goats. We don't know. I repeat: "In case" and "randomly", as you can read above. So nothing has changed, except knowledge that behind door 3 there was the only one there "given" goat. Two goats are never "given" in a pair of two doors. In case you hint at "not randomly opened door" then please make that evident and do not hesitate to confess that. -- Gerhardvalentin (talk) 17:18, 15 February 2010 (UTC)Nijdam (talk) 19:33, 15 February 2010 (UTC)

again

No, the issue at stake was: the answer is not simply: 1/3 + 2/3 + 0/3, because for each door the chances are 1/3. And hence there is a contradiction, I hope you agree. And from here on it is your turn again. Nijdam (talk)` —Preceding undated comment added 20:48, 14 February 2010 (UTC).

I agree. The answer is 1/3 = 1/3 until something changes it. Martin Hogbin (talk) 21:05, 14 February 2010 (UTC)

And 5=5.Nijdam (talk) 21:36, 14 February 2010 (UTC)

The prior probability that the originally chosen door hides the car is 1/3. No event changes this. Therefore the posterior probability, after any event is 1/3. Martin Hogbin (talk) 21:49, 14 February 2010 (UTC)

Also after the event {the car is behind another door}?Nijdam (talk) 21:54, 14 February 2010 (UTC)

We agree that the probability that the car is behind the door opened by the host is changed to 0. But the probability that it is behind the door originally chosen by the player must be unchanged. Martin Hogbin (talk) 22:44, 14 February 2010 (UTC)

Let me be of help(?) Let us use some familiar notation (assuming door 1 is picked). P(C=1)=P(C=2)=P(C=3)=1/3. There we agree. I hope you also mean to say: P(C=3|H=3)=0. I.e. not only has it changed into a conditional probability, its value is also changed. What else is there to say? Your turn. Nijdam (talk) 22:57, 14 February 2010 (UTC)

If the question were, does the probability that the car is behind door 3 change when the host opens door 3 to reveal a goat then the answer is clearly 'yes, of course', the probability that the car is behind door 3 becomes 0 on the condition that the host opens door 3 to reveal a goat. If we start with the standard rule that the host must open an unchosen door to reveal a goat and ask the question what is the probability that the car is behind door 3, then this problem is clearly one of conditional probability, the condition being which door the host actually opens. The door opened by the host is clearly a significant condition for that problem, thus that problem is one of conditional probability.

But that is not the question asked. It is not the problem to be solved. The probability that the car is behind door 3 is not the probability of interest, it is obviously 0 in the question as asked. The problem that we need to answer is what is the probability that the car will be behind door 1, given that door 3 has been opened to reveal a goat. It can be shown that this probability is independent of the door opened by the host. Or using your notation P(C=1) must be identically equal to P(P=1|H=3) and P(C=1|H=2). Martin Hogbin (talk) 22:38, 15 February 2010 (UTC)

You need a lot of words, but I distilled from it: The problem that we need to answer is what is the probability that the car will be behind door 1, given that door 3 has been opened to reveal a goat. Im my notation: P(C=1|H=3). Did I understand you correctly? We'll talk about the rest later on. Nijdam (talk) 10:50, 16 February 2010 (UTC)

Yes what you say in words is OK but the notation is superfluous. Martin Hogbin (talk) 15:25, 16 February 2010 (UTC)

It doesn't harm, does it? Do you also agree, that in general P(C=1|H=3)+P(C=2|H=3)+P(C=3|H=3)=1, and because P(C=3|H=3)=0, it follows that P(C=1|H=3)+P(C=2|H=3)=1.Nijdam (talk) 16:25, 16 February 2010 (UTC)

Yes, that is a perfectly good way of representing the situation, But you have chosen to use unnecessary notation. P(C=1|H=1) for example may be simply written as P(C=1) as there is no relevant condition. We know P(C=1) is independent of P(H=3) given the rules of the game. If you want to use H=3 as a condition you can of course but there is no need to. Martin Hogbin (talk) 16:46, 16 February 2010 (UTC)

No, keep calm. I assume you mean P(C=1|H=3). We know P(C=1)=1/3, but what is the value of P(C=1|H=3)? (What you say about independence, I have no idea what you mean. Events may be independent of each other, but P(C=1)=1/3, hence "independent" of anything.) So tell me how you derive the value of P(C=1|H=3). This will give you the opportunity to show your arguments. Nijdam (talk) 21:11, 16 February 2010 (UTC)

My argument is that P(C=1|H=3) = P(C=1) because at no time is any information revealed that might change the probability that the car is behind door 1. No information is revealed because the door chosen by the host is either determined by the dandom position of the car or the random decision of the host which legal door to open. Martin Hogbin (talk) 21:30, 16 February 2010 (UTC)

That's a tricky argument. The host has to act randomly. And a lot of people will also say no info is revealed in the case the host has some preference, purely because he is always able to show a goat. It is the other way around. We can prove (in different ways) that P(C=1|H=3) = P(C=1), and from this we may decide that apparently no information is revealed.

Yes, it is a complicated argument and some people may argue that, ' no info is revealed in the case the host has some preference'. Whether they are correct or not is a question that goes right to the heart of the meaning of the word probability and which would be interesting to discuss further, but it is undoubtedly, and not just apparently, true that no information is revealed if the host chooses randomly.

But ... before you start arguing, did you notice that you yourself stated: The problem that we need to answer is what is the probability that the car will be behind door 1, given that door 3 has been opened to reveal a goat. Just the only thing Rick and I have been said all along, and what you did fight. Nijdam (talk) 00:17, 17 February 2010 (UTC)

Yes, I have been looking for some form of words that might satisfy us both. The point that I am trying to make is that, using the term I used when I first came here, the condition that under the standard rules and with a host who chooses randomly, that fact the host opens door 3 is a null condition. It is a condition that does not affect the probability of interest and that does not invalidate the simple solution. I full accept that to treat the door opened by the host as a condition of the problem is a good way to tackle the problem because it covers the case that the host has a known preference, thus Morgan's suggestion that this is the best way for students of probability to address the problem is correct, but you seem to have accepted that by using alternative, but 'tricky' argument the symmetrical problem can be solved without the formal use of conditional probability, even though an English statement of the required probability may appear to asl for a conditional probability.

Is there something we can agree on here? Martin Hogbin (talk) 09:51, 17 February 2010 (UTC)

Yes, we did already: The decision is based on the conditional probability P(C=1|H=3). Even, and that is important, in the symmetric, etc. case. I don't know what a null condition is, it is a term invented by you.

Yes its is. I thought it might be useful.

But I guess you mean to say: P(C=1)=P(C=1|H=3).

Yes given the game rules and a random legal door choice by the host.

And not because the events {C=1} and {H=3} are independent they are thus related, but the other way around: because they have this relation, {C=1} and {H=3} are independent.

Tossing a coin

Yes, I agree that we can deduce {C=1} is independent of {H=3} from the fact that P(C=1)=P(C=1|H=3) but can we not say that {C=1} is independent of {H=3} because {H=3} is based on a random decision. Suppose, for example that the host secretly tosses a coin when {C=1} to decide which door to open. How could it be that {C=1} (which we know =1 in this case) is dependent on whether the host tosses a head or a tail?

What if the host throws a die and decides for door 2 if the outcome is 1 and for door 3 otherwise. Your question would also be: How could it be that {C=1} (which we know =1 in this case) is dependent on whether the host gets 1 or something else?

The short answer is that in this case the hosts legal door choice is not uniform at random, but this actually goes right to the heart of the meaning of probability. Several other editors have made this same point. In the MHP the only relevant meaning of the word 'probability' is the Bayesian meaning of a state of knowledge. Do you agree with this? If not please state the meaning that you give to the word. Martin Hogbin (talk) 23:16, 19 February 2010 (UTC)

(a) I don't understand; you still may ask the same question. (b) I think most (ordinary) people consider probability in the MHP frequentistic.Nijdam (talk) 21:16, 22 February 2010 (UTC)

The host will reveal a goat with certainty so that tells us nothing. Whether {H=2} or {H=3} depends only on the random placement of the car or the random toss of a coin. P(C=1) must therefore be independent of this choice, we do not need to show that from the fact that (C=1)=P(C=1|H=3).

This is what it is about. What you called the "symmetric solution", but what is a calculation of P(C=1|H=3) using the symmetry, or if you wish, the proof based on the symmetry that P(C=1)=P(C=1|H=3).

But notice {C=2} and {H=3} are not independent.

That is true but I cannot see why it is important.

Only to show that you can't speak of "null condition".

What is a significant condition and what is insignificant (which I call a null condition) must depend on the probability to be calculated. A condition might affect on probability but not another. Martin Hogbin (talk) 23:01, 19 February 2010 (UTC)

And the decision is always based on the conditional probability P(C=1|H=3), although it may be calculated without using Bayes'theorem. That's what I have said. I hope you are not reluctant to admit this. It is not that because P(C=1)=1/3 also P(C=1|H=3)=1/3. This needs, although very little, proof. We cannot on forehand know {C=1} and {H=3} are independent. This is not a given fact in the problem. Nijdam (talk) 21:42, 17 February 2010 (UTC)

I would say that because the choice between {H=2} and {H=3} is random and a goat is revealed with certainty, P(C=1) must be independent of P(H=3).
Martin Hogbin (talk) 11:01, 18 February 2010 (UTC)

You have to say {C=1} must be independent of {H=3}, the events are independent, but the "must" needs a little proof.Nijdam (talk) 11:19, 18 February 2010 (UTC)

Of course, I did not say that this is easy, just that it is possible. Martin Hogbin (talk) 09:13, 19 February 2010 (UTC)

I noticed a loose end here. I have not the slightest clue what you mean by "easy" and "possible".Nijdam (talk) 20:32, 1 March 2010 (UTC)

Again combined doors

I hope you understand now, from the analysis this far, the objection Rick, Kmhkmh and I have against the combined doors solution, which just states that P(C=1)=1/3 and hence, without considering P(C=1|H=3), concludes that P(C=2|H=3)+P(C=3|H=3)=2/3.Nijdam (talk) 13:12, 18 February 2010 (UTC)

No, we can show that P(C=1|H=3) must equal P(C=1) if the host chooses a legal door randomly, as I say above.

I don't follow you. Why the no? You yourself say we can show..., and that's the only thing we are concerned about. It has to be shown, how simple the proof may be. Nijdam (talk) 21:08, 22 February 2010 (UTC)

No, I do not understand the objection. The proof is that no information about the likelihood of the car being behind door 1 is revealed by the door number opened by the host, because this is random information. If you disagree, please tell me what information is revealed by the door number opened by the host that relates to the probability that the car is behind door 1. Martin Hogbin (talk) 22:18, 22 February 2010 (UTC)

It doesn't work this way. If you say no info is revealed, then either this is a given fact, or you have to prove your statement.Nijdam (talk) 23:06, 22 February 2010 (UTC)

Regarding the probability that the car is behind door 1, no information is given by the fact that a goat is revealed, because this is done with certainty. The only other information revealed is the number of the door behind which the goat has been revealed. The door opened is random, either decided by the random initial placement of the car or the random choice of the host. This random number contains no information.

Regarding the probability that the goat is behind door 3, door 3 is not opened with certainty, thus the above argument does not apply. Martin Hogbin (talk) 10:18, 25 February 2010 (UTC)

No Martin, whether info is revealed or not, cannot just be "shown" in thus way of arguing. Your argumens also apply in the case of a host with preference, and then info is really revealed. Hence your arguing is wrong. Nijdam (talk) 10:06, 26 February 2010 (UTC)

If the host has a preference, the choice of door opened is not random thus the number of the door opened may give information. The case that host chooses randomly is a special case in that it results a door being opened uniformly at random. It is a generally accepted fact of information theory that a random value contains no information. The 'information in number of the door shown', as Morgan put it, simply does not exist, there is none. Martin Hogbin (talk) 14:06, 26 February 2010 (UTC)

It seems you don't understand what sound, logical reasoning means.Nijdam (talk) 16:31, 26 February 2010 (UTC)

Well I though I did. Perhaps you can help me. What is illogical about the absence of informaytion in a random number? Martin Hogbin (talk) 22:44, 26 February 2010 (UTC)

The assumption that the host could give away some kind of information if he has two goats to choose from is not the MHP. It's a modified variant, may be created for students of probability theory for training purpose. Can be helpful there. But it's not the MHP. -- Gerhardvalentin (talk) 19:45, 26 February 2010 (UTC)

Yes, I describe it as an academic diversion. Interesting for those who want to study probability but not the MHP. Martin Hogbin (talk) 10:26, 27 February 2010 (UTC)

Yep. Glkanter (talk) 17:14, 27 February 2010 (UTC)

Public coin toss

Suppose we have the standard game with the standard rules but the host publicly tosses a coin to decide which door to open. After the player chooses a door, the host says that he will toss coin to decide which door to open but, if only one door hides a goat he will ignore the coin toss and open that door. Only the host knows which door is heads and which is tails when there is a choice. So tell me, when is information about the whereabouts of the car revealed in this sequence of events.

1. The player chooses a door.
2. The host tosses a coin
3. The coin toss results in a head
4. The host says that he must now open door 3
5. The host opens door 3 to reveal a goat

When is the information revealed and what is it? Martin Hogbin (talk) 18:14, 1 March 2010 (UTC)

After the choice of the player info is revealed, and also after the host opens door 3. The information revealed is the restriction of the sample space.Nijdam (talk) 20:24, 1 March 2010 (UTC)

Another question. Assuming the car was initially placed uniformly at random, the probabilty of the car being behind door 1 is 1/3 after 1 above. After 5, is the car more or less likely to be behind door 1? Martin Hogbin (talk) 19:13, 1 March 2010 (UTC)

Before step 1 the probability of the car being behind door 1 is 1/3. After step 1 and before step 2 the conditional (!) probability of the car being behind door 1 is 1/3. After step 5 the then valid conditional probability of the car being behind door 1 is 1/3. is this what you're looking for? Nijdam (talk) 20:29, 1 March 2010 (UTC)

Not really. It is a generally accepted fact that the probability of an event can only change if information about the likelihood of that event is revealed. For example, in the event that the host has a known (or presumed), non-uniform, door opening policy, the number of the door opened reveals information about the likelihood of the car being behind door 1. If the host does not coose a legal door randomly then information could be revealed when he opens a specific door.

In the case above, what information (that might change the probability of the car being behind door 1) is revealed, and when? Martin Hogbin (talk) 11:51, 2 March 2010 (UTC)

Notice your problem with logic. The probabilities didn't change in value, hence no conclusion can be made from your statement.Nijdam (talk) 12:20, 2 March 2010 (UTC)

I see no problem with my logic. Let us suppose that we have not yet calculated any probabilities, so we do not know whether the conditional probability P(C=1|H=3) is equal to the original unconditional probability P(C=1) or not. My argument shows, without calculating anything, that P(C=1|H=3) must be equal to P(C=1). Thus we know we can use the unconditional value in place of the conditional value in any future calculation. Martin Hogbin (talk) 09:43, 3 March 2010 (UTC)

You have to look better. You stated: change in prob. ==> info revealed. And then you ask: what info is revealed? To be sure info is revealed, there must be a change in prob. But there is no change.Nijdam (talk) 11:13, 3 March 2010 (UTC)

Of course, all sorts of irrelevant information is revealed when the host opens a door to reveal a goat, but none of it is information relating to the probability that the car is behind door 1. Martin Hogbin (talk) 11:27, 3 March 2010 (UTC)

(1) Did you notice the flaw in your logic? (2) How do you know no info is revealed? (3) You can't know, than info about the relevant sample space is actually revealed (I told you this, didn't I?). (4) It turns out this info does not influence the value of the probability for the car to be behind door 1, but it reveals info about the position of the car all along. Nijdam (talk) 13:00, 3 March 2010 (UTC)

No, I have not noticed the flaw. As I said of course some information is revealed, but not informatoon that could affect the probability that the car is behind door 1. Only two pieces of information are revealed when the host opens a door. Firstly that a goat is revealed, this can be done with certainty and therefore tells us nothing. Secondly a door number is revealed. This is a random number and can therefore tell us nothing about the probability that the car is behind door 1 (although it clearly tells us that the car is not behind door 3). If you believe that information relating to the probability that the car is behind door 1 is revealed then please tell me what this information is. Does the opening of door 3 make it more or less likely that the car is behind door 1 (note that in the case that the host has a known door preference this question does have an answer). Martin Hogbin (talk) 23:13, 6 March 2010 (UTC)

Remaining door

Another erroneous way of looking at the MHP is introducing the variable R = the remaining closed door (after a door has been chosen and one opened by the host). With C=door of car, X=chosen door and H=door opened by host, one may argue that X, H and R are all different and because P(C=X)=1/3 and P(H=C)=0, P(R=C)=2/3, hence the simple solution should be correct. The error lies in the use of P(R=C) to base the decision on, instead of P(R=C|X=1,H=3).Nijdam (talk) 18:01, 28 February 2010 (UTC)

The Answer

If one door is opened and it has to be a gat and you started with 2 goats and one car then you are down to one goat and one car so am i t wrog in saying if you switch you have a 50% chance of getting the car and staying on the current door also gives you a 50% chance? I will now graph out the problem so you can get a better idea.

door 1 car goat|goat

door 2 car goat|goat

____________

door 3 car goat goat

The line shows what happens when you remove one goat door — Preceding unsigned comment added by 2601:246:800:E060:71CD:7254:74BA:4312 (talk) 16:43, 27 October 2019 (UTC)