Monty Hall Problem -- Questions

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• Jeffdef
In summary, the Monty Hall problem deals with a game show scenario where a contestant initially chooses one of three doors in hopes of winning a car, but then is given the option to switch doors after one of the remaining doors is opened to reveal a goat. While some argue that there is a 1/3 chance of winning initially and a 2/3 chance after switching, this is not entirely accurate as the initial choice is not affected by the contestant's bias and the car has an equal chance of being behind each door. Ultimately, the best strategy is to switch doors, as this increases the chances of winning the car to 2/3.
Jeffdef
Monty hall theory(extra).
Hi,

Im not really a math guy here but more of a logical thinker and this question really confused me because of the way people explain it.
Namely i do agree with the fact that taking a wrong door away after the contestant has chosen initially, rises the chance that the contestant will pick the right door by switching.
However: people say there is a one third chance initially and a two third chance after. Maybe I am digging into deep but let me explain with my next example.

Me and two of my friends John and Mike used to play a lot of cards.
We would play a simple game which only took 3 cards from the deck. The 1 clubs. the 2 clubs and the 3 clubs.
We played as followed: John would shuffle the 3 cards and then would deal each of us a random card from the top of the deck. The one who got the highest card would win (It was a really exciting game).

Now let me ask this: what are the odds each game Mike will get the highest card?
Well if you ask me it isn't 1 in 3.

Which one will be getting the highest card is really a matter of how the deck is getting shuffled. If John shuffles in a way which makes the 3 Clubs go on the top a lot time and he deals to Mike first then it wouldn't be 1 in 3 would it?

What I am saying is that it really depends on the coincidence of the way the card is getting shuffled and therefore you can never predict (no matter the amount of games) how often (not even approximately) you will get the highest card.

Now what if there would be 2 “3 clubs” and 1 “1 clubs” does that mean there now is a 2/3 of a chance of getting the high card? The way I look at it the chance has indeed increased but the ratio still isn't 2/3.

The only way of saying what the ratio is of the highcard falling is is by testing each case uniquely. And then afterwards you could say what the ratio is of the amount of times you try d in that case. and not before hand where coincidence and random human decision making has way to much influence on the outcome.

Its the same thing with coinflip.

The ratio the coin is falling heads or tails depends on the way the person throws the coin.

If the person throws the coin in a similar way every time, it will also fall on the same side everytime.

In case of the Monty Hall scenario the the chance of knowing the ratio to the winning door is impossible to know because it depends on the random choice of the contestant and the random doors the gameshow put the goats at.

Still i agree that if you switch you will win more often, but that is only because randomness will always try and go toward switching between 3 possible options(from which 2 in this case are the right ones).

However there is no way to know for certain where the 2 goats will switch and which door the contestant will choose.

Makes sense right?

In the Monty Hall problem, it is assumed that everything is done completely randomly. Not 100% true to reality because as you said, the goats could be shuffled around in a non-random manner. However, a bias of the contestant cannot affect the game provided that the goats are placed in a completely random manner. By definition, a completely random manner is one such that the probability of choosing a goat will be 2/3, and the probability of getting a car will be a third. Unless the contestant has x-ray vision or something, they cannot have enough information to affect their choice in any significant way. Even if they have a bias, they can't actually
be biased in such a way so as to significantly impact the game. Of course, when offered to swap their bias will affect their chances of winning, but there is a difference to be made. In the case of a bias, the contestant's chances of swapping doors is affected but the probability that a car lies behind the other door is something that depends only on how the goats are shuffled, which is random by assumption.

The reason the car's winning chance doubles post-swap is because you initially have only a third chance to pick the car. So it is more likely that the door you chose has a goat behind. When Hall opens another door to reveal a goat, it becomes more likely that the other door contains the car. There still is a 2/3rds chance that the door you chose has a goat behind because once you choose, you can't go back in the past. But the chance of winning a car must have doubled by opening that door. Thus you get a 2/3rds chance to win the car.

The car has equal chance to be placed behind each door. This is one of the requirements of the Monty Hall problem.

It does not matter (=there is no good or bad strategy) which door you choose initially - the car will be behind this door with 1/3 probability.

Jeffdef said:
Makes sense right?

No. The way your friend shuffles the cards has no bearing on the Monty Hall problem. Have you ever tried tossing a coin so that it lands Heads every time?

Albertrichardf said:
In the Monty Hall problem, it is assumed that everything is done completely randomly. Not 100% true to reality because as you said, the goats could be shuffled around in a non-random manner. However, a bias of the contestant cannot affect the game provided that the goats are placed in a completely random manner. By definition, a completely random manner is one such that the probability of choosing a goat will be 2/3, and the probability of getting a car will be a third. Unless the contestant has x-ray vision or something, they cannot have enough information to affect their choice in any significant way. Even if they have a bias, they can't actually
be biased in such a way so as to significantly impact the game. Of course, when offered to swap their bias will affect their chances of winning, but there is a difference to be made. In the case of a bias, the contestant's chances of swapping doors is affected but the probability that a car lies behind the other door is something that depends only on how the goats are shuffled, which is random by assumption.
Thanks for the replies,

I understand that the bias of the person doesn't affect the game itself, but it should affect the chance of picking the right door. assuming the choice of the contestant isn't completely random.
So i think Monty hall is a bad example for the equation and might be why so many people get confused.
Randomly means it has a chance to be behind everydoor but it does not say how often.
If something is realisticly random, the car might as well be behind door number 3 90% of the time.
But the more something gets randomized the bigger the chance the ratio will be more even. And in most cases it will be shuffled in such a way the outcome will be close to perfectly random. In most games of chance you would be more likely to end up with a ratio of for example: door1: 30% of the time a car, door2: 30% of the time a car, door3: 40% of the time a car.
And that is why it seems 1/3 of the time.

If the contestant now makes a random choice on top of the the random place the car is.
Which could be depending on their current mood for example: 50% on door 3. It would be a random choice on top of a random winning door.
In that case the contestants biased choice would have effect right?

btw i do agree that your chances will get higher after he shows 1 goat.
But i don't see how it would necessarily double the chance of 3 on 10 when it might as well be a chance of 4 on 10 created by the random biaz of the contestant to begin with. which then wouldn't be doubled but increased to 6 out of 10.

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Jeffdef said:
So i think Monty hall is a bad example for the equation and might be why so many people get confused.

Monty Hall is a good example of probability theory because most people get confused!

If you watched the show and came to the conclusion that the car was behind door #3 40% of the time, what would be your strategy?

PeroK said:
No. The way your friend shuffles the cards has no bearing on the Monty Hall problem. Have you ever tried tossing a coin so that it lands Heads every time?
PeroK said:
Monty Hall is a good example of probability theory because most people get confused!

If you watched the show and came to the conclusion that the car was behind door #3 40% of the time, what would be your strategy?

That is true, ill take that part back.
But they way it is often explained is with an example of 1/3 which isn't realistic and can be confusing at first.

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Jeffdef said:
But they way it is often explained is with an example of 1/3 which isn't realistic and can be confusing at first.
It is realistic. Getting a uniform distribution for the car position is trivial. Making it non-uniform adds unnecessary complexity to th problem.

Jeffdef said:
Randomly means it has a chance to be behind everydoor but it does not say how often.
If something is realisticly random, the car might as well be behind door number 3 90% of the time.

And then Michael Larson will play the game...

Jeffdef said:
Randomly means it has a chance to be behind everydoor but it does not say how often.

If you are saying that the probability of a repeated event is different from the actual frequency with which it happens, you are correct. Probability theory deals with probability. There is no mathematical connection between probability theory and observed frequency except that probability theory can tell you the probability of a given frequency actually being observed.

However, the analysis of the Monty Hall problem is an analysis of probability. If people are confused by the fact that observed frequencies are different from probabilities, they are confused about any situation that involves analyzing probabilities.

A connection between probabilities and actual frequencies is often made by people who apply probability to real world situations. They may choose to assume that actual frequencies will be equal to probabilities. However, that is not guaranteed by the mathematical theory of probability. It must be justified by the particulars of the real world problem and/or the belief that having "average luck" is guaranteed.

Stephen Tashi said:
If you are saying that the probability of a repeated event is different from the actual frequency with which it happens, you are correct. Probability theory deals with probability. There is no mathematical connection between probability theory and observed frequency except that probability theory can tell you the probability of a given frequency actually being observed.

However, the analysis of the Monty Hall problem is an analysis of probability. If people are confused by the fact that observed frequencies are different from probabilities, they are confused about any situation that involves analyzing probabilities.

A connection between probabilities and actual frequencies is often made by people who apply probability to real world situations. They may choose to assume that actual frequencies will be equal to probabilities. However, that is not guaranteed by the mathematical theory of probability. It must be justified by the particulars of the real world problem and/or the belief that having "average luck" is guaranteed.

Oh ok, well I am glad we got that straight.

Jeffdef said:
Randomly means it has a chance to be behind everydoor but it does not say how often.
We have been naturally casual in terminology. If we want to be formal, we should have said "uniformly random", but that has been assumed throughout. If you are being really formal, "randomly" without "uniformly" does not say there is a chance to be behind every door. It just says that there is a random process, which may always exclude a specific door, for picking the door for the car.

Sorry to resurrect, but this is a pet peeve, and almost never addressed.

Albertrichardf said:
The reason the car's winning chance doubles post-swap is because you initially have only a third chance to pick the car. So it is more likely that the door you chose has a goat behind. When Hall opens another door to reveal a goat, it becomes more likely that the other door contains the car. There still is a 2/3rds chance that the door you chose has a goat behind because once you choose, you can't go back in the past.
The Monty Hall problem is indeed a wonderful example of how Conditional Probability confuses people. And the reason it continues to be controversial, is because the usual explanation of the correct answer, repeatable, doesn't use conditional probability. So although it gets the right answer of 2/3, it doesn't actually explain why the other answer is wrong, and as a solution is technically less correct than the solution that produces that answer.

After the contestant picks a door - and without loss of generality we can call it #1 - there are four possible ensuing outcomes:
1) The car is behind #3, and the host opens #2.
2) The car is behind #2, and the host opens #3.
3A) The car is behind #1, and the host opens #2.
3B) The car is behind #1, and the host opens #3.

The probabilities of these are 1/3, 1/3, 1/6, and 1/6, respectively.

Most people actually use a naive application of conditional probability. If door #3 is opened, they know case #1 is ruled out, Reasoning that cases #2 and #3 (combined) must still be equally likely, they update the 1/3 probabilities to 1/2. This is a correct approach, with one incorrect interpretation that I'll get to soon.

The above solution does not use conditional probability at all; it says the probability of case #3 can't possibly change, which is untrue (again, I'll get there soon), so whatever is left must be 2/3.

The error of interpretation is how to apply the information we have as a conditional probability. If door #3 is opened, case #3A is also ruled out. Case #2 is now twice as likely as what is left of case #3. The same approach taken by the naive solution now gives the right answer.

The solution I quoted is technically incorrect, because it implies that the probability of case #3 is fixed, regardless of any information gained. This is true only if cases #3A and #3B are the same. That turns out to be so, but the implication is incorrect.

JeffJo said:
because it implies that the probability of case #3 is fixed
It is, if the information which door is opened is uncorrelated with the event "it is behind door 1". And that is indeed the case.

mfb said:
It is, if the information which door is opened is uncorrelated with the event "it is behind door 1". And that is indeed the case.
And I said as much. But the solution is still incorrect if the assumption of this state of no correlation is left out. Nor is there an explanation of why it affects the answer.

The reason people are almost never convinced by explanations like that, is because they know that their solution is closer to the technically correct approach. It just makes a simple mistake, and that explanation does nothing that would point it out.

Think about it: say you had two students who gave different answers to the same problem. Would you just say "Susan is right," or explain why John is wrong?

JeffJo said:
Think about it: say you had two students who gave different answers to the same problem. Would you just say "Susan is right," or explain why John is wrong?
Both, and most explanations I see do that properly.

I don't see why conditional probability is needed: If the person chose a door randomly, 2/3 of the time s/he will have selected a goat, and only one third of the time s/he would have selected the car.

It's an easy application of Bayes' Rule, which is all about conditional probabilities. So the proof is easy. The thing that trips people up is the intuitive feeling that the probability of both unopened doors are equally affected by opening the third door.

JeffJo said:
Think about it: say you had two students who gave different answers to the same problem. Would you just say "Susan is right," or explain why John is wrong?

In this portion of the thread, we haven't heard any Susan who gave a complete and coherent answer. Susan should state the probability space, state the conditional probability to be computed and demonstrate how Bayes rule is applied. Until that is done, we'll have to be content with John-type answers.

mfb said:
Both, and most explanations I see do that properly.
Since the two options were mutually exclusive, you can't do both. And not to pick on WWGD, but the following is a perfect example of what I meant.

WWGD said:
I don't see why conditional probability is needed: If the person chose a door randomly, 2/3 of the time s/he will have selected a goat, and only one third of the time s/he would have selected the car.
This is a form of the typical answer. If the host always opens the highest-numbered door he can after the contestant chooses door #1, it is true that s/he will win by switching 2/3 of the time. But that will be all of the time when the host opens door #2, which comprises 1/3 of all games, and half of the time when he opens door #2, which comprises 2/3. Since s/he knows which case she is in, the 2/3 result WWGD arrived at is meaningless.

Now, I'm not suggesting this bias exists, I'm saying that you can point out the error in the 1/2 answer only by addressing it with conditional probability, like that solution does incorrectly.

Stephen Tashi said:
In this portion of the thread, we haven't heard any Susan who gave a complete and coherent answer. Susan should state the probability space, state the conditional probability to be computed and demonstrate how Bayes rule is applied. Until that is done, we'll have to be content with John-type answers.
I'm going to assume you reversed the names, so John gave the correct answer.

And I'm saying that (my) Susan's is less complete, and technically less coherent, than John's. It's only virtue is that it gets the right answer, which turns out to be the average of the two conditional cases.

But considering a probability space is a good idea. Neither does this explicitly. Both imply one where the initial probability the car is behind door #1 is 1/3.

John's incorrect solution goes on to imply that it is 1/3 for each door, that one door is eliminated, so the conditional probability for the remaining two is found by dividing by sum of the remaining initial probabilities. THIS IS CORRECT TECHNIQUE. It just missed one random variable.

Susan's correct answer ignores any other part of the probability space, implying, without explicitly saying, that it doesn't matter. My point is that John can see that he used correct technique, even if it was incomplete. And that Susan didn't. So why should John accept her answer?

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JeffJo said:
And I'm saying that (my) Susan's is less complete, and technically less coherent, than John's. It's only virtue is that it gets the right answer, which turns out to be the average of the two conditional cases.

In recent posts, I don't find a coherent argument that is correct and I don't see a coherent description of an argument that could be called wrong. You are alluding to some sort of argument that you consider wrong, but what is that argument?. If the wrong argument is actually an argument, there should be some unambiguous way to state it that uses the standard terminology such as probability spaces and Bayes rule. If the wrong argument is some verbal presentation that makes no reference to probability theory then its wrong-ness is that it is incoherent.

JeffJo said:
Since the two options were mutually exclusive, you can't do both. And not to pick on WWGD, but the following is a perfect example of what I meant.This is a form of the typical answer. If the host always opens the highest-numbered door he can after the contestant chooses door #1, it is true that s/he will win by switching 2/3 of the time. But that will be all of the time when the host opens door #2, which comprises 1/3 of all games, and half of the time when he opens door #2, which comprises 2/3. Since s/he knows which case she is in, the 2/3 result WWGD arrived at is meaningless.
EDIT : I don't see your point. 2/3 of the time, s/he ( contestant) will choose a goat. It will then make sense to switch. Only 1/3 of the time s/he will choose the car. How is the result meaningless? By what measure/account?
It seems that just because _you_ , hardly an expert in probability Mathematics (as you have , yourself, stated) disagree, it automatically means everyone else is "confused" or " wrong".

EDIT 2: At any rate, the argument still stands: when choosing randomly, 2/3 of the time the contestant will select a goat. Your argument is too imprecise and informal to make a clear point

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WWGD said:
EDIT : I don't see your point. 2/3 of the time, s/he ( contestant) will choose a goat.
And assuming s/he chose #1, half of that time (1/3 of the total), the car will be behind #2 and the host must open #3. And in the other half, the car will be behind #3 and the host must open #2.

In the remaining 1/3 of the time, s/he will have chosen the car, and the host must choose between #2 and #3. The fact is that the answer is determined entirely by the probability of this choice. If the probability the host will choose #3 is called Q, then the probability that switching will win after #3 gets opened is 1/(1+Q). I realize that we should assume Q=1/2 so this is 2/3, but it is not because the chances of picking a goat were 2/3. It is because the host is twice as likely to open #3 if the car is behind #2, than if it is behind #1.

Only 1/3 of the time s/he will choose the car. How is the result meaningless?
Because the principles of Conditional Probability say that, when you given event A, the conditional probability Pr(B¦A)=Pr(B&A)/Pr(A). Here, A is "Host opens #3" with probability 1/3 + Q/3, and B&A is "car behind #2 and host opens #3" with probability 1/3. Notice that the probability of choosing a goat does not appear in the mathematically-correct solution.

Update:
[QUOTE="JeffJo, post: 5784156, member: 177881" <Snip>Because the principles of Conditional Probability say that, when you given event A, the conditional probability Pr(B¦A)=Pr(B&A)/Pr(A). Here, A is "Host opens #3" with probability 1/3 + Q/3, and B&A is "car behind #2 and host opens #3" with probability 1/3. Notice that the probability of choosing a goat does not appear in _A_ mathematically-correct solution.[/QUOTE]

Does not appear in _A_ Mathematically-correct solution. This is not pure Mathematics; there are additional assumptions/conditions And, of course, we do assume/know, the host will open a door with a goat in it, etc.

JeffJo said:
And assuming s/he chose #1, half of that time (1/3 of the total), the car will be behind #2 and the host must open #3. And in the other half, the car will be behind #3 and the host must open #2.

In the remaining 1/3 of the time, s/he will have chosen the car, and the host must choose between #2 and #3. The fact is that the answer is determined entirely by the probability of this choice. If the probability the host will choose #3 is called Q, then the probability that switching will win after #3 gets opened is 1/(1+Q). I realize that we should assume Q=1/2 so this is 2/3, but it is not because the chances of picking a goat were 2/3. It is because the host is twice as likely to open #3 if the car is behind #2, than if it is behind #1.
If the contestant chooses the car, won't she lose with certainty when she switches? I don't see how you're getting a non-zero probability of winning if she chose the door the car is behind in the first place and then switches.

WWGD said:
Does not appear in _A_ Mathematically-correct solution. This is not pure Mathematics; there are additional assumptions/conditions And, of course, we do assume/know, the host will open a door with a goat in it, etc.
That's right, it does not appear in a/any mathematically correct solution. The solution, no matter how you derive it in a mathematically correct way, is that the probability of winning by switching is 1/(1+Q), where Q is the prior probability that the host would choose the door to open that you observed after you opened the door you did. Since this is *A* mathematically correct solution, every mathematically correct solution must produce this answer, and it does not reference the probability that the first choice is a goat.

vela said:
If the contestant chooses the car, won't she lose with certainty when she switches? I don't see how you're getting a non-zero probability of winning if she chose the door the car is behind in the first place and then switches.
[/QUOTE]Yes, she will. What's your point?

In fully 1/3 of the games, the car will be behind door #1when she chooses it, and she loses by switching.

But in only Q/3 of the games will the host open, say, door #3. The other (1-Q)/3 games, he opens door #2, which contradicts the condition that we saw him open #3. he solution, as I stated before but seems to be getting lost, can only consider the games where #3 gets opened.

You wrote, "In the remaining 1/3 of the time, s/he will have chosen the car, and the host must choose between #2 and #3. The fact is that the answer is determined entirely by the probability of this choice." The way I interpret what you wrote here is that "this choice" refers to the host's decision on whether to choose #2 or #3 in the case that the contestant chose door #1 which has the car behind it. Apparently, you meant something different even though that's not what you wrote.

JeffJo said:
In fully 1/3 of the games,

If the point of the recent posts is to illustrate correct reasoning in probability theory, we should not confuse frequencies with probabilities. We don't know how often a probabilistic event will actually happen in a set of games.

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I find it amazing how far people will digress to argue against what they don't want to accept.
vela said:
You wrote, "In the remaining 1/3 of the time, s/he will have chosen the car, and the host must choose between #2 and #3. The fact is that the answer is determined entirely by the probability of this choice." The way I interpret what you wrote here is that "this choice" refers to the host's decision on whether to choose #2 or #3 in the case that the contestant chose door #1 which has the car behind it. Apparently, you meant something different even though that's not what you wrote.
I may have been hurried when I wrote that (on a phone at the airport waiting for a flight that eventually got canceled), so it may not have been well worded, but there is nothing unclear except what it is that you mean.

The antecedent of the expression "this X" should be the last "X" that was mentioned, which is the host's choice between opening door #2 or door #3 after the contestant chose door #1. That's what I meant. I have clearly described the value Q as being the probability that the the host will open door #3 in this circumstance, plus the car being behind #1. I have clearly explained why the mathematical solution for the conditional probability that door #2 now has the car is 1/(1+Q), and is not a function of the probability that the car is behind #1. So maybe you should explain what you think you mean here.

Stephen Tashi said:
If the point of the recent posts is to illustrate correct reasoning in probability theory, we should not confuse frequencies with probabilities. We don't know how often a probabilistic event will actually happen in a set of games.
And you should also not confuse observed frequencies with predicted and/or expected frequencies that I was talking about, which are indeed related to probabilities. We do know what we expect will happen in a set of games, but not what actually will.

+++++

Assuming the door numbers are rotated so that the contestant's door is always #1 (an invariant assumption), the solution via Bayes Theorem is:

Pr(Car=2|Open=3) = Pr(Open=3|Car=2)*Pr(Car=2)/Pr(Open=3)

1) The incorrect answer 1/2 is obtained when the solver misinterprets the fact that the event "Open=3" occurred, with its (prior) probability being the same as the probability of not choosing door #3. Which is 2/3. It should be Pr(Open=3)=(1+Q)/3.
1A) The incorrect answer, while usually arrived at in a naive manner, is a correct application of this formula with the mistaken value.
2) The correct result is Pr(Car=2|Open=3) = (1)*(1/3)/[(1+Q)/3] = 1/(1+Q); and yes, Q should be 1/2.
2A) Yes, this assumes that the prior probability for the car to be behind each door is 1/3. But note that the value cancels out under this required assumption; and that not making it makes the following "usual explanation" even more incorrect.
3) The usual explanation is that Pr(Car=2|Open=3)=Pr(Car=2|Open=2) = 1-Pr(Car=1). This has no basis in mathematical theory, for the question that was asked.
3A) Those who use this explanation are really solving a different question: "What is the probability you will win the car if you are allowed to switch to the better prize from the two behind the other two doors." All revealing a door in this case does, is eliminate a same-or-lesser valued prize.
3B) It is also the expected value of the probability of winning by switching, when averaged over all the possible choices the host could make.
3C) It turns out to get the correct answer if we can't expect the result to depend on the choice, which is true. It would be a correct explanation - but still not a mathematically correct solution - if this were pointed out.

While they may not know exactly why, most people who make the error in (1) can feel that the solution in (1A) is a correct application, and can easily intuit the fallacy in (3A) even if they can't explain it. So once again, why should they accept (3)?

.[/QUOTE] In no other formulation _you can conceive off_ which is not exhaustive: A=Win, B=Switch, then your probability of winning
JeffJo said:
That's right, it does not appear in a/any mathematically correct solution. The solution, no matter how you derive it in a mathematically correct way, is that the probability of winning by switching is 1/(1+Q), where Q is the prior probability that the host would choose the door to open that you observed after you opened the door you did. Since this is *A* mathematically correct solution, every mathematically correct solution must produce this answer, and it does not reference the probability that the first choice is a goat.
So you believe you can conceive every single possible correct solution? You know of every single possible Mathematical solution? And you're telling us you are no expert? Hmm... Here, to show you how you are wrong: A=Win B=Switch and switch all the time. Then P(A|B) =2/3. No need for your ##\frac {1}{1+Q} ## Just because _you_ cannot conceive of other solutions, does not mean they don't exist. I thought this would be obvious , but maybe not so.
I am out of here, enjoy your trip back to the imaginary world where you know and can conceive of every thing.

WWGD said:
So you believe you can conceive every single possible correct solution?
No.

To defend that statement, which I thought I did pretty clearly, all I need to "conceive of" is one. By definition, to be "correct" another other solution must ultimately arrive at the same correct answer,1/(1+Q). If the contestant has no knowledge of where the car is, this is the most general, and correct, answer. Another solution can't both be correct, and get a different answer. (Note that 1/(1+Q) includes the less general, but correct, assumption that Q=1/2 so the probability is 2/3.)

Otherwise, we'd really have a paradox. So maybe you should go back to your imaginary world where mathematics doesn't have to be consistent.

1. What is the Monty Hall Problem?

The Monty Hall Problem is a famous probability puzzle named after the host of the game show "Let's Make a Deal". It involves three doors, behind one of which is a prize. The contestant is asked to choose one door, and then the host reveals one of the other doors which does not contain the prize. The contestant is then given the option to switch their choice to the remaining door. The question is, is it better to stick with the original choice or switch to the other door?

2. What is the correct answer to the Monty Hall Problem?

The correct answer to the Monty Hall Problem is to switch to the other door. This may seem counterintuitive, as it seems like the contestant has a 50/50 chance of choosing the correct door. However, by switching, the contestant actually increases their chances of winning to 2/3.

3. Why is the correct answer to the Monty Hall Problem to switch?

The reason why switching is the correct answer is because when the contestant makes their initial choice, they have a 1/3 chance of choosing the correct door. This means that the remaining two doors have a combined probability of 2/3. When the host reveals one of the other doors, this probability does not change. Therefore, by switching to the other door, the contestant is essentially choosing the 2/3 probability over the 1/3 probability.

4. Is the Monty Hall Problem a real-life scenario or just a theoretical puzzle?

The Monty Hall Problem is a theoretical puzzle that is used to demonstrate the concept of conditional probability. However, it can also be applied to real-life scenarios, such as choosing between two job offers or selecting a door in a game show with multiple prizes.

5. Are there any variations of the Monty Hall Problem?

Yes, there are variations of the Monty Hall Problem, such as having more than three doors or having the host randomly choose which door to reveal. In these variations, the correct answer may change, but the underlying concept of conditional probability remains the same.

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