B The Monty Hall paradox/conundrum

[hutchphd]

Shall we call this the Ignorant Monty Version??
It is useful to know that this version does require the winning probability to go to either 0 or 1/2 but only after the reveal. And that is obviously true whether you choose to switch or not.

 

[FactChecker]

Shall we call this the Ignorant Monty Version??
It is useful to know that this version does require the winning probability to go to either 0 or 1/2 but only after the reveal. And that is obviously true whether you choose to switch or not.

The "ignorant one" is essentially the same as the one that people erroneously intuit that there is no benefit to switching. It is meant to illustrate the error of their intuition by showing that their intuition gives the same result as a scenario that is obviously wrong.

I see that this example is not adding anything constructive to the discussion.

 

[hutchphd]

I guess I concur with your evaluation of utility but as #151 on the hit parade one can hardly be shocked that it breaks no new ground!

 

[sysprog]

I think that the error that is commonly made is in thinking that, once there are only 2 doors, the chances are 50:50, instead of understanding the 1/3 versus 2/3 sticking versus switching consequences of the facts that Monty won't open the contestant's door or the prize door before offering the option to switch. Everyone knows in advance that there's at least 1 non-prize door for Monty to open. So Monty's opening of a door doesn't change the chance to 50:50. The other 2 doors continue to hold 2/3 of the prize chance. Switching gets the contestant in effect both of the other doors.

 

[DaveC426913]

I see that this example is not adding anything constructive to the discussion.

It is. I understand your point.
I think the problem is that it is no less complex to intuit for the novice.

 

[sysprog]

It is. I understand your point.
I think the problem is that it is no less complex to intuit for the novice.

It's not complex -- it's dirt easy -- the contestant starts out with 1 of 3 chances to get the prize, and is offered the option to trade his chance for the other 2 chances -- showing that 1 of the other 2 isn't the winner is meaningless -- the contestant already knew that . . .

 

[hutchphd]

Switching gets the contestant in effect both of the other doors.

Yes but that must be coupled with the assurance that Monte is smart and not ignorant. If Monte is ignorant it is again a 50-50 shot that Ignorant Monte will choose the prize door leaving contestant out of luck. So I am missing your point?

 

[phinds]

Yes but that must be coupled with the assurance that Monte is smart and not ignorant. If Monte is ignorant it is again a 50-50 shot that Ignorant Monte will choose the prize door leaving contestant out of luck. So I am missing your point?

Yes, you are missing the point of the game. Monty NEVER picks the door with the car.

 

[hutchphd]

Sorry we redefined the Ignorant Monty game while your back was turned...don't worry this is clearly out of control! No Mas.

 

[sysprog]

Yes but that must be coupled with the assurance that Monte is smart and not ignorant. If Monte is ignorant it is again a 50-50 shot that Ignorant Monte will choose the prize door leaving contestant out of luck. So I am missing your point?

In the version of the game that you partially posited, what happens if Monty prematurely reveals the prize? Does he cancel the game? Does he award the prize? Does he drive away in the car? What does Monty do (the version to which you're apparently alluding is sometimes called the 'Monty Fall' problem).

 

[lavinia]

It's important that you understand the game first.
https://en.wikipedia.org/wiki/Monty_Hall_problem

1. The host must always open a door that was not picked by the contestant.
2. The host must always open a door to reveal a goat and never the car.
3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

I understood the original game after watching the show.

But I thought that a different game was being asked and was not clear what it was exactly. But it seemed to require that Monty pick a door at random rather than a door he knew was a goat. So what then is the game?

One possibility is that if he randomly picks the car then he just gives it to the contestant. In this game the contestant has a two thirds chance of getting the car if he abandons his original door. This just because 2/3 of the time the first door he chose was the wrong door.

Another game would be to try again if Monty picks the car so one ends up with the original game and the probability is still 2/3.

A third game is the Monty doesn't reveal the door in which case the probability of getting the car is 1/3 whether the contestant switches or not.

Maybe the question was what if Monty doesn't reveal the door but by chance always selects a goat.
This seems to be the same as asking what the distribution is of a finite sequence of plays of the original game and this depends on the number of plays. The more plays the more tight the distribution is around its expected value of 2/3. But this didn't seem right either.

So all of this confused me and I thought I didn't understand what the question was.

 

[hutchphd]

what happens if Monty prematurely reveals the prize?

You don't win. That's all that matters for my analysis. Sorry I'm officially done.

 

[Orodruin]

One possibility is that if he randomly picks the car then he just gives it to the contestant. In this game the contestant has a two thirds chance of getting the car if he abandons his original door. This just because 2/3 of the time the first door he chose was the wrong door.

While this is the overall probability in that game given that tactic, it is not the probabilities in front of the contestant at the time of choosing. At the time of choosing, the contestant has already lost the possibility of Monty opening up the car. At the time of choosing, the probability is 1/2 so the probability is independent of your tactics in this game.

This can be seen as follows: The game is equivalent to randomly selecting one door to open. If this door contains the car (1/3) you win. Otherwise (2/3) you get to choose one of two equiprobable doors (1/2) for an overall winning chance of 1/3 + (2/3)(1/2) = 2/3. However, once Monty has revealed the goat, the probability has gone down to 1/2 because we have excluded the outcome that the car was in Monty’s random door.

 

[PeroK]

Shall we call this the Ignorant Monty Version??

It's already got a name. It's called the Monty "fall" problem, where Monty accidentally falls against one of the two remaining doors!

 

[sysprog]

While this is the overall probability in that game given that tactic, it is not the probabilities in front of the contestant at the time of choosing. At the time of choosing, the contestant has already lost the possibility of Monty opening up the car. At the time of choosing, the probability is 1/2 so the probability is independent of your tactics in this game.

This can be seen as follows: The game is equivalent to randomly selecting one door to open. If this door contains the car (1/3) you win. Otherwise (2/3) you get to choose one of two equiprobable doors (1/2) for an overall winning chance of 1/3 + (2/3)(1/2) = 2/3. However, once Monty has revealed the goat, the probability has gone down to 1/2 because we have excluded the outcome that the car was in Monty’s random door.

Only new arrival persons who don't know the original conditions ever experience the 1/2. For the contestant, no, it's not 1/2 -- switching wins 2/3 of the time -- this is something that's been proven in 84 computer languages at https://rosettacode.org/wiki/Monty_Hall_problem $\dots$

 

[PeroK]

I understood the original game after watching the show.

But I thought that a different game was being asked and was not clear what it was exactly. But it seemed to require that Monty pick a door at random rather than a door he knew was a goat. So what then is the game?

One possibility is that if he randomly picks the car then he just gives it to the contestant. In this game the contestant has a two thirds chance of getting the car if he abandons his original door. This just because 2/3 of the time the first door he chose was the wrong door.

Another game would be to try again if Monty picks the car so one ends up with the original game and the probability is still 2/3.

A third game is the Monty doesn't reveal the door in which case the probability of getting the car is 1/3 whether the contestant switches or not.

Maybe the question was what if Monty doesn't reveal the door but by chance always selects a goat.
This seems to be the same as asking what the distribution is of a finite sequence of plays of the original game and this depends on the number of plays. The more plays the more tight the distribution is around its expected value of 2/3. But this didn't seem right either.

So all of this confused me and I thought I didn't understand what the question was.

If we label the doors $C$ for the contestant's first choice, $M$ for the door Monty opens and $R$ for the remaining door, then we have two games:

1) The actual game show. Here the probabilities for the location of the car are:

$C = 1/3,\ M = 0, \ R = 2/3$

In the game show Monty never reveals the car because he knows where it is.

2) Then there is the alternative "fall" game where Monty opens a door at random. Here the probabilities are:

$C = 1/3, \ M = 1/3, \ R = 1/3$

In which case, we have the two conditional cases.

2a) If Monty reveals the car, the probabilities change to:

$C = 0, \ M = 1, \ R = 0$.

2b) If Monty does not reveal the car:

$C = 1/2, \ M = 0, \ R = 1/2$

These can easily be checked by a computer simulation if need be.

 

[sysprog]

@PeroK nicely stated.

 

[Orodruin]

Only new arrival persons who don't know the original conditions ever experience the 1/2. For the contestant, no, it's not 1/2 -- switching wins 2/3 of the time -- this is something that's been proven in 84 computer languages at https://rosettacode.org/wiki/Monty_Hall_problem $\dots$

Please read the actual posts I was replying to instead of just assuming that I am wrong and posting in bold face. I was not discussing the Monty Hall problem. I was discussing the case where Monty opens a random of the remaining doors. If the car is there the contestant wins. If it is not then the contestant can choose to switch. This is just the Monty Fall problem with the addition that if Monty opens the door with the car then the contestant wins.

 

[sysprog]

Please read the actual posts I was replying to instead of just assuming that I am wrong and posting in bold face. I was not discussing the Monty Hall problem. I was discussing the case where Monty opens a random of the remaining doors. If the car is there the contestant wins. If it is not then the contestant can choose to switch. This is just the Monty Fall problem with the addition that if Monty opens the door with the car then the contestant wins.

I didn't mean to be discourteous, @Orodruin, I very much admire you and your deeply insightful writing.

I think that I didn't assume that you were wrong, and I acknowledge that yes, I did use boldface in my declaration that the probability doesn't go to one half in the situation in which you seemed to me to be stating that it does, and I did fully read your post, and the posts to which you were replying.

While this is the overall probability in that game given that tactic, it is not the probabilities in front of the contestant at the time of choosing. At the time of choosing, the contestant has already lost the possibility of Monty opening up the car. At the time of choosing, the probability is 1/2 so the probability is independent of your tactics in this game.

This can be seen as follows: The game is equivalent to randomly selecting one door to open. If this door contains the car (1/3) you win. Otherwise (2/3) you get to choose one of two equiprobable doors (1/2) for an overall winning chance of 1/3 + (2/3)(1/2) = 2/3. However, once Monty has revealed the goat, the probability has gone down to 1/2 because we have excluded the outcome that the car was in Monty’s random door.

I think that unless Monty could have revealed the car the revelation of the goat is inconsequential, because we already knew that he was going to do that. Maybe I should have seen that such a modified scenario was what you were replying regarding. I didn't mean to be presumptuous.

 

[sysprog]

I know that and that is why I rejected this. It is also a stupid game.

While we're imagining changing of the rules, how about this: Monty always opens the contestant's door, and then offers him the opportunity to switch for the other 2 doors -- if the prize is behind the opened door, the contestant probably won't switch, and if it isn't, he probably will switch . . .

 

[lavinia]

While this is the overall probability in that game given that tactic, it is not the probabilities in front of the contestant at the time of choosing. At the time of choosing, the contestant has already lost the possibility of Monty opening up the car. At the time of choosing, the probability is 1/2 so the probability is independent of your tactics in this game.

This can be seen as follows: The game is equivalent to randomly selecting one door to open. If this door contains the car (1/3) you win. Otherwise (2/3) you get to choose one of two equiprobable doors (1/2) for an overall winning chance of 1/3 + (2/3)(1/2) = 2/3. However, once Monty has revealed the goat, the probability has gone down to 1/2 because we have excluded the outcome that the car was in Monty’s random door.

I understood that. So the game seemed pretty stupid if it was Monty picks a door at random and if it is the car he just gives it to the contestant. But then I thought yeah but maybe @FactChecker was asking 'But what if by pure chance Monty picks a goat every time. What then were the overall chances of getting the car by switching doors?' The answer would seemingly depend on how many times the game was played and the distribution of finite strings of goats (I always thought that it was a "lousy pepperoni" not a goat but maybe that was the $50000 Pyramid). If he played the game infinitely many times and always at random got a goat, the probability of switching would again have been 2/3. But then that seemed wrong also. Sigh.

 

[FactChecker]

I understood that. So the game seemed pretty stupid if it was Monty picks a door at random and if it is the car he just gives it to the contestant. But the I thought yeah but maybe @FactChecker was asking 'But what if by pure chance Monty picks a goat every time. What then were the chances of getting the car by switching doors?' The answer would seemingly depend on how many times the game was played and what the distribution of finite strings of goats (I always though that it was a "lousy pepperoni" not a goat but may that was the $50000 Pyramid). But then that seemed wrong also. Sigh.

As in any other conditional probability, the probabilities are only in reference to the condition being satisfied. One can arrange that in several ways. For instance, suppose Monty picks a door with the prize. That can be rejected and ignored and he randomly picks again. (With only two doors for him to randomly pick from, the repeated random selection seems dumb. But it would not seem so dumb if there were more doors.)

 

[DaveC426913]

It's not complex -- it's dirt easy -- the contestant starts out with 1 of 3 chances to get the prize, and is offered the option to trade his chance for the other 2 chances -- showing that 1 of the other 2 isn't the winner is meaningless -- the contestant already knew that . . .

That's only part way to the solution.

 

[sysprog]

That's only part way to the solution.

In the original problem the contestant is confronted with a meaningless revelation -- the following pseudocode is in my opinion an example of a full solution:

if chosen=prize then switch=lose else switch=win

The fact that Monty reveals a losing door does not change the chances from 1/3 for sticking versus 2/3 for switching -- it's never 50:50 unless you change the rules . . .

 

[Nick-stg]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not. What has changed is that you are given the opportunity to select another door, whereas in the beginning there were 2 doors now there is only one and thus the probability is 2/3 that the car is behind the remaining door.

Simple.

Note: I didn't read through the 7 pages of posts, just the first and last, so I apologize if this explanation has already been given by others, I imagine it must have.

The joke is on you, because despite knowing how to "calculate" the probability, in our society nobody cares if you made the "rational" selection or not. People only care that you are the winner, the contestant that choses the door with the car gets all the praise, even if the car was in the first door selected. In fact the contestant likely gets more praise because he/she "knew" the car was there from the beginning.

 

[sysprog]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not. What has changed is that you are given the opportunity to select another door, whereas in the beginning there were 2 doors now there is only one and thus the probability is 2/3 that the car is behind the remaining door.

Simple.

Note: I didn't read through the 7 pages of posts, just the first and last, so I apologize if this explanation has already been given by others, I imagine it must have.

The joke is on you, because despite knowing how to "calculate" the probability, in our society nobody cares if you made the "rational" selection or not. People only care that you are the winner, the contestant that choses the door with the car gets all the praise, even if the car was in the first door selected. In fact the contestant likely gets more praise because he/she "knew" the car was there from the beginning.

Your post is apparently in agreement with mine one post earlier.

 

[FactChecker]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not.

This is where I can never resist a comment because I think you have skipped an essential point. Your statement here is only true because it is guaranteed ahead of time that Monty will not open a door with the prize. If he accidentally bumped a door and it opened, showing a goat, then this is wrong. In that case, the probabilities of both your door and the remaining closed door each increase to 1/2.
In the standard puzzle, it is the intentional avoidance of opening a door with the prize that increases the remaining closed door probability to 2/3 while your door remains at 1/3.

 

[hutchphd]

.
I bid 8 quatloos on door number 3....

 

[sysprog]

I bid 8 quatloos on door number 3...

That reminds me of the Steve Goodman song (further popularized by Jimmy Buffet) that includes the line "my whole world lies waiting behind door number three".

 

[WWGD]

My take: Since doors are opened/chosen randomly, 2/3 of the time , a door with a goat will have been chosen, in which case switching wins. 1/3 of the time a car has been chosen in which case switching loses.

 

[FactChecker]

My take: Since doors are opened/chosen randomly, 2/3 of the time , a door with a goat will have been chosen, in which case switching wins. 1/3 of the time a car has been chosen in which case switching loses.

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete. There never was a side in this argument that thought it was bad to switch. When you say that switching wins, the argument has been whether switching is actually better versus switching just breaks even. IMO, the correct answer to that question must specifically take into account that Monty is forced to select a door without the prize. That is the essential difference between switching being better versus switching just breaking even.

 

[Merlin3189]

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete.

My experience is that, in probability there are even more ways to intuit falacies!

 

[WWGD]

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete. There never was a side in this argument that thought it was bad to switch. When you say that switching wins, the argument has been whether switching is actually better versus switching just breaks even. IMO, the correct answer to that question must specifically take into account that Monty is forced to select a door without the prize. That is the essential difference between switching being better versus switching just breaking even.

Edit:True, I am assuming that in that if a goat is chosen first then a change implies a win ( edit) because Hall will present a goat after the first door has been chosen.

 

[PeroK]

My experience is that, in probability there are even more ways to intuit falacies!

The fundamental issue in this case is to target your thinking towards the right problem. Whether by calculations or "intuition" it's always possible to solve the wrong problem.

The wrong solution to this problem is often the right solution to the different "fall" problem.

 

[FactChecker]

Edit:True, I am assuming that in that if a goat is chosen first then a change implies a win ( edit) because Hall will present a goat after the first door has been chosen.

Yes. This is a good example of why I am a big fan of methodically applying Bayes' Rule on problems like this. It forces one to think systematically of the probability of the condition (probability=1 that Monty picks a door with a goat) and the probabilistic consequences of the condition. I think that helps to point out where some people's intuition goes wrong.

 

[supernova1054]

First rebuttal is experiment. This has been tested and the standard analysis is correct. The switching strategy does in fact measurably increase your odds in the amount predicted.

The second rebuttal is that your analysis is incomplete. You need to include the cases where the goat is behind 2 and where the goat is behind 3. And look at the probabilities for each case.

Think of this. Suppose you could pick 2 doors. Your odds are 2/3. That is the same as the host necessarily picking an empty door if you switch doors.

 

[gmax137]

EDIT
Sorry, I misunderstood your post. Never Mind.

 

[Frodo]

There are many ways to convince people unfamiliar with probability calculations.

The best is undoubtedly to draw up the table of every possibility, where the car is placed behind each door in turn, and where the player chooses each door in turn. There are nine possible outcomes as in the table below. The player who keeps his choice wins three times (1/3) and loses six times (2/3). If the player swaps his choice, he wins six times (2/3) and he loses 3 times (1/3).

A simple way to convince people (it may have been said before bit I haven't waded through all the posts) is to re-order the game slightly. Remember the game is "You choose a door. I will open one of the remaining two doors which I know does not have the car. Do you now want to swap?" So play it as:
1. Choose your door.
2. I ask "Do you want to keep your choice? Or do you want to swap and have both the other doors?"
3. I will now open one of the two doors which I know does not have the car.

Everyone chooses to swap at Step 2.

 

[sysprog]

Another honesty version:

 

[BrassOctopus]

Is this a situation where the probability is being adjusted in the middle of an operation? Can we do that in mathematics?

(General question - I know very little about math)

 

[etotheipi]

Is this a situation where the probability is being adjusted in the middle of an operation? Can we do that in mathematics?

You could analyse it using Bayes' theorem, yes. That theorem will tell you how the probability of a certain event updates, given some new evidence (the new evidence being, that you now know a door behind which there is definitely no car).

 

[sysprog]

You could analyse it using Bayes' theorem, yes. That theorem will tell you how the probability of a certain event updates, given some new evidence (the new evidence being, that you now know a door behind which there is definitely no car).

I think that you don't need to do such analysis if you simply recognize that given the rules Monty always offers the 2 other door chances in exchange for your original 1 door chance.

 

[etotheipi]

I think that you don't need to do such analysis if you simply recognize that given the rules Monty always offers the 2 other door chances in exchange for your original 1 door chance.

Yes. I don't want to re-start this thread since the topic has been discussed to death, but by far the easiest way to think about it is just to notice that if your strategy is to switch then if you initially pick the goat you get the car, and if you initially pick the car you get the goat. The $2/3$ probability falls right out.

But I think @BrassOctopus was asking about a Bayesian analysis, and the answer is that yes, you can do such an analysis if you like.

 

[BrassOctopus]

But I think @BrassOctopus was asking about a Bayesian analysis, and the answer is that yes, you can do such an analysis if you like.

Yeah, I looked this up and, well, it's a bit out of my intellectual reach. But I was just curious if it could be done.

 

[sysprog]

Yeah, I looked this up and, well, it's a bit out of my intellectual reach. But I was just curious if it could be done.

I agree with @etotheipi that the Bayesian approach would work in this instance; however, it's a more subtly principled approach than is actually required in this case: if you want the easy answer it's in Monty's speech bubble in post #189.

 

[sysprog]

Oh and @BrassOctopus and anyone else who might be reading, you might be pleasantly surprised to learn how much more of mathematics than you had previously supposed would be, turns out to indeed be, within your intellectual reach, if not yet within your firm grasp, upon visiting https://www.khanacademy.org/ $-$ it might take some time and effort $\cdots$

 

[JeffJo]

In case anybody, like me, doesn't want to read through all of this? But still wants to see it resolved correctly (which very seldom happens)? I decided to add a response at the end of the thread. Hopefully it will be closed soon.

Let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

Instead, let's say I will roll a special six-sided die. The sides are painted either red or green. Two are painted one of those colors, and four are painted the other color. I tell Alice which color is on four sides, but not Bob. I ask both to bet on which color will come up.

If Alice bets on the color that I told her is on four of the sides, she has a 2/3 chance of winning her bet. Bob has no idea which color predominates, and has to guess. So he has a 50:50 chance of guessing right. If he does, he has a 2/3 chance like Alice. If he guesses wrong, he has a 1/3 chance. His overall chances are thus:

(2/3)*(1/2) + (1/3)*(1/2) = 1/2.​

This is exactly analogous to cmb's scenario. According to the usual interpretation, you (like Alice) have information that says one choice has a 2/3 chance. Your quiz partner (like Bob) does not, and must guess. This is how probability works, and it makes perfect sense if you understand probability.

But this doesn't say why your information means that one choice has a 2/3 chance. The usual explanation is that your original chances, 1/3, can't change. That's wrong. They in fact can change, but nothing you know allows this change. To see why, let's examine a slightly modified version of cmb's table:

This isn't the way it should be displayed - we should fix the player's choice and let the location of the car vary - but it works the same as long as we assume the choice is random. I added the probability of reaching each state at two points in time - before a door is opened, and after. Then, under "Stick" and "Swap," I carried over the probability that each choice was a winner. As you can see, the chances that swapping wins total to 2/3.

What cmb did was count the number of rows where Sticking, or Swapping, would win. That only works if there is no difference between the rows. My table shows that there is a difference. But that isn't the only mistake in this table. A better one is this, where I do switch to fixing the player's pick at door #1 while letting the car placement vary:

But I also added a new variable: Q is the probability, when the Host can open either Door #1 or #2, that he chooses #1. The chances for Staying and Swapping still sum to 1/3 and 2/3, respectively, but one piece of information is ignored. The player knows which door was opened. If it was #2, we have to update the probabilities like this:

The point is that the chances when you stay can change, but only if you know that Monty Hall chooses with a bias. If Q=1/2, which means an unbiased choice, the probabilities evaluate to 1/3 and 2/3.

 

[sysprog]

Reiterating:

 

[Moes]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not. What has changed is that you are given the opportunity to select another door, whereas in the beginning there were 2 doors now there is only one and thus the probability is 2/3 that the car is behind the remaining door.

Simple.

In other words , the host, by showing you which of the 2 doors not to choose, he showed you which of the 2 to choose. So you are basically choosing 2 doors with the host just showing you which one to point to, since even if you were able to choose 2 doors from the beginning there would be one with a goat behind it. If she was able to choose 2 doors from the beginning the host showing her that one door she chose had a goat behind it obviously wouldn’t change her odds from 2/3.

Yes, it’s simple.

 

[JeffJo]

Reiterating:

View attachment 294328

Reiterating - this ignores information that the contestant has. Plus, it adds information that the contestant does not. The contestant sees what door is opened, which is what it ignores. And the contestant does not know where the car is, so your "If initial guess..." conditional cannot be evaluated by the constestant.

Maybe numbers will help Play the game 30 times. Initially, pick door #1 each time:

• 10 times, the car is behind #3 and the host opens #2.
• 10 times, the car is behind #2 and the host opens #3.
• 10 times, the car is behind #1 and the host has a choice.
  • He opens #3 N times, where 0<=N<=10.
  • He opens #2 the remaining 10-N times.

Now consider three cases:

1. If #3 is opened:
   • Staying wins N out of (10+N) times.
   • Switching wins 10 out of (10+N) times.
2. If #2 is opened:
   • Staying wins (10-N) out of (20-N) times.
   • Switching wins 10 out of (20-N) times.
3. If Monty pushes #2 and #3 together so that they are one door, goes behind it without letting you see, and comes back leading a goat on a leash:
   • Staying wins 10 out of 30 times.
   • Switching wins 20 out of 30 times.

Either case #1 or case #2 is what happened. So the correct solution has to consider something between 10 and 20 possible games, not all 30. The probability for door #1 does not have to be "1/3 because that is what it was when you initially chose door #1." But unless you have some reason otherwise, you should assume N=5, and the probability for door #1 is 1/3 because that is 5/15.

Your solution is for case #3, and it is an incorrect solution (because case #3 did not happen) that gets the right answer (because you can only assume N=5).