The Monty Hall paradox/conundrum

[sysprog]

Right, but what is not so clear is that a random opening of doors, all without the prize, WILL change your odds to 50:50. And that makes all the difference.

In the original question, the chance goes from 1/3 to 2/3 -- your first chance doesn't change if you don't switch, and it changes to the complement if you do, and the door-opening stuff is meaningless sideshow. Monty's essentially letting the contestant swap his one door for two. If you know that Monty's not going to open your door, or the door that he knows to be the prize door, and you know that he's going to leave only one door other than yours unopened, then you know that the only remaining door has all of the chance of being the prize door that your first-pick door does not have.

 

[FactChecker]

It seems to me that if Monty picks the door at random then the odds are still 2/3.

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

 

[lavinia]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

I don't see that.

 

[lavinia]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

What if the new door does have the prize?

 

[FactChecker]

What if the new door does have the prize?

What do you mean by "new door"? If you mean the one that Monte chose to open, then the game is over and both your door and the remaining door odds go equally to zero. Of course, this is not how the game is actually played. Monte avoids opening a door with the prize.

 

[lavinia]

What do you mean by "new door"? If you mean the one that Monte chose to open, then the game is over and both your door and the remaining door odds go equally to zero.

Yes I meant the door that Monty opens.

Sorry. I didn't understand your point. You are saying what if Monty chooses randomly and the cases where he randomly picks the car are excluded from the game? I guess I don't understand what the game is.

 

[DaveC426913]

I guess I don't understand what the game is.

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.

 

[sysprog]

Given that he picked the door at random AND there was no prize behind it, your odds change. The new information that that door had no prize changes your door odds just as much as it changes the odds of the other remaining door.

Monty chooses at random only when your door conceals the prize. If the prize is behind one of the other 2 doors, he can open only the non-prize door. It's really as simple as this: if chosen has prize (1/3 chance) then switch loses, else switch wins (2/3 chance) -- there is no 50:50 chance ##\dots##

 

[FactChecker]

Monty chooses at random only when your door conceals the prize. If the prize is behind one of the other 2 doors, he can open only the non-prize door. It's really as simple as this: if chosen has prize (1/3 chance) then switch loses, else switch wins (2/3 chance) -- there is no 50:50 chance ##\dots##

In the real game that is true. My statement was intended to show the difference between the real game, in which the expected gain is better if you switch, and another variation, where the 50:50 odds result always holds. In that other variation, Monte is not influenced by any knowledge of where the prize is and makes a random pick of the door to open.

 

[sysprog]

In the real game that is true. My statement was intended to show the difference between the real game, in which the expected gain is better if you switch, and another variation, where the 50:50 odds result always holds. In that other variation, Monte is not influenced by any knowledge of where the prize is and makes a random pick of the door to open.

So in such a variation, Monty could reveal the prize before offering the switch, which would make the switch offer meaningless.

 

[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.