Monty Hall Problem

[articulett]

Articulett: the odds can change when Monty opens a door, depending on what new information can be inferred from what is revealed behind the door. CurtC summed up the different possibilities here: http://www.internationalskeptics.com/forums/showpost.php?p=3906664&postcount=418

Marilyn vos Savant (who is known to be very smart indeed and an expert on this particular problem) also explained the difference between the version where Monty chooses a door at random and the original problem, where he knows where the car is. The odds really do change to 50/50 when Monty opens a door at random, happening to reveal a goat. In the original version, where Monty always reveals a goat because he knows where the car is, the chance of the prize being behind your chosen door stays at 1/3, (while that of it being behind the remaining door is 2/3.

You can read this in the Wikipedia article, but for convenience I'll quote vos Savant again here. The question is: is it in the contestant's interest to switch doors after the host has revealed the goat? The answer is: it depends on what information the host had when choosing the door. As Marilyn vos Savant puts it:

Yes... I agree... with the alternate scenarios instead of the basics (where the host always reveals the goat and gives you an option to switch) that whether it's better to switch or not IS altered. However, your odds of having picked the correct door the first time don't change... your information as to whether to switch or not is changed by alternating host scenarios... but your odds of having picked correctly on the first choice remain 1 in 3... in the random scenario the host has the odds of picking a goat 2/3 of the time... and if he does... then the remaining door has an equal chance of being the car as the one you have. But even still-- in this case if you played it again and again-- you'd still have only picked the correct door the first time in 1/3 of the cases... and in 1/3 of those cases Monty will reveal a car in his blind guess... whether he offers you a chance to switch or not changes the game.

But my claim is that no matter how you fix the game or what the host does... your odds of having chosen the correct car on the first choice is 1 in 3-- whether you are allowed to switch or not or whether he reveals a car or a goat blindly or on purpose or only offers you a chance to switch if you are wrong doesn't change the fact that your original choice has a 1 in 3 chance of being right. That doesn't change... what the host does or doesn't offer you and his motives for doing so certainly could change whether you switch or not-- but your odds of having chosen correctly on the first choice don't change.

Yes, if he only offers to switch when you are a winner (that would be in 1/3 of the cases), then you should never switch. If it's blind, then it's just as good to stay as switch if he reveals a goat-- if he reveals the car (1/3 of the time) and he lets you switch-- you switch 100% of the time. If what he offers a person is based on whether they have a car or a goat--then the offering will be different in 1/3 of the cases. But these are all side stories. None of these change the fact that you first choice has a 1/3 chance of having been correct. Whether it's correct or not may change what the host offers you and whether you should switch... but it doesn't change the fact that your first choice has a 1 in 3 chance of being correct. It's never more than that. You can increase your odds by switching. But the odds that you have chosen correctly the first time remain at 1 in 3.

I agree that it's not always best to switch given the alternating scenarios. However I still maintain that your original odds for having chosen right the first time are not more than 1 in 3. Claus contended that your odds of having chosen correctly changed. In fact, that's what this derail is about. That isn't the case. Your information regarding whether to switch or not IS altered by these differing scenarios. Having more information doesn't change the fact that your first choice has a 1 in 3 chance of being correct.

(And because of this, if given the opportunity to switch-- without knowing anything else-- the odds are better on average if you do switch... the whole "lesson" in this little scenario.)

 

[sol invictus]

Whether it's correct or not may change what the host offers you and whether you should switch... but it doesn't change the fact that your first choice has a 1 in 3 chance of being correct. It's never more than that.

I agree that it's not always best to switch given the alternating scenarios. However I still maintain that your original odds for having chosen right the first time are not more than 1 in 3. Claus contended that your odds of having chosen correctly changed. That isn't the case. Your information regarding whether to switch or not is altered by these differing scenarios.

So let me get this straight.

You now agree that (in the alternate scenario where Monty is "clueless" and has opened a goat door) your odds of winning are not improved by switching; i.e., that the odds the car is behind your door are equal to the odds that it's behind the other unopened door. But at the same time you continue to insist that the odds the car is behind your door are 1/3 and never change. Those two claims imply that the probability that the car is behind any of the three doors is 2/3.

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat .

 

[CFLarsen]

But my claim is that no matter how you fix the game or what the host does... your odds of having chosen the correct car on the first choice is 1 in 3-- whether you are allowed to switch or not or whether he reveals a car or a goat blindly or on purpose or only offers you a chance to switch if you are wrong doesn't change the fact that your original choice has a 1 in 3 chance of being right.

Why do you keep on with this? Who has disputed it?

 

[Vorpal]

I agree with Sol. Earlier, I calculated (post #400) that a clueless Monty that opens a random door that is not the player's pick gives equal chances of winning the game regardless of keep or switch. I thought this was fairly obvious, but since it appears that some people won't be convinced without a program and I've a few minutes to spare, here's an ugly MATLAB script:

trials  = 2e5; cars = 0; valid_trials = 0; for n=1:trials
 car    = 1+floor(3*rand); player = 1+floor(3*rand);  %
 monty  = player;           % Monty guesses until his pick
 while ( monty == player )  % is different from the player's
  monty = 1+floor(3*rand);  % (equivalent to picking a random
 end                        %  door that is not the player's)
 if ( monty ~= car )        % Car not revealed
  valid_trials = valid_trials+1;
  % Player switches to the other door
  for k=1:3 if (k~=player & k~=monty) guess2 = k; end; end
  if ( guess2 == car ) cars = cars+1; end
 end
end; valid_trials, cars/valid_trials

The result was 133172 trials in which Monty did not accidentally reveal the car, 49.82% of the player won by switching.

 

[Herzblut]

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat .

Men do those nasty things!! "Violation of unitrinity", pshaw! Physicists like yourself are the devote, stupid apologists of the evil male! Pigs! Right, articulett?

 

[boooeee]

Monty Hall is a lucky man. Marilyn Vos Savant has granted him cultural immortality. In three hundred years, people will will still be arguing about this problem. Just extrapolate the progression of this zombie thread.

 

[AntiTelharsic]

I think I found the source of some of the confusion here.

From Vorpal's program, it looks like the trial isn't counted when Monty, choosing randomly, reveals the car. That's really important.

Consider the following cases:

1. If Monty chooses randomly and you always switch to the remaining, unopened door (even if he revealed a car), you will win 1/3 of the time.

2. If Monty chooses randomly and you always switch to his door if he revealed a car and to the unopened door otherwise, you will win 2/3 of the time.

3. If Monty chooses randomly and you always switch to the remaining, unopened door, but it doesn't count when he reveals a car, you will win 1/2 of the times that count.

In the first case, half of Monty's would-have-been goat revelations in situations where you'd chosen a goat are now car revelations, so half of your winning switches have been turned to losses. The second case is identical to the standard problem. The third case is just like the first, but since some of the trials have been invalidated your winning choices make up a larger proportion of the total number of trials.

Hope that helps.

 

[Scott1972]

3. If Monty chooses randomly and you always switch to the remaining, unopened door, but it doesn't count when he reveals a car, you will win 1/2 of the times that count.

It might be more clear to say if Monty reveals the car the game ends and if Monty reveals a goat a new game, with new probabilities, begins.

 

[AntiTelharsic]

It doesn't have to end, though.

I pick a door; Monty opens a random other door; if Monty's door hid a goat, I switch to the third door; otherwise this iteration isn't valid.

As the number N of valid iterations of the above goes to infinity, the number of times I switched and got the car approaches 1/2 N. I will win 1/3 of the total number of attempts, but the number of valid attempts will be 2/3 of the total number.

 

[Scott1972]

It doesn't have to end, though.

I pick a door; Monty opens a random other door; if Monty's door hid a goat, I switch to the third door; otherwise this iteration isn't valid.

As the number N of valid iterations of the above goes to infinity, the number of times I switched and got the car approaches 1/2 N.

I understand what you are saying. I just think it's more clear to say the probability changed because a new, different game began rather than the probability changed because we choose to ignore the instances in which Monty reveals the car.

 

[sol invictus]

I think I found the source of some of the confusion here.

From Vorpal's program, it looks like the trial isn't counted when Monty, choosing randomly, reveals the car. That's really important.

Yes, I said that several times already in trying to explain this to articulett.

The statement of this alternative version is that Monty does not know where the car is, and that in the instance in question he has opened a door with a goat. We want to know what the odds are given that data. To compute odds for that situation we should only consider games in which he opens a door and reveals a goat - not ones where something else happens - and then ask in what fraction of those games the car is behind the player's door.

 

[Skeptical Greg]

How many times to we have to reiterate the OP ?

He opens a door and REVEALS A GOAT ..

Really folks, this is not rocket science ...

 

[CFLarsen]

How many times to we have to reiterate the OP ?

He opens a door and REVEALS A GOAT ..

...because he has to:

The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.Source

That's the whole point of The Monty Hall Problem.

This is not about the 1/3rd chance of what the first choice is - or bigoted misandry, for that matter.

This is about the counter-intuitive choice that one has to make, when the additional, inside information is revealed.

 

[Scott1972]

Yes, I said that several times already in trying to explain this to articulett.

The statement of this alternative version is that Monty does not know where the car is, and that in the instance in question he has opened a door with a goat. We want to know what the odds are given that data. To compute odds for that situation we should only consider games in which he opens a door and reveals a goat - not ones where something else happens - and then ask in what fraction of those games the car is behind the player's door.

I agree that everything you wrote is clear and correct.

I think articulett's argument is that the 1/3 probability for each door at the beginning doesn't change, the game changes. At least that is what I got from her posts. By only considering games in which Monty randomly reveals a goat you have essentially created a new game with two doors, 1 goat, 1 car, and no knowledge about what is behind each door. Unless I'm way off base, I don't think anyone is claiming the probability of finding the car behind each door is not 50% for this scenario.

 

[sol invictus]

Unless I'm way off base, I don't think anyone is claiming the probability of finding the car behind each door is not 50% for this scenario.

Articulett claimed that the odds the car is behind the door you pick were always 1/3 and never change, even in this scenario. That is false. The odds are perfectly well-defined, and if the door is opened and a goat is revealed they change to 1/2. If the door is opened and a car is revealed they change to 0.

There is no need to discard the games where the car is revealed, but including them has no effect on the odds in cases where a goat is revealed.

 

[articulett]

Why do you keep on with this? Who has disputed it?

You did, Claus-- in post 535--in your everlasting quest to prove me wrong by knocking down a straw man version of what I actually said.

but your odds don't suddenly change when you place your bet or pick your door...

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

I think anyone can see that if you always stick with your current choice--you will have 1 in 3 chance of winning-- the only way to increase those odds is to change your choice after a goat is shown. The host can always show a goat. The host does not always need to offer you the switch. However, in the scenario--with no added information-- your odds increase by switching. Moreover, even if you add information-- you always have 1 in 3 chance of winning by not switching--that doesn't change with the added information. In your made up scenarios you are only switching whether the host is going to be offering you a switch or not.

Because the only information we have as the scenario is laid out is that he's shown a goat and offered a switch. That is the only information in the scenario. But no matter how much you add information, you are never increasing the odds over 1 in 3 by staying. If the host only offers a switch if you have the car... then the information you get when if he asks you to switch, is that you have the car. But that means that in 2/3 of the cases, he'll never ask you to switch. And because you don't know this is the case, it's more likely that he's offering you to switch because he always does (the original scenario) or because he blindly chose a goat... (meaning that if he blindly chose the car... he'd not be offering you the chance to switch.)

The salient part of this exercise is that your odds if you stay are not more than 1 in 3-- ever. The only way to increase those odds is to switch if given the opportunity. (Of course 1 in 3 isn't bad odds)... and even with the 2/3 increased odds with a switch-- you still lose 1/3 of the time.

 

[articulett]

Articulett claimed that the odds the car is behind the door you pick were always 1/3 and never change, even in this scenario. That is false. The odds are perfectly well-defined, and if the door is opened and a goat is revealed they change to 1/2. If the door is opened and a car is revealed they change to 0.

There is no need to discard the games where the car is revealed, but including them has no effect on the odds in cases where a goat is revealed.

Wrong.

Your odds of having picked the car on the first choice are always 1 in 3. The host can always reveal a goat. Revealing a goat doesn't change the fact that you have a 1 in 3 chance of being a winner if you stay with your first choice. The only way to increase these odds is to switch if offered the choice to switch and you have no further information.

 

[RecoveringYuppy]

Can I clarify something articulett?

If you accept that there is only one car behind three doors and one door is open revealing a car, what are the odds that there is a car behind either of the two closed doors?

 

[articulett]

So let me get this straight.

You now agree that (in the alternate scenario where Monty is "clueless" and has opened a goat door) your odds of winning are not improved by switching; i.e., that the odds the car is behind your door are equal to the odds that it's behind the other unopened door. But at the same time you continue to insist that the odds the car is behind your door are 1/3 and never change. Those two claims imply that the probability that the car is behind any of the three doors is 2/3.

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat .

That is only true if you KNOW he's chosen blindly... this presumes however that if he'd accidentally revealed the car, he'd change whether you could switch or not. What would change is whether you are offered a choice to switch. Your odds don't go up of having chosen correctly the first time... It's just that the odds of rechoosing a car go down. And/or whether you are offered the choice to switch or not go down. You cannot change the odds that staying with your first choice every time gives you a 1 in 3 chance of winning the car. You can however increase your odds of getting a car by changing your choices depending on how many extra conditions you want to add to the equation. Does Monty ALWAYS give a choice to switch? Does he always reveal a goat? Do you know of ulterior motives or statistics which show the percentage of times he offers to switch and why?

The added information can only tell you whether you are increasing your odds by switching-- it cannot make "staying with your original choice" better than 1 in 3. What your changing is whether you are offered that choice to stay or not.

 

[Scott1972]

Just to be clear articulett, if Monty randomly opens a door and it reveals a goat and he ALWAYS gives you the option to switch do agree that the probability of the car being behind your originally chosen door is now 1 in 2?

Giving Monty the choice of offering or refusing a switch obviously changes the problem.

Edit: The last sentence is incorrect. The odds are 1 in 2 regardless. I confused myself by thinking about Monty having the choice to either open a door or not.