Monty Hall Problem... For Newbies

[GlennB]

I thought it was this,

If you maintain your original choice, door 1 (1:3 that its the car) then its 2:3 that one of the other two doors is the car,

once you have seen the goat behind one of the other two doors, then its still 2:3 that one of the other two doors is the car, but there's only one door left to swap to.

so by swapping, you are swapping from a position of 1:3 that its the car, to a position of 2:3 that its a car, making it more likely to get the car by swapping than not swapping.

makes sense to me.

Yes (and Hi, welcome to the madhouse )

 

[fromdownunder]

Putting it another way, you originally have a one in three chance of being correct with three closed doors.

What if Monty opens a goat door before you pick a door? You move to a one in two chance of winning the car.

The odds shift from 33% to 50%, because Monty is always going to open a goat door, whether or not you pick a door before he shows you a goat.

Norm

 

[bruto]

I'm curious about how many people pick the correct answer to the Monty Hall problem when they first hear about it, so only answer the poll if you hadn't heard about the Monty Hall problem until now (and no looking up the answer before you respond).

There are a couple of threads about this problem already (here, and here), but none of them have polls attached.

The Situation:

You're on a game-show trying to win a new car. The car is behind one of three doors. Behind the other two doors are goats.

You pick one of the three doors at random and the presenter (who knows which door the car is behind) opens a door that you didn't pick, revealing a goat. (This is standard procedure for the show.)

The presenter then offers you a chance to change your mind and pick the remaining door instead.

The Problem:

Are you more likely to win the car if you stick with your original choice?
Are you more likely to win the car if you change your mind?
Does it even make a difference?

Now you know the problem, let us know what you think the right answer is in the poll above.

Obligatory XKCD reference
[qimg]http://imgs.xkcd.com/comics/monty_hall.png[/qimg]
"A few minutes later, the goat from behind door C drives away in the car."​

When I first heard about this problem very long ago, I was occasionally playing with programming in BASIC, so I figured I would make a very simple program to test this. As I was trying to find a way to avoid complexity, it occurred to me how really simple this problem is. If you initially picked right, then change, you will always end up wrong. If you initially picked wrong, then change, you will always end up right. The odds of your original choice being wrong are reversed in your favor if you change your mind.

 

[ddt]

Actually, it doesn't matter - for the probabilities - if Monty Hall knows where the car is or not. The probabilities stay the same if he randomly opens another door.

In the canonical setup, you precisely get the car in case your initial pick is a goat-door. That remains the case if Monty picks a random other door. Suppose your initial pick is a goat-door and Monty randomly opens another door. Then either he picks the other goat-door, and you switch to the third door - which has the car - or he picks the car-door and you obviously switch to the (now opened) car door.

I disagree. If Monty RANDOMLY picks his door and shows a goat then it increases the likelihood of your original choice being correct in the sense that he COULD have revealed the car, but didn't. This is more likely to have happened if he was choosing from two goats (i.e.you chose the car) than if he were picking from one goat and one car. Applying Bayes theorem would show it now becomes 50/50.

No. There's a major error in your thinking. Monty opening a door doesn't change your a priori probability of having picked the right door. Monty opening a goat door - when he knows which door hides the car - doesn't give you extra information, so that doesn't increase your a posteriori probability either.

What most astounds me about the problem is that it is basically very simple. The simplest and most basic strategy in probability theory is simply counting. Count the total number of possible scenarios and count the successful ones. Of course, you have to be careful that all those scenarios are equally likely. But really, that strategy is the simplest and it works in this case. No amount of waffling - of which I see a lot in these threads - can replace that. Mentioning Bayes without actually using it doesn't cut it either.

So, to lay this case to rest, let's count in this alternate scenario:
1) contestant chooses a door
2) Monty randomly (say, with a coin flip) opens one of the other two doors
3) contestant can switch or not

So, what does a switching strategy do? Of course, in this case "switching" means to switch to the open door if the car is revealed, and switch to the remaining door if a goat is revealed.

Let's call the doors A, B, and C, and fix the car behind door "C". That gives 6 scenarios:

First choice|Monty opens|Second choice|Success
A|B|C|yes
A|C|C|yes
B|A|C|yes
B|C|C|yes
C|A|B|no
C|B|A|no

And again, we have a probability of success of 4/6 or 2/3. QED.

 

[ddt]

Putting it another way, you originally have a one in three chance of being correct with three closed doors.

What if Monty opens a goat door before you pick a door? You move to a one in two chance of winning the car.

The odds shift from 33% to 50%, because Monty is always going to open a goat door, whether or not you pick a door before he shows you a goat.

Norm

No, your odds shift from 33% to 67%. I wonder how many of the people who answered the poll with the right answer - that your odds increase - do so with the wrong calculation of those odds.

 

[fromdownunder]

No, your odds shift from 33% to 67%. I wonder how many of the people who answered the poll with the right answer - that your odds increase - do so with the wrong calculation of those odds.

How does this follow? If you choose one of three doors, the odds are 1 in 3.

If there are only two doors (since one door has been eliminated before you even choose), as per my scenario, it is no different than only having two doors in the first place. 50:50.

Norm

 

[UKBoy1977]

No. There's a major error in your thinking. Monty opening a door doesn't change your a priori probability of having picked the right door. Monty opening a goat door - when he knows which door hides the car - doesn't give you extra information, so that doesn't increase your a posteriori probability either.

What most astounds me about the problem is that it is basically very simple. The simplest and most basic strategy in probability theory is simply counting. Count the total number of possible scenarios and count the successful ones. Of course, you have to be careful that all those scenarios are equally likely. But really, that strategy is the simplest and it works in this case. No amount of waffling - of which I see a lot in these threads - can replace that. Mentioning Bayes without actually using it doesn't cut it either.

So, to lay this case to rest, let's count in this alternate scenario:
1) contestant chooses a door
2) Monty randomly (say, with a coin flip) opens one of the other two doors
3) contestant can switch or not

So, what does a switching strategy do? Of course, in this case "switching" means to switch to the open door if the car is revealed, and switch to the remaining door if a goat is revealed.

Let's call the doors A, B, and C, and fix the car behind door "C". That gives 6 scenarios:

First choice|Monty opens|Second choice|Success
A|B|C|yes
A|C|C|yes
B|A|C|yes
B|C|C|yes
C|A|B|no
C|B|A|no

And again, we have a probability of success of 4/6 or 2/3. QED.

If Monty randomly chooses a goat then the probability of our having the car now becomes the conditional probability of our having the car GIVEN that Monty has randomly picked a goat. Therefore we discount your 2nd and 4th outcomes above (where Monty picked the car) and ask 'What is the probability we have the car given that Monty has picked the goat?' which we can see is 2 out of the 4 remaining outcomes, or 1/2.

 

[SezMe]

I thought the same when I first heard the problem too.

Me three.

It's how the gameshow stays profitable. Most players think their odds changed and stay with their door, meaning they lose and the show gives away fewer prizes.

Are the prizes really a major cost component of a game show production? I would GUESS not. Anybody know?

 

[ddt]

How does this follow? If you choose one of three doors, the odds are 1 in 3.

If there are only two doors (since one door has been eliminated before you even choose), as per my scenario, it is no different than only having two doors in the first place. 50:50.

Norm

Sorry. My apologies. I totally misread your post. There's nothing wrong with the probabilities in the scenario you sketch. I mistook yours for one of the many others in this thread which claimed a 50% probability upon switching in the canonical scenario.

 

[bobdroege7]

I voted how I would've when I first heard the problem, which I now know was incorrect.

I am voting the same way as well.

At first I thought it would be 50-50.

that is incorrect.

 

[ddt]

If Monty randomly chooses a goat then the probability of our having the car now becomes the conditional probability of our having the car GIVEN that Monty has randomly picked a goat.

There's no such thing as randomly picking a goat. Either Monty knows where the car is and on purpose picks a goat; or Monty randomly picks a door and it happens to turn out to be a goat.

Therefore we discount your 2nd and 4th outcomes above (where Monty picked the car) and ask 'What is the probability we have the car given that Monty has picked the goat?' which we can see is 2 out of the 4 remaining outcomes, or 1/2.

I'm puzzled why you use the royal we. And I'm not even going to delve into your question, because it is not relevant. My claim was that Monty randomly opening a second door does not change your chance for success, and you disagreed with that. Show your work for that, or concede I am right.

 

[UKBoy1977]

And now, in an attempt to 'cut it', I will use Bayes formula to demonstrate what I said in my last post.

Bayes Formula : P(A/B) = P(B|A).P(A) / (P(B|A).P(A) + P(B|not A).P(not A))

Let event A = We have chosen the car
Let event B = Monty randomly chooses a goat

then

P(A) = 1/3 (our prior probability)
P(not A) = 2/3
P(B|A) = 1 (as if we have chosen the car, Monty is certain to choose a goat)
P(B|not A) = 1/2 (as he is choosing from one car and one goat)

So from Bayes formula

P(We have car | Monty chose goat) = (1 x 1/3) / ((1 x 1/3) + (1/2 x 2/3))

= 1/3 / 2/3

= 1/2

 

[UKBoy1977]

There's no such thing as randomly picking a goat. Either Monty knows where the car is and on purpose picks a goat; or Monty randomly picks a door and it happens to turn out to be a goat.

OK, when I say 'randomly picking a goat' I mean 'Monty randomly picks a door and it happens to turn out to be a goat'.

 

[ddt]

Are the prizes really a major cost component of a game show production? I would GUESS not. Anybody know?

My estimation too would be that the prizes are a minor cost component. This link claims that "1 vs. 100" cost $700k per episode in the USA. And if the US version is like the original Dutch format, it's a low-budget one for a game show: no entertainers, no costly changing decors, no elaborate games; just a second-rate presenter, an idiot as a candidate, and 100 more idiots in the audience who have to answer the same questions as the candidate and undoubtedly do so for not more than travel expenses and a free lunch.

 

[Brian-M]

However, if you swap, you are now picking from a 1:2 or fifty - fifty option, thereby improving your chances of winning the car.

Now it's 50-50 right? Clearly you would change your door.

Once he reveals a goat, the odds of you having the winning door are now 50%. If you change, your odds are also 50%.

I'm amazed at how many people think you should switch because the odds are now 50%.

When I first heard the problem I mistakenly thought the odds had become 50% too, but then concluded that this meant it didn't matter whether you switched or not.

If you only have a choice between two remaining doors, and there's a 50% chance of the car being behind one door then this means that there's a 50% chance of the car being behind the other door too, so it wouldn't matter which door you picked because the odds are the same.

(It's only when I realized that the odds weren't 50% of it being behind the other door that I changed my mind.)

But why isn't just as likely to be behind the door you originally picked? 50% one door, 50% the other. An equal chance at the start that the car is behind X, and an equal chance after seeing a goat that the car is behind X.

Because when you picked that door, there was a 33% chance of it being behind that door and a 67% chance of it being behind a different door.

Monty opening a door that doesn't have the car behind it doesn't change the odds of your original choice at all. There's still a 67% chance that the car is behind a different door... and there's only one other door left to choose. Therefore if you change your choice to the remaining door, you have a 67% chance of winning.

If you didn't make your pick until after he opened the door, then the odds would be 50%, because you're picking between two equally likely options.

 

[UKBoy1977]

And I'm not even going to delve into your question, because it is not relevant

It's completely relevant and the fact you would say this suggests to me you don't know what conditional probability is. We have an initial probability, something new happens and we then ask ourselves what the probability is now given this new information. If Monty knows where the car is then the probabilities don't change (and plugging the relevant numbers into my calculation above would show this). If he chooses randomly and happens to get a goat, it does change as I showed above.

 

[Jrrarglblarg]

Monty Haul problem.

[nitpick]

The term is Monty Hall. The term Monty Haul was coined in the days of AD&D 1st ed to refer to dungeons with an excessive amount of loot.

[/nitpick]

 

[ddt]

And now, in an attempt to 'cut it', I will use Bayes formula to demonstrate what I said in my last post.

Bayes Formula : P(A/B) = P(B|A).P(A) / (P(B|A).P(A) + P(B|not A).P(not A))

Let event A = We have chosen the car
Let event B = Monty randomly chooses a goat

then

P(A) = 1/3 (our prior probability)
P(not A) = 2/3
P(B|A) = 1 (as if we have chosen the car, Monty is certain to choose a goat)
P(B|not A) = 1/2 (as he is choosing from one car and one goat)

So from Bayes formula

P(We have car | Monty chose goat) = (1 x 1/3) / ((1 x 1/3) + (1/2 x 2/3))

= 1/3 / 2/3

= 1/2

And how does this affect my claim that in a scenario that Monty randomly opens a second door, you still have a 2/3 probability for success? You started your post #77 with saying you disagree with that claim. I'm still waiting for an explanation for that. Only looking at a subset of the possible scenarios doesn't cut it.

 

[ddt]

It's completely relevant and the fact you would say this suggests to me you don't know what conditional probability is. We have an initial probability, something new happens and we then ask ourselves what the probability is now given this new information. If Monty knows where the car is then the probabilities don't change (and plugging the relevant numbers into my calculation above would show this). If he chooses randomly and happens to get a goat, it does change as I showed above.

You sneakily added the highlighted part. Review what I wrote, and how you reacted to it. This is the scenario I described:

Actually, it doesn't matter - for the probabilities - if Monty Hall knows where the car is or not. The probabilities stay the same if he randomly opens another door.

In the canonical setup, you precisely get the car in case your initial pick is a goat-door. That remains the case if Monty picks a random other door. Suppose your initial pick is a goat-door and Monty randomly opens another door. Then either he picks the other goat-door, and you switch to the third door - which has the car - or he picks the car-door and you obviously switch to the (now opened) car door.

And your reaction started with:

I disagree.

And the rest of your post is not relevant. If you disagree, look at the whole scenario I described and don't try to narrow the focus to a subset of possible outcomes. You set up a strawman there. And I resent your claim I wouldn't know about conditional probabilities. But they're not relevant here at all.

 

[UKBoy1977]

And how does this affect my claim that in a scenario that Monty randomly opens a second door, you still have a 2/3 probability for success? You started your post #77 with saying you disagree with that claim. I'm still waiting for an explanation for that. Only looking at a subset of the possible scenarios doesn't cut it.

My post IS the explanation for that. It uses Bayes' theorem to show how our initial probability of having the car (1/3) increases to 1/2 if Monty randomly chooses a door which happens to have a goat behind it. If you are unable/unwilling to follow the proof then we have reached an impasse.