Monty Hall Problem... For Newbies

[Michael C]

No. the remaining two doors are 50/50 to anybody.

The probability is not 50/50 for anybody. The probability of an event is not an absolute value: it depends on what information the observer has.

- For an observer who arrives after Monty has opened a door and only knows that there is a prize behind one of the closed doors, the probability of the prize being behind either door is 1/2.

- For an observer who has watched the game and has the same information as the contestant, the probability of the prize being behind the first chosen door is 1/3 and the probability of it being behind the other door is 2/3.

- For an observer who knows where the prize is (Monty, for instance), the probability of it being behind a certain door is 1 and the probability of it being behind the other door is zero.

 

[Rasmus]

The probability is not 50/50 for anybody. The probability of an event is not an absolute value: it depends on what information the observer has.

- For an observer who arrives after Monty has opened a door and only knows that there is a prize behind one of the closed doors, the probability of the prize being behind either door is 1/2.

- For an observer who has watched the game and has the same information as the contestant, the probability of the prize being behind the first chosen door is 1/3 and the probability of it being behind the other door is 2/3.

- For an observer who knows where the prize is (Monty, for instance), the probability of it being behind a certain door is 1 and the probability of it being behind the other door is zero.

Can someone explain to me why we're wasting valuable electrons on a discussion of specimens one and three?

 

[JohnnyG]

No. the remaining two doors are 50/50 to anybody.

Let's be clear on what exactly has 50/50 probability. It is the probability that an observer will win the car if they chose between the two remaining doors randomly (meaning without knowledge of the door chosen prior to the reveal).

The probability that the car is behind the door originally chosen is 1/3 and for the other door it is 2/3. Since the second chooser doesn't know which one was previously chosen, the choice of doors is random, and probability of winning the car is the sum of the probability of choosing the door times the probability the car is behind it...1/2*1/3+1/2*2/3 = 1/2.

ETA: However, all of this is a tangential because in the Monty Hall Problem the second chooser is the first chooser.

 

[jt512]

@fromdownunder:

I take it, based on your protestations, that this is what you meant:

The odds of an observer who has no idea which door the host wasn't allowed to open guessing which of the two remaining doors has the car behind it is 1 in 2. (Because this observer is effectively just flipping a coin to decide.)

Unfortunately, that is not what you wrote. The statement you wrote was false, as has been explained to you by numerous others, but especially well in this post by Michael C.

 

[jbm]

People seem to be missing or not understanding what I said. So I need to say it again and even more simply. Using three doors with one open, the odds of the car being behind either door to an independent observer is 50/50. The odds that the contestant who picked the first door are 33/67.

Using 100 doors with 98 open, the odds of the car being behind either door to an independent observer is 50/50. The odds that the contestant who picked the first door are 1/100

Yes, and if then the contestant goes to the independent observer and tells him "by the way, I chose door x at the start of the game", then the independent observer knows that door x only has a 1/100 chance of winning. This is the whole point of the Monty Hall problem being discussed here, that the chance is not 50/50 if you know which door was first chosen, and which door was subsequently offered by Monty, because then that second door has a (total-1)/total chance of being the correct one.

But yes, if person A chooses a door, leaves the room, Monty does his thing and then person B walks in without having seen anything and has to choose between the two doors, then it's a 50/50 chance of winning. It's just not what this thread is about.

 

[Paul C. Anagnostopoulos]

Along with the Irrelevant Goat, it's worth considering the Inevitable (Irrelevant) Car. Of course, when it's time to open the doors the car becomes very relevant, but when the problem is posed it's as inevitable and irrelevant as the goat. There is a car - we have no more idea where it is than we do the location of the Inevitable Goat, but it's just as inevitable. It's also irrelevant - because the odds never change on account of it.

I'm not sure why you keep saying that the goat is irrelevant, since Monty will definitely open a door with a goat. He does not open a random door. Perhaps I misunderstand what you mean by "irrelevant."

You pick a door. Is the car more likely to be somewhere else? Yes.

Monty shows a goat. Is the car more likely to be somewhere else? Still yes. Nothing has changed.

No numbers necessary .

So we agree that picking the other two doors is better than picking the original single door. But since you are only allowed to pick one door, Monty fixes the problem by opening one of the two doors with a goat. Then you get to "pick both of the other two doors." What has changed is your ability to pick them. I agree that probabilities haven't changed.

If, instead of opening a door, Monty asked if you'd just like to switch to the other two (or 99) doors, the answer would be obvious.

~~ Paul

 

[ynot]

I never mentioned choices.

How about you re read all of my post instead of jumping in like a bull at a gate, and see what I actually said in the second paragraph. We are talking about two different things.

1. The odds of which door the car contains: With two doors it is 50/50

2. The odds that the contestant chose the correct door. This is, as I clearly stated in my post completely dependent on the number of doors. Do you need me to repeat it. I can if you are not willing to reread my post.

Do people on JREF sometimes read 1/5th of a post then jump in and base conclusions on what was said in the first sentence, and totally ignore context and the rest of the post? It would seem so.

Norm

Why is it common for many people to falsely conclude that not agreeing means not reading and understanding? (and do so with such emotion)

 

[ynot]

I'm not sure why you keep saying that the goat is irrelevant,
~~ Paul

The goats are red-herrings .

 

[ynot]

ETA: However, all of this is a tangential because in the Monty Hall Problem the second chooser is the first chooser.

Exactly - Why refer to the Monty Game then ignore the first part of the Monty Game? Perhaps the contestant has short term memory loss?

 

[Dr. Keith]

Exactly - Why refer to the Monty Game then ignore the first part of the Monty Game? Perhaps the contestant has short term memory loss?

That's a good premise for a short story.

Well, if you are flexible in your definition of "good".

 

[jiggeryqua]

I'm not sure why you keep saying that the goat is irrelevant, since Monty will definitely open a door with a goat. He does not open a random door. Perhaps I misunderstand what you mean by "irrelevant."

He does not open a random door. But whatever you do, whichever door you choose, inevitably (and so irrelevantly) Monty can show you a goat. Seeing a goat is not new information. It is not useful information (except insofar as it reduces the choice to 2 doors, but that whole '50/50' thing is the point of the problem and the goat is irrelevant in that regard).

So we agree that picking the other two doors is better than picking the original single door. But since you are only allowed to pick one door, Monty fixes the problem by opening one of the two doors with a goat. Then you get to "pick both of the other two doors." What has changed is your ability to pick them. I agree that probabilities haven't changed.

If, instead of opening a door, Monty asked if you'd just like to switch to the other two (or 99) doors, the answer would be obvious.

~~ Paul

Largely unargued. But... Monty doesn't offer you both doors, and I can only speak from my conceptual framework, but claiming he's offering you both the goat and the (2/3 car)-door doesn't wash. He's also, at the same time and equally ('50/50') offering you your door plus the goat door. You can change to the other door and see a goat, or keep your door and see a goat. 50/50, right? Wrong - so, with the best will in the world, the explanation must be wrong. Right?

 

[ynot]

Because other people on this thread had already suggested that the final choice was always a 50/50. I was trying to refute this, but apparently my point is invisible to some people.

Norm

Your lack of ability to clearly explain what you mean is not the fault of others.

Perhaps your point would be more visible if your posts weren’t so contradictory . . .

the odds of the car being behind either door to an independent observer is 50/50.
Norm

No. the remaining two doors are 50/50 to anybody.
Norm

 

[Dr. Keith]

Your lack of ability to clearly explain what you mean is not the fault of others.

Perhaps your point would be more visible if your posts weren’t so contradictory . . .

I feel like you are pulling the bow around and heading directly down wind on this one. You know you can't go any faster if you are heading directly down wind, don't you?

 

[ynot]

I feel like you are pulling the bow around and heading directly down wind on this one. You know you can't go any faster if you are heading directly down wind, don't you?

Ahhh . . . the good old days . . . Being wrong isn’t as bad as not admitting you were .

 

[ynot]

The probability is not 50/50 for anybody. The probability of an event is not an absolute value: it depends on what information the observer has.

- For an observer who arrives after Monty has opened a door and only knows that there is a prize behind one of the closed doors, the probability of the prize being behind either door is 1/2.

- For an observer who has watched the game and has the same information as the contestant, the probability of the prize being behind the first chosen door is 1/3 and the probability of it being behind the other door is 2/3.

- For an observer who knows where the prize is (Monty, for instance), the probability of it being behind a certain door is 1 and the probability of it being behind the other door is zero.

And this thread/poll only considers the Monty Game from the perspective of the contestant.

People that don’t understand the Monty Game think that the odds of the two final doors are 50/50 from the perspective of the contestant, not some irrelevant “independent observer” or "anybody".

 

[Turgor]

<snippage>“Do you want to open 1 or 2 doors to find the car?”.

This is the most pithy description of the Monty Hall problem I have ever read.

 

[Brian-M]

Seeing a goat is not new information. It is not useful information

The useful information lies in which door he opens. Whenever the door you picked has a goat behind it there's only one door he's allowed to open (because the other door you didn't pick has the car behind it). Knowing which door he was allowed to open (by the fact that he did open it) tells you with 100% certainty that the car is behind the remaining door.

(Of course, this only applies when you pick door with a goat behind it. The one-third of the time when you pick the door with the car behind it he's free to open either door, and so in this situation his choice to open a particular door provides no useful information.)

 

[jiggeryqua]

The useful information lies in which door he opens. Whenever the door you picked has a goat behind it there's only one door he's allowed to open (because the other door you didn't pick has the car behind it). Knowing which door he was allowed to open (by the fact that he did open it) tells you with 100% certainty that the car is behind the remaining door.

(Of course, this only applies when you pick door with a goat behind it. The one-third of the time when you pick the door with the car behind it he's free to open either door, and so in this situation his choice to open a particular door provides no useful information.)

(of course, your post is utterly wrong, and then you skip over the problem of explaining the problem in parentheses. Since you never know when he has one goat or two to choose from, we're back where we started.)

 

[ynot]

The purpose of revealing the goat door is to reduce the door choices to two and guarantee that the car is behind one of those two doors. Same thing applies if there were a million doors. I think the game would make a better point if red-herrings were used rather than goats.

 

[ynot]

The advantage of switching in the Monty Game can be quickly and easily demonstrated in practice using two black and one red playing cards. As Monty you place the cards randomly facedown and knowing their positions. You ask a “contestant” to guess which card is red by pointing to it. You then turn over one of the other cards you know to be black and ask the contestant if they want to stay with their first choice or switch to the other. Rinse and repeat and keep track of the contestant’s wins and losses. The advantage of always switching will become self-evident by the empirical results and convoluted intellectual arguments will become moot (Yep – I’m an optimist).