Monty Hall Problem

[Rolfe]

You can only tell whether or not a coin is weighted by tossing it lots of times. You can only tell which way the Monty Hall choice is set by running the scenario multiple times.

If you do this, and every single time Monty reveals a goat at step 2, then indeed, you double your chances if you switch. If, however, he reveals the car at step 2 1/3rd of the time, then switching confers no advantage.

Please tell me you get it!

If indeed the problem is set so that Monty always reveals a goat, then he is choosing to avoid the car, and so transferring all the 2/3rds probability that the car is behind one of the doors you didn't pick, on to the other closed door. Just as is illustrated by the 100-doors-and-one-car version. Switch.

If, however, the problem is set so that Monty picks a door at random, then there is no advantage to switching. In that event, 1/3rd of the times the scenario is run, he reveals the car at step 2.

You can't tell which rule Monty is operating on without repeated runs of the problem, just as you can't tell whether and in what direction a coin is weighted unless you toss it multiple times.

Rolfe.

 

[jsfisher]

If Monty picks a door at random after we pick a door (pretty miuch at random, too), then there are 6 possible aggregate selection possibilities, all with probability 1/6 = 1/3*1/2.

1. We pick door #1, and Monty picks #2.
2. We pick door #1, and Monty picks #3.
3. We pick door #2, and Monty picks #1.
4. We pick door #2, and Monty picks #3.
5. We pick door #3, and Monty picks #1.
6. We pick door #3, and Monty picks #2.

Without loss of generality, assume the car is behind door #1.

So, now, if we find ourselves staring at a goat and Monty asking us if we want to switch, it must be one of the cases 1, 2, 4 or 6. Cases 3 and 5 have been eliminated because in them, the car was revealed.

The conditional probability of cases 1, 2, 4, and 6 given we can see a goat behind Monty's door is 1/4 for each case.

For cases 1 and 2, if we switch we lose; for cases 4 and 6, if we switch we win.

It is, therefore, a 50/50 proposition.

 

[Rolfe]

If Monty picks a door at random ....

If.

There you go again, setting conditions that are not in the original scenario.

Nobody stipulated that he was picking his door at random.

Rolfe.

 

[Vorpal]

This is exactly why I explicitly defined the different cases back in post #400.
-- If Monty picks a random goat, which could be behind the player's door, then the probability of winning via switching is indeed 1/2.
-- If Monty picks a random goat, but never the player's door (as per the OP), then this is the standard interpretation of switching giving 2/3 probability. And yes, this is equivalent to the scenario of any goat not behind player's first guess being revealed, regardless of Monty's knowledge (as Monty's knowledge is used only to guarantee that this condition obtains in the first place).

 

[jsfisher]

If.

There you go again, setting conditions that are not in the original scenario.

Nobody stipulated that he was picking his door at random.

Rolfe.

The conditional was to join in an existing sub-discussion.

That aside, the real point is that we do not know Monty's strategy, but Monty's strategy does affect the final probabilities. Without the game Monty is playing, we cannot conclude anything about the advantage of switching doors.

 

[GreyICE]

If Monty reveals anything but a goat (car, gorilla, Elton John, ..) the trial is invalid and discarded. Monty cannot reveal a car in any valid trial per definition.


Discarding invalid trials (car, Madonna, Fidel Castro ..) makes Monty reveal a goat in every valid trial. Each trial is independent from all others, it's a random process of repeated non-correlated trials.


How so?

The gambler's fallacy is that you know the outcome of one random event based on another random event.

Like the gambler, you are saying that because Monty's coin came up goats, your coin is due to come up cars.

Lets examine the cases again.

Car is behind door number 1 - 33%
You lose.

Car is behind door number 2, Monty picks door number 3 - 16.5% (33% chance on the car, 50% chance Monty picks door number 3)

Car is behind door number 3, Monty picks door number 2 - 16.5% (same odds)

This leaves a 50/50 gain from switching.

The fallacy is as follows. It is true that if you flip two coins, and one comes up heads, it is 66% likely that the other one is tails. HOWEVER, this is not the probability we are in. MONTY'S specific coin came up heads. The fact that we have knowledge of the outcome of Monty's coin flip in no way influences your coin flip.

The probabilities involved in Monty's specific coin in no way influence your coin. You are still 50% likely to have flipped heads, and 50% likely to have flipped tails.


To approach this from yet another direction, imagine that Monty picks first, and picks a goat at random. The chances of you picking the car are now 50/50.

 

[gnome]

*sigh*

I'm headed for the mall, and I'm barricading both entrances when I get there. Who else wants to survive the zombie thread?

 

[Mobyseven]

No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

I feel like I should step in here and defend billydkid, because he's actually quite right.

There is no need to include Monty in the problem, and indeed sometimes it can help people understand it better if he is not included. People often project human motives onto Monty (hell, this thread is a prime example) when really he is only even in the problem for one reason - to open up a door that will reveal a goat.

There are a few ways to think about this. First, imagine a situation with no Monty. You have three doors, and you pick one. Another door mysteriously opens and reveals a goat. You are now given the option to stay with your current door, or swap to the other unopened one. You will benefit by swapping in this situation, just as in the original.

Another situation is that there are three doors, and you pick one. No other doors are opened and you are then asked if you would like to stay with your current door, or switch to both of the two remaining doors. This situation is for all purposes identical to the problems stated above - in this situation, however, the reason you should swap becomes much more obvious.

Bringing Monty into the mix really only serves to confuse things most times, given that he is really only there to play one specific part. But to explain his part, you have to give him certain pieces of knowledge, and that makes people give him a motive and a personality, which just confuses the matter. The knowledge that 'Monty' has in the Monty Hall problem isn't actually his at all...he is just an agent of the system, and it's easy to just eliminate him entirely from the problem.

 

[Mobyseven]

*sigh*

I'm headed for the mall, and I'm barricading both entrances when I get there. Who else wants to survive the zombie thread?

Twelve pages of thread, much of which can be summed up as, "But what if the Monty Hall problem isn't the Monty Hall problem?"

 

[jsfisher]

There is no need to include Monty in the problem, and indeed sometimes it can help people understand it better if he is not included. People often project human motives onto Monty (hell, this thread is a prime example) when really he is only even in the problem for one reason - to open up a door that will reveal a goat.

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

As the problem is normally stated (and was so in the opening post), we cannot assume Monty would (a) always have opened a door, (b) there'd be a goat behind it, and (c) we'd have an opportunity to change our door choice.

All we know is that Monty did open a door this particular time, and there was a goat behind the door this particular time, and we now have an opportunity to change our choice.

Monty must have some strategy. Depending on what it is, the probability of a change in our choice being beneficial to us can be 0 or 1 or anything in between.

There are a few ways to think about this. First, imagine a situation with no Monty. You have three doors, and you pick one. Another door mysteriously opens and reveals a goat. You are now given the option to stay with your current door, or swap to the other unopened one. You will benefit by swapping in this situation, just as in the original.

Did the door open by accident? Was a non-goat door guaranteed to have opened? What was the probability the door would have opened exactly as it did...and that's equivalent to asking about Monty's behavior.

Another situation is that there are three doors, and you pick one. No other doors are opened and you are then asked if you would like to stay with your current door, or switch to both of the two remaining doors. This situation is for all purposes identical to the problems stated above - in this situation, however, the reason you should swap becomes much more obvious.

If you guarantee me you will always follow this strategy when given the option to switch choices, well, then, I have a little game I'd like to play with you.

You have a built-in assumption that you would have been given the option to switch no matter what. All you really know is that you have been given the option for the particular case at hand.

Bringing Monty into the mix really only serves to confuse things most times, given that he is really only there to play one specific part. But to explain his part, you have to give him certain pieces of knowledge, and that makes people give him a motive and a personality, which just confuses the matter. The knowledge that 'Monty' has in the Monty Hall problem isn't actually his at all...he is just an agent of the system, and it's easy to just eliminate him entirely from the problem.

Again, you have made an assumption about Monty's behavior. You have assumed that Monty will aways open a door with a goat behind it after you choose. There is nothing in original problem statement to justify that assumption.

 

[Herzblut]

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

No. A purpose, a reason, a role in the game. The same role can be played by a chimpanzee or a mechanical unlocking mechanism.

What you say is complete nonsense. You fantasize about an "always" which is pure imagination etc. etc.

May I ask, what is your profession? Is it distantly related to mathematics?

 

[bobdroege7]

If I regurgitated the thread about whether an airplane on a conveyor belt would take off would anyone join me?

 

[Mobyseven]

Again, you have made an assumption about Monty's behavior. You have assumed that Monty will aways open a door with a goat behind it after you choose. There is nothing in original problem statement to justify that assumption.

Holy hell. What do you mean that there's nothing in the original problem statement to justify that assumption? That is the original problem!

Look, every time you play the game, Monty will open a door with a goat behind it. Every time. He will never, ever reveal a car. That's the nature of the problem. Deal with it.

 

[Mobyseven]

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

Also, this confuses me. I never gave Monty a motive. I pointed out that there's only one reason he's included in the system at all...but that's not a motive, that's a description of the system.

 

[sol invictus]

What you say is complete nonsense. You fantasize about an "always" which is pure imagination etc. etc.

What precisely do you find so difficult to understand about this discussion? The problem, as stated in the OP, simply does not give enough information to conclude that switching gives you a 2/3 chance of winning. You need to make some additional assumptions - which were almost certainly intended by the OP, as they are part of the standard Monty Hall problem - to come to that conclusion. That's all that's being pointed out here.

It's really not very difficult.

 

[lionking]

Holy hell. What do you mean that there's nothing in the original problem statement to justify that assumption? That is the original problem!

Look, every time you play the game, Monty will open a door with a goat behind it. Every time. He will never, ever reveal a car. That's the nature of the problem. Deal with it.

Spot on moby, this is the intent of the problem and what makes it most interesting. Monty is offering a two-for-one deal, and the odds are 2/3, even if it doesn't appear so initially. Too many others are complicating a very simple problem. If anyone wants to pose a problem where Monty sometimes opens a door with a car because he didn't know what was behind the door, let them do so. In the OP, he opens a door with a goat.

 

[AntiTelharsic]

Why would Monty open the door with the car behind it and give you the choice to switch from your obvious goat to the other obvious goat? It's ridiculous.

"You've lost. Would you like to switch to the other, equivalently losing position or just stay where you are? It doesn't matter in the least, so I'm not sure why I'm giving you this opportunity..."

The Monty Hall Problem has the contestant choosing one of three doors, where one door conceals a win and each of the other two conceals a loss; the host opening another of the doors, to reveal a loss; and the contestant being given the chance to either stay with his original choice or change his choice to the remaining door. The problem is assumed to be identical if repeated, because there is no information given to suggest otherwise -- if I take part in the scenario, and then you do, and then I do again, there is nothing in the statement of the problem to suggest that those scenarios may differ. This situation gives a 2/3 probability of winning if you switch. If you're contriving circumstances where switching doesn't give you a 2/3 chance of winning, then you're not talking about The Monty Hall Problem as it is usually defined.

If someone mentions "The Monty Hall Problem", three doors, two goats, and a car, this is what they're talking about. People who want to discuss other conditions should specify how their conditions differ from the standard; as long as they do that there should be no confusion.

 

[lionking]

Better said than me Anti-Telharsic

 

[Ivor the Engineer]

<snip>

Supposing I have a cunningly weighted coin. It will come down heads only 1/3rd of the time. I toss it once, and get heads. I ask you what the probability of that is. You say, 50/50. I say no, the coin is weighted, there was only a 1/3rd chance that was going to happen.

You can't prove me wrong by saying, but it came up heads and that is entirely consistent with what would have happened if you'd used a normal coin!

You can only tell the coin is weighted, and how, by multiple tosses. After you've tossed it 100 times, you'll know whether heads is a 50/50 shot, or just 1/3rd.

<snip>

Or the experiment was one of the 0.044% performed with a fair coin which produce a result as or more extreme.

 

[pgwenthold]

Spot on moby, this is the intent of the problem and what makes it most interesting.

This may be the intent, but it usually isn't stated as such.

Interestingly, many posters are saying, "This is what the problem says" but we are the ones who go back and quote the OP and ask, "Where does it say that he always opens a door to reveal a goat?"

Given just the problem, as written in the OP, there is not enough information to answer the question. If you claim it is the "intent" of the problem, then you are basing your answer on more than just the problem, as written in the OP.