3 Door Logic Problem

[XXX]

I have read a version of the questin before (called "the angel's gift) and I always thought it was stupid.

Here's my question. You are saying that if you pick a door, and the guy opens another door that didn't have the prize your chances of winning don't change. 1 out of 3, unless you make the swap and now it's 1 out of 2. So you are saying that the fact that he opened an empty door doesn't change the odds that you have picked the right door, which remain 1 out of 3.

BUT, if you pick door 1 and the host opens door 3, AND door 3 DOES have the prize behind it, your odds of having the prize in your box aren't still 1 out of 3! They are 0. Only and idiot would say otherwise. So the fact that you have elimitanted a door DOES in fact change the remaining odds that you have picked the right box.

At least that's how I see it.

 

[pgwenthold]

so despite the fact that this very well known logic puzzle is based on a specific and fixed set of rules (and is even CALLED the Monty Hall puzzle by most people), you're going to go with someone's incomplete phrasing of the puzzle?

You know, if this was the only time this issue came up, it would be one thing. I would just point out that the poster has given an incomplete version of the problem.

However, this is the usual problem with this problem, and, in fact, was the origin of a lot of the complaints to Marilyn (there were some who didn't get it, either, but that is a different story). People pointed out that there are assumptions that need to be made in order to draw the conclusion that is drawn.

I don't have any problems with those assumptions, just as I don't have any problems with the parallel postulate. However, we have to remember neither positions are required, and that assumption you are implicitly making has a huge impact on the solution you will get.

 

[Rasmus]

I have read a version of the questin before (called "the angel's gift) and I always thought it was stupid.

Here's my question. You are saying that if you pick a door, and the guy opens another door that didn't have the prize your chances of winning don't change. 1 out of 3, unless you make the swap and now it's 1 out of 2. So you are saying that the fact that he opened an empty door doesn't change the odds that you have picked the right door, which remain 1 out of 3.

BUT, if you pick door 1 and the host opens door 3, AND door 3 DOES have the prize behind it, your odds of having the prize in your box aren't still 1 out of 3! They are 0. Only and idiot would say otherwise. So the fact that you have elimitanted a door DOES in fact change the remaining odds that you have picked the right box.

At least that's how I see it.

What you describe is not possible according to the rules of the game. The situation simply cannot occur.

Introducing this option would seriously alter all odds, just like introducing a 7-sided die in an ordinary game of monopoly would.

 

[XXX]

OK, so the reason it made no sense to me is because the versions of the question I have read aren't accurate.

That makes sense.

 

[Lamuella]

pgwenthold, you have a point, and I apologize if I was getting argumentative. I was assuming the rules of the version I first heard. My apologies.

 

[Mr. Scott]

I have read a version of the questin before (called "the angel's gift) and I always thought it was stupid.

How does "The Angel's Gift" go?

There are an infinite number of ways to describe the problem, for example:

You decide, on a warm clear day, to go to the beach. You resolve to go to an island that has three beaches. You recall that one of the beaches is nice, and two are badly polluted, but can't remember which is the nice one. You decide randomly on one of the beaches, but halfway there, you recall clearly one particular beach you didn't choose that was polluted. Should you change your choice of beaches, or continue to the one you've already decided upon? Assume you have no hint of any unrecovered memory.

The Coin Puzzle was originally presented as two parents of different families, each of whom have two children, who reveal piecemeal information about their childrens' genders. I believe if you reduce the math of the coin and Monty Hall problems sufficiently, you'd find them to be identical.

 

[Luke T.]

The Monty Hall problem:

First of all, you are not being given 50/50 odds when Monty opens one door.

If you always pick door number 1 in 100 tries, without knowing what is behind any door, you will get the prize 1/3 of the time. Right?

That means doors 2 and 3 will have the prize behind one of them 2/3 of the time.

Now, if Monty is good enough to tell you which door of the two doors that have a 2/3 chance of the prize is the one that doesn't have the prize, then he is giving you a 2/3 chance of winning with the remaining door. Or you can stick with door number 1 all 100 times and stick to your 1/3 chance.

 

[pgwenthold]

pgwenthold, you have a point, and I apologize if I was getting argumentative. I was assuming the rules of the version I first heard. My apologies.

I used to be the same way, until I realized that, yeah, you really do have to make an assumption. That's why I don't ever consider it a bad thing, just important to recognize that is what it takes.

However, I question whether that was ever really established in the first version you heard. Even if it is called the Monty Hall problem, has it ever been established for a fact that he did ALWAYS open a door? I don't know that to be true.

(that's back to tsg's response)

I do not know that it was ever established in the original statement of the problem that this is to be taken as a given. As I said, I was always under the impression that it was the origin of the dispute (anytime you hear arguments about Monty's "motives" that's what they are talking about - it took me a long time to catch on to that)

 

[Luke T.]

The original statement of the problem in the OP was:

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

It is a one-time thing. Doesn't matter if the host opens it every time. In this case, he does. So is it to your advantage to switch your choice?

And yes, it is. You go from a 1/3 chance of winning to a 2/3 chance of winning by switching.

There's a 1 in 3 chance of the car being behind door number 1. There's a 2 in 3 chance of it being behind door number 2 or door number 3. When the host opens door number 3, there is still a 2 in 3 chance the car is behind door number 2.

 

[hodgy]

I would quite like to win a goat - how does that affect the equation?

 

[hodgy]

I've got room up the garden for it (the goat).

 

[pgwenthold]

The original statement of the problem in the OP was:

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

It is a one-time thing. Doesn't matter if the host opens it every time.

Yes it does, because it matters WHY he opened it for you.

If he only opened it because he knew you picked the right door initially, then switching gives you a 0% chance of winning. Given the information provided in the initial question, you don't know that this isn't what he is doing.

 

[GreedyAlgorithm]

so despite the fact that this very well known logic puzzle is based on a specific and fixed set of rules (and is even CALLED the Monty Hall puzzle by most people), you're going to go with someone's incomplete phrasing of the puzzle?

The only assumption I'm making is that the Monty Hall puzzle is the Monty Hall puzzle.

Can someone find Articulett (the original source of the quote) and get him to clarify, please?

Articulett quoted from http://www.cut-the-knot.org/hall.shtml, entitled "The Monty Hall Dilemma", saying:

This is an embarrassingly common human fallacy. Just because there are two choices (god or no god) doesn't mean they are equally likely. You can see this readily if you insert Zeus or Xenus into the equation as your preferred deity. One of my favorite logic problems points out this error nicely

so I'm pretty sure he really was talking about the Monty Hall puzzle.

 

[tsg]

The original statement of the problem in the OP was:

"Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?"

It is a one-time thing. Doesn't matter if the host opens it every time. In this case, he does. So is it to your advantage to switch your choice?

You must know that he will open another door regardless of whether you picked the car. Otherwise, it could be he is offering you the choice to switch because you have already picked the car.

 

[RandFan]

You must know that he will open another door regardless of whether you picked the car. Otherwise, it could be he is offering you the choice to switch because you have already picked the car.

That would only have value if you knew that he would offer you the choice only if you had chosen the car. Otherwise it makes no difference. You must assume that it is regardless.

Great problem BTW, it was very instrumental in my seriously questioning my own intuition. I have given the problem to many people who simply refuse to accept the facts even after I offer them the explanation and the various ways of understanding the problem. I call these folks believers.

 

[Dogdoctor]

Hey didn't Monty Hall used to mix up the odds a bit by offering "Or what Carol Merrill has in the box."

 

[Luke T.]

Hey didn't Monty Hall used to mix up the odds a bit by offering "Or what Carol Merrill has in the box."

Yep.

 

[Luke T.]

You must know that he will open another door regardless of whether you picked the car. Otherwise, it could be he is offering you the choice to switch because you have already picked the car.

Nope. Doesn't matter.

If the only reason he offers a switch is because you picked the car, then why would you switch? Therefore, why would he make the offer if you knew he did it because you picked the car?

 

[tsg]

Nope. Doesn't matter.

If the only reason he offers a switch is because you picked the car, then why would you switch? Therefore, why would he make the offer if you knew he did it because you picked the car?

Obviously he wouldn't. But the contestant wouldn't necessarily know for sure that is the only time he is going to be offered the switch in a one-shot game. Unless he is told explicitly, which would be pointless, he can't know for sure. The contestant then has to assume there is some probability that the offer of the switch is the result of having picked the car correctly in the first place, which reduces the chances that it is behind the remaining door.

In order for the contestant to be able to deduce the probability is 2/3 in favor of switching, the offer to switch must be automatic and he must be aware that it is.

 

[Paul C. Anagnostopoulos]

It's supposed to be worded so that Monty knows what's behind the doors and opens one specifically with a goat behind it.

Here's the easy way to understand it: What if Monty offered you both of the other doors? Obviously you'd go for it. Well, that's what he did by opening one and offering you the other.

~~ Paul