Monty Hall Problem

[Robin]

Yes, when people communicate via human language, they make assumptions about meanings of words and phrases based on shared cultural experiences. If you want to engage in pedantic nitpickery you can uncover all kinds of bogus ambiguity.

Who didn't know, unquestionably, that the picking of the door was for receiving the prize behind that door? Raise you hands. I can understand that some puzzles are constructed with tricky loopholes hidden in the precise construct of the question. No reason to think that this is such a puzzle. This is a puzzle about probability.

You are utterly determined to miss the point aren't you? The point I was making is that we have to assume.

I originally made the assumption that the second choice was part of the rules of the game and everyone jumped all over me and said "but it depends on the intentions of the host" because the problem does not explicitly state that Monty was obliged to offer the choice.

I had assumed that all information required answer the puzzle had been supplied and if the offer of second choice was only optional that would have been stated.

So my assumption that Monty had to offer the second choice is being called unreasonable.

But the alternative assumption that Monty did not have to offer the second choice is not regarded as unreasonable.

So my question is what is the criteria for any assumption being reasonable? That was the point I was making.

This is a puzzle about probability.

Incidentally and let's be crystal clear about this:

It is a probability problem if and only if Monty was obliged to reveal the goat and offer the second choice

In which case it was settled long ago and why are we arguing?

 

[hgc]

...

In which case it was settled long ago and why are we arguing?

I have no idea. I think it had something to do with hitting 150 posts in this thread in a single day.

 

[TeaBag420]

To sum up: Cabbage presents a twist on the Monty Hall problem by introducing the shell game (which is a notorious con; check with Penn if you think this assertion is wrong), but with a twist. Robin then asserts that the shell game is exactly the same as the Monty Hall problem. He is incorrect about this in the real world, but it's not his puzzle so I ignore him. Then Cabbage says the grifter in the shell game IS playing fair in that the pea is really there somewhere, so I accept the redefinition of "shell game" solely for this discussion.

Then CurtC offered his game theory interpretation, which even he admitted was a blind alley (my phrase). CurtC then demonstrates that he doesn't understand that the OP was for one trial, and that the problem states what Monty actually DOES, not what he's required to do.

Cabbage clarifies that his version of the shell game is Platonian in its refinement, in that not only can the con not cheat by removing the pea, the player (mark) can't try to keep track of the location of the pea. Which makes it pointless to introduce this new game where trying to win is cheating. Oh well.

Now Cabbage explicitly told us on page five of this thread that there are two circumstances under which you, the passerby could win or lose. One, if the player (mark) switches and win: you win. Two, if the player switches and loses: you lose. If he doesn't switch, you neither win nor lose. No matter whether he does or not. Any redefinition of the bet for the "not switching" case or for the odds (you're betting even money--$50 to win, $50 to lose) at this point will go down in the book as puzzle changing, and puzzle changers are just one step above kid touchers in my book.

So, if you're inclined to gamble, you take the bet, because the chances of winning or breaking even are better than 50/50. Of course if the guy who offered you the bet wins the first time and walks away, you're screwed.

Class dismissed. What's next, Mrs. Landingham?

 

[TeaBag420]

You are utterly determined to miss the point aren't you? The point I was making is that we have to assume.

I originally made the assumption that the second choice was part of the rules of the game and everyone jumped all over me and said "but it depends on the intentions of the host" because the problem does not explicitly state that Monty was obliged to offer the choice.

I had assumed that all information required answer the puzzle had been supplied and if the offer of second choice was only optional that would have been stated.

So my assumption that Monty had to offer the second choice is being called unreasonable.

But the alternative assumption that Monty did not have to offer the second choice is not regarded as unreasonable.

So my question is what is the criteria for any assumption being reasonable? That was the point I was making.



Incidentally and let's be crystal clear about this:

It is a probability problem if and only if Monty was obliged to reveal the goat and offer the second choice

In which case it was settled long ago and why are we arguing?

No, it's a probability in cases where, as the OP states, Monty does in fact r.t.g.a.o.t.s.c. Otherwise it's one out of three. But since the original puzzle does in fact state that Monty does in fact r.t.g.a.o.t.s.c. it's a probability problem.

 

[Cabbage]

To sum up: Cabbage presents a twist on the Monty Hall problem by introducing the shell game (which is a notorious con; check with Penn if you think this assertion is wrong), but with a twist. Robin then asserts that the shell game is exactly the same as the Monty Hall problem. He is incorrect about this in the real world, but it's not his puzzle so I ignore him. Then Cabbage says the grifter in the shell game IS playing fair in that the pea is really there somewhere, so I accept the redefinition of "shell game" solely for this discussion.

Then CurtC offered his game theory interpretation, which even he admitted was a blind alley (my phrase). CurtC then demonstrates that he doesn't understand that the OP was for one trial, and that the problem states what Monty actually DOES, not what he's required to do.

Cabbage clarifies that his version of the shell game is Platonian in its refinement, in that not only can the con not cheat by removing the pea, the player (mark) can't try to keep track of the location of the pea. Which makes it pointless to introduce this new game where trying to win is cheating. Oh well.

Now Cabbage explicitly told us on page five of this thread that there are two circumstances under which you, the passerby could win or lose. One, if the player (mark) switches and win: you win. Two, if the player switches and loses: you lose. If he doesn't switch, you neither win nor lose. No matter whether he does or not. Any redefinition of the bet for the "not switching" case or for the odds (you're betting even money--$50 to win, $50 to lose) at this point will go down in the book as puzzle changing, and puzzle changers are just one step above kid touchers in my book.

So, if you're inclined to gamble, you take the bet, because the chances of winning or breaking even are better than 50/50. Of course if the guy who offered you the bet wins the first time and walks away, you're screwed.

Class dismissed. What's next, Mrs. Landingham?

I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

 

[Robin]

No, it's a probability in cases where, as the OP states, Monty does in fact r.t.g.a.o.t.s.c. Otherwise it's one out of three. But since the original puzzle does in fact state that Monty does in fact r.t.g.a.o.t.s.c. it's a probability problem.

No, and this is absolutely crucial to the misunderstanding:

It is a probability problem if and only if Monty has to reveal the goat offer the choice.

If all you know is that on this occasion he revealed the goat and offered the choice then it is not a probability problem.

Incidentally neither is it a matter of Monty's intentions, not at least until we know whether or not it he had to make the offer.

 

[Paul C. Anagnostopoulos]

rppa said:
That's right. But in order to assess expectation values, you need to come up with a probability model for how Monty will behave under all possible scenarios, and all you have is a single data point: whatever prize I have behind my door, Monty is showing me a goat.

I agree that if this were the statement of a puzzle for which we were offering a $1 million prize for a solution, it would be important to explicitly state all the parameters of the problem. I'm simply saying that some of us are taking it a bit too seriously. It is generally the case with logic problems that unstated conditions can be assumed to be absent. Otherwise you could conjure up dozens of possible conditions: What if Monty uses different procedures on different days of the week?

You've got to make SOME assumption about his behavior, or there's no way to calculate the probabilities. There is absolutely more than a hint of a conditional nature to the problem. The assumptions about the conditional nature are key to your calculation of expectation value.

Where is the hint about any conditions?

Sorry, this is a strawman.

And a joke, too. Well, except for my remark that there are many conditions we could assume if we wanted to.

What makes you think the information given is sufficient to know what Monty does under all circumstances?

It's not, if we are being formal about it.

Here's Craig Whitaker's letter to "Ask Marilyn":

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, always 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?

I have edited it to eliminate the ambiguity. Well, unless this is only the procedure on Tuesday.

For someone who is merely objecting to our taking this too formally, I'm being awfully serious, ain't I?

~~ Paul

 

[TeaBag420]

I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

No. If he doesn't offer the switch you break even. So if he offers the switch once you have a 2/3 chance of winning $50. If he offers it twice you have a 1/9 chance of losing $100, a 4/9 chance of winning $100, and a 4/9 chance of breaking even.

Because of the nature of the bet, you don't even have to consider cases where he doesn't offer the switch. If he never offers the switch, it's as if you never made the bet.
As long as he offers a switch at least once, the odds are in favor of your at least breaking even. If he never offers the switch at all, you lose nothing. It doesn't matter what his plan or strategy is, nor what you know or don't know about it; this paragraph encompasses all possibilities.

Unless, of course, there's collusion between him and the man you're betting with. In which case we're back to this being a con.

Let me reiterate: Given the problem as presented, there is no strategy the grifter could employ short of collusion that would give an advantage to the man with whom you are betting. It does not matter.

 

[TeaBag420]

No, and this is absolutely crucial to the misunderstanding:

It is a probability problem if and only if Monty has to reveal the goat offer the choice.

If all you know is that on this occasion he revealed the goat and offered the choice then it is not a probability problem.

Then what field of mathematics would it fall under?

 

[Cabbage]

It is generally the case with logic problems that unstated conditions can be assumed to be absent. Otherwise you could conjure up dozens of possible conditions:

In fact, that's exactly what I've been trying to say!

The problem never stated the condition that a door is always revealed; therefore, that conditioned is assumed to be absent.

The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.

The one claiming Monty may or may not always open a door are operating without assumptions. Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case.

 

[Cabbage]

No. If he doesn't offer the switch you break even. So if he offers the switch once you have a 2/3 chance of winning $50. If he offers it twice you have a 1/9 chance of losing $100, a 4/9 chance of winning $100, and a 4/9 chance of breaking even.

But here you are assuming that you know how the man operates the shell game. Maybe he likes to be tricky, offering the switch only when the initial guess is correct, just to tempt them. It's still a perfectly fair game--The person playing the game has every right to choose not to switch, and therefore win the game. But if the man operates the game this way, the player will lose every time he switches

Because of the nature of the bet, you don't even have to consider cases where he doesn't offer the switch. If he never offers the switch, it's as if you never made the bet.
As long as he offers a switch at least once, the odds are in favor of your at least breaking even. If he never offers the switch at all, you lose nothing. It doesn't matter what his plan or strategy is, nor what you know or don't know about it; this paragraph encompasses all possibilities.

Unless, of course, there's collusion between him and the man you're betting with. In which case we're back to this being a con.

Let me reiterate: Given the problem as presented, there is no strategy the grifter could employ short of collusion that would give an advantage to the man with whom you are betting. It does not matter.

There's no collusion between the shell game man and the man you're betting with; he's every bit in the dark as you are as to how the shell game man plays his game. However, the fact that there is no collusion doesn't mean the odds are clear cut--That, once again, depends on how the man operates his shell game.

 

[Robin]

Then what field of mathematics would it fall under?

It does not come under any field of mathematics because it is incomplete.

For example look at the following:
<pre>
1. x=4
2. y=3
3. z=redistribute(x,y)
</pre>
What is the value of z?

This looks like a mathematical problem, but is not because we don't know what the hell line 3 means.

 

[Robin]

The one claiming Monty may or may not always open a door are operating without assumptions.

No, they are operating with one less assumption.

Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case. [/B]

It makes it unsolvable as a probability problem at least.

 

[Robin]

The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.

The ones claiming that the results depend on the hosts intentions are also making an assumption, ie that Monty optionally opens a door.

 

[Cabbage]

The ones claiming that the results depend on the hosts intentions are also making an assumption, ie that Monty optionally opens a door.

No, we're not.

I certainly agree it's possible that Monty always opens a door. I am also aware of the possibility that Monty may open a door only occasionally. I'm not assuming, I'm simply admitting that I don't know.

 

[Robin]

No, we're not.

I certainly agree it's possible that Monty always opens a door. I am also aware of the possibility that Monty may open a door only occasionally. I'm not assuming, I'm simply admitting that I don't know.

Exactly my point.

If you say that the result depends on the intentions of the host then you are assuming that Monty has a choice.

If you don't know whether Monty has a choice then you don't know if it depends on the behaviour of the host.

For example you said:

Actually, the intentions of the host do matter

(your emphasis) which makes the assumption that he has a choice. More correctly you should have said.

Actually, the intentions of the host may matter

(edited for clarity)

 

[Cabbage]

Exactly my point.

If you say that the result depends on the intentions of the host then you are assuming that Monty has a choice.

If you don't know whether Monty has a choice then you don't know if it depends on the behaviour of the host.

For example you said:



(your emphasis) which makes the assumption that he has a choice. More correctly you should have said.



(edited for clarity)

I'm afraid I don't see the distinction.

The point I'm trying to make is that the host's intentions always matter.

If he always opens a door, then his intentions matter, simply because his intention is to always open a door.

If he sometimes opens a door, then his intentions matter, simply because his intention is to sometimes open a door.

Any way you look at it, his intentions matter.

 

[TeaBag420]

In fact, that's exactly what I've been trying to say!

The problem never stated the condition that a door is always revealed; therefore, that conditioned is assumed to be absent.

The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.

The one claiming Monty may or may not always open a door are operating without assumptions. Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case.

The OP makes none of these claims. It also doesn't claim that Monty discriminates against black people. It is about a ONE-TIME incident where you, the contestant are offered a choice. Nothing else. These other lines of questioning are not part of the original problem, which I posted. The original submitter of the problem has posted here... nope, got the name wrong... Whitaker?I was thinking CFLarsen (or maybe I misunderstood).

 

[TeaBag420]

I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

First, I don't think it matters how many games are played, so let's agree to throw the stipulation out.

Second, and I'm starting to get this, if the grifter has the option of only offering the switch when you've got the right shell, then in those cases switching gives you a zero chance of winning. If the grifter isn't required to follow rules that you, the bettor know, then this isn't solvable for any arbitrary number of games, which is what I suspect is what you were getting at. Unless you consider it a solution to say "Don't take the bet". Which gets us back to it being a con, and was my original solution.

Because if the grifter doesn't follow known rules, why would someone offer you the bet? And if he doesn't follow known rules, why offer you the bet, absent collusion?

It's a con.

Robin's assertion that the shell game is the same as the Monty Hall problem appears defective under this analysis and I will let you take that up with him.

 

[TeaBag420]

It does not come under any field of mathematics because it is incomplete.

For example look at the following:
<pre>
1. x=4
2. y=3
3. z=redistribute(x,y)
</pre>
What is the value of z?

This looks like a mathematical problem, but is not because we don't know what the hell line 3 means.

I have rephrased your example:

<pre>
1. x=4
2. y=3
3. z=Idon'tknow(x,y)
</pre>

Equally helpful, don't you think? Your argument seems to boil down to "I don't know." But in a bad way.

If it's not a math problem, maybe it should be moved to another board. Why don't you liaise with the moderators on that?