Philosophy Now Forum

Obvious Leo wrote:

Scott Mayers wrote:If you disagree, what do you think is actually wrong with the enumerated events above?

What is wrong with the enumerated events above is that they have nothing to do with the fucking problem. you are just plain WRONG.

Are you high or something? Read it first. How does this NOT represent the problem?

Am I missing something? Aren't you saying that changing the guess will make no difference to the outcome?

Obvious Leo wrote:Am I missing something? Aren't you saying that changing the guess will make no difference to the outcome?

Something is off about you interpreting me here. I lose no matter how I answer your question depending on what precisely you are meaning.

The diagram above shows all possibilities laid out such that the first choice the guest makes is a "Pick". It is also the same as the option if you "Stayed". Thus to stay, rather than switch, where you see "Pick or Stay" are equivalent.

The green background is where the car is in each option. I put all the sub-events into each row. So the "Pick or Stay" indicates the first choice by the contestant (guest). Then the host reveals the goat (illustrated wherever the background is blue) as indicated by the "Host Reveal". The host is not allowed to pick and reveal the car. So if the guest picks the car (even though he is unaware), the host is able to select one the other doors as two distinctly different acts. This is because there is a goat behind both remaining doors. He may arbitrarily pick one of the two. But each option, though they have a similar value (that they will both be goats), have to be counted in the probability since each are real possibilities.

In Option 1 set, when the contestant picks Door 1, the host can reveal Door 2 as one possibility OR Door 3 as the second possibility. Note that the distinct Options are only redundant but I included it because dionisus seemed to prefer trying to include them.

I'm guessing it throws you off for the similar reason you have with trying to understand what a nothing (or zero) means. At first it appears as though the first two rows are equal because the value of the outcomes are identical in meaning with respect to the wins or loses. But they are still two separate events because in the first row case, when you switch, you choose Door 3; In the second case, when you switch, you choose Door 2. These are two separate acts. While the host may know that they mean no different, to you as a guest, you cannot determine this and so must see these two as distinctly separate options.

Scott Mayers wrote:Something is off about you interpreting me here. I lose no matter how I answer your question depending on what precisely you are meaning.

How can there be any ambiguity about the question I'm asking? The Monty Hall puzzle is perhaps the most famous logic puzzle in modern history and incidentally one which my father taught me when I was only about 6 years old. Small children are far more likely to get it right and I tested my own kids and grandkids on it as well.

Are you saying that changing one's pick makes no difference?. YES or NO.

Obvious Leo wrote:

Scott Mayers wrote:Something is off about you interpreting me here. I lose no matter how I answer your question depending on what precisely you are meaning.

How can there be any ambiguity about the question I'm asking? The Monty Hall puzzle is perhaps the most famous logic puzzle in modern history and incidentally one which my father taught me when I was only about 6 years old. Small children are far more likely to get it right and I tested my own kids and grandkids on it as well.

Are you saying that changing one's pick makes no difference?. YES or NO.

YES, and it is apparent from my illustration above if you mean that 'changing one's pick' means to always choose to switch regardless of which DOOR the Host opts to show you when he has them . Thus the red colored "Switches" represent those options per each row.
NO, if you interpret "changing one's pick" as in simply the act to always assume that row one or row two are equivalent by result (the value you place on the car as opposed to the goats.)

Knowing you, you do not distinguish between the first two rows as different because you base your preference to what the host knows that both doors remaining are goats (losers). This blinds you to perceive them as one because you think that the result is all that matters. But with respect to the player, being unaware of this fact, the question the puzzle asks is whether you (as the Guest) have an advantage to switch. But because you are unaware from your perspective of the result, you cannot conclude that switching improves your odds since the first two rows above demonstrate that if you switched regardless of which TWO optional remaining doors the Host reveals leads you to two distinctly events where you lose.

I get where you and others are coming from and even thought the same when I first encountered it too. But you can't simply base your judgement that those two events are the same. As such, the perception of the player can only depend on all possibilities based on the perception of the doors, not the result (what is certainly behind the doors).

What problem are you trying to answer? You appear to be trying to make this up as you go along and then you're shifting the goalposts as you're going. In the classical Monty Hall problem the host knows where the car is and is then required to open a door on a goat before offering the contestant a chance to change his mind. I presume you're aware of the fact that in THIS scenario the contestant will DOUBLE his chances of winning if he does indeed change his mind. I presume you're also aware of the fact that this has been PROVEN countless times.

Leo, I'll give an example of the type of error that is being made that you and I discussed elsewhere using zeros and ones.

I will say that 1 = 01 = 001 = 0001 = ...0001 (infinite times) This is the convention that a computer would equivocate but distinguish separately depending on how many bits they have to illustrate the number one. You might say the zeros are useless and I would agree if we are talking about the efficiency in practice for us writing these down on paper.

Nevertheless, if we represent these numbers depending on place, each zero to the left of the first place is a 0 times some power of the base (base ten for our regular use). For instance, 001 = (0 x 10²) + (0 x 10¹) + (1 x 10º) = 1. We obviously think it would be redundant to place zeros before numbers we write because they do not matter because they all mean the same. However, by the perspective a computer that has to use a collective set of bits for every number, what is in those spaces ALL matter. For instance, a "byte" is eight "bits" or positions that act always as a whole. And if the construction of the memory space has one byte, you have to name the number 1 as 0000001. If you only attend to what is in the right-most bit, that memory space may have some '1' in one of the other spots to the left of the first bit and would change the result.

So, by perspective of the computer, those positions MATTER!

The illusion is the same with respect to this Monty Hall Problem. The claim is that you should switch because the odds are 2/3 better for doing so. But this is only if you could read the mind of the Host (his perspective) that the two options in rows one and two in my illustration are the same in result. But the question asks about what the Guest should do and has to be based on his own perspective. From his perspective of NOT knowing that two possibilities lead to the same result, to switch has to recognize each door choice as distinct and separate events. Therefore, he MUST recognize that his odds are even to Switch OR Stay as you include those two possibilities where you may have picked the original correct door making the Host's capacity to choose variable from your perspective as a guest.

Obvious Leo wrote:What problem are you trying to answer? You appear to be trying to make this up as you go along and then you're shifting the goalposts as you're going. In the classical Monty Hall problem the host knows where the car is and is then required to open a door on a goat before offering the contestant a chance to change his mind. I presume you're aware of the fact that in THIS scenario the contestant will DOUBLE his chances of winning if he does indeed change his mind. I presume you're also aware of the fact that this has been PROVEN countless times.

The problem asks what the Guest should do from his perspective. So you cannot include what the Host knows. I'm not shifting the goal (result) but you are shifting the perspective as you think in terms of the route (door) to the goal without recognizing the distinction.

This is NOT proven appropriately as the proofs all assume the same ignorance of perspective. In fact, even their claims to do this by statistical counting through a computer depends also on the computer program. And no doubt, the input program also ignores the difference between those two distinguishing events (in each Option where the car is placed).

I give up, Scott. All I'm seeing is a man making a fool of himself. Do the experiment yourself and see what happens. Those contestants who change their minds DOUBLE their chances of winning, as has been shown over and over again. You can shove your zeros where the sun doesn't shine, mate, because mathematics is NOT the same thing as logic.

Obvious Leo wrote:I give up, Scott. All I'm seeing is a man making a fool of himself.
Do the experiment yourself and see what happens. Those contestants who change their minds DOUBLE their chances of winning, as has been shown over and over again. You can shove your zeros where the sun doesn't shine, mate, because mathematics is NOT the same thing as logic.

Put down the mirror and look at me, Grandpa!

Scott Mayers wrote:Before even beginning with the math, let us enumerate for all possibilities graphically here to at least show the error. Below is a diagram I created to show all possibilities absolutely. Each option represents the Host placing the car in the door and is shaded as green. Each row represents a possible event. For each place one picks where the car actually is leaves two possibilities in which the host is allowed to reveal what is empty (or a goat). Each original pick is also represented as the option one picks if they stay. The red printed the switches.

Please look over this and assure me if you either agree or not to these possibilities. If you disagree, state why.

Mont_Hall_1.jpg

If you agree to these as all the possible events, count up the events where you switch AND win. If you get 6/12, this is 1/2. Likewise, if you look at all the switches in which you lose, count them up too. Do you see 6/12 = 1/2 here too? How about the Stays? 1/2 too, correct? This demonstrates graphically that whether you switch OR stay, the odds are 1/2 no matter what you could figure prior to using probability. If the math using probability does not match, then this must assure you that it would have to be the math you use to which is causing the trouble. Is this correct?

I totally agree with your diagram, thanks to make it, it is very similar to the cars example i give to you, but it is well presented. (and i am happy you finally come to use this "12")
And you do exactly the error i thought you was doing, the problem is not the diagram, but one of the assumption you do to draw your conclusion.
You believe each case of the diagram are equiprobable, you believe that P(option1 ligne 1) = P(option1 line 3), but it is not the case.
because it is not the case, you can’t just count the number of time you see a switch and a win to know the probability.
Please, go back to my example with the car, and particularly the example i make with a dice roll, if you read it carefully and honestly, i think you will agree with me.

Obvious Leo wrote:I give up, Scott. All I'm seeing is a man making a fool of himself. Do the experiment yourself and see what happens. Those contestants who change their minds DOUBLE their chances of winning, as has been shown over and over again. You can shove your zeros where the sun doesn't shine, mate, because mathematics is NOT the same thing as logic.

It is a little off topic, i agree that math is not only logic, but it include pure logic in about all of its steps.
And the problem here, is totally a mathematical problem. (and also a logical problem, because mathematical problems are logical problems)

Scott Mayers wrote: In fact, even their claims to do this by statistical counting through a computer depends also on the computer program. And no doubt, the input program also ignores the difference between those two distinguishing events (in each Option where the car is placed).

Did you look at the program that test it by statistical counting ? Where did you see a mistake ?
I assure you the program distinguish between these two different events.

You give the colored cars problem, to force people to explicitly distinguish between these two events, i was ok, and resolved the colored cars problems, showing you i took all the events into account, and that your error was to think all those event was equiprobable.
I believe you read it way too rapidly for the simple reason i used a "12" you didn’t see where it come from, but that had no importance in my argumentation.

Obvious Leo wrote:To understand this problem correctly you need to forget about the contestant and think about the host of the game. He knows which door the car is behind and he knows which door the contestant has chosen and THEN he opens a door to reveal a goat. The reason why very few contestants switch doors is that they forget about the prior knowledge of the host, who already knows he's going to reveal a goat.

No, the host knowledge is not important here. For the host, the only unknown is if the player will switch or not, there are certainty about where the car and the goats are, it is a very different problem from the host perspective.
What is important is the rule of the game, the host should reveal a goat, when the host know it, doesn’t matter, and that the host know it, doesn’t matter. (outside the material fact it should know it to do it)

The first problem is that people doesn’t clearly differentiate between the different kind of cases in probabilistic problems.
The second problem, is that most of people, when facing with a paradox, will get their answer to be right, and then try to contradict the opposite answer, until they finally arrive at something enough complex for them to make a mistake, and think they have solved the paradox.
And the person could be very comfortable with the matter, it will not help, because he will just do whatever complex and obscure reasoning he need for the reasoning to include a mistake.
It is exactly what Scott Mayers did when he answered to your simple demonstration, he just made a enough obscure argument, for his reasoning to break at some point, and then he though you didn’t understand it enough well.
It is a pretty vicious psychological mechanism, if you really try to prove something that is false, you will inevitably reach this subtle step between when you know what you do, and when you don’t know what you do, because this is exactly where you will believe you know what you do, when you don’t, this is exactly where you will do a mistake that permit you to do the proof.

And i think it is exactly why children are better at this problem, it is only because they only see the solution you give them, they understand the solution, but they doesn’t see the paradox at all, and then they have no motive to contradict the solution you give them.
Scott Mayers have a big motive to contradict what you said, and this motive is that he see a apparently totally clear but different solution to this problem.
Now i think the only way to make him reasonable, is to show him the mistake in his own reasoning, not the reasoning he use to contradict the others solutions (because it will naturally be the point where is reasoning break, like i explained), but contradict the real reason he believe other solutions are wrong in the first place.

dionisos wrote:

Scott Mayers wrote:Before even beginning with the math, let us enumerate for all possibilities graphically here to at least show the error. Below is a diagram I created to show all possibilities absolutely. Each option represents the Host placing the car in the door and is shaded as green. Each row represents a possible event. For each place one picks where the car actually is leaves two possibilities in which the host is allowed to reveal what is empty (or a goat). Each original pick is also represented as the option one picks if they stay. The red printed the switches.

Please look over this and assure me if you either agree or not to these possibilities. If you disagree, state why.

Mont_Hall_1.jpg

If you agree to these as all the possible events, count up the events where you switch AND win. If you get 6/12, this is 1/2. Likewise, if you look at all the switches in which you lose, count them up too. Do you see 6/12 = 1/2 here too? How about the Stays? 1/2 too, correct? This demonstrates graphically that whether you switch OR stay, the odds are 1/2 no matter what you could figure prior to using probability. If the math using probability does not match, then this must assure you that it would have to be the math you use to which is causing the trouble. Is this correct?

I totally agree with your diagram, thanks to make it, it is very similar to the cars example i give to you, but it is well presented. (and i am happy you finally come to use this "12")
And you do exactly the error i thought you was doing, the problem is not the diagram, but one of the assumption you do to draw your conclusion.
You believe each case of the diagram are equiprobable, you believe that P(option1 ligne 1) = P(option1 line 3), but it is not the case.
because it is not the case, you can’t just count the number of time you see a switch and a win to know the probability.
Please, go back to my example with the car, and particularly the example i make with a dice roll, if you read it carefully and honestly, i think you will agree with me.

Good morning, dionisos,

I underlined the above because I figured you made a mistake. Did you mean "P(option 1, line 1) = P(option 1 line 2)"?
I've been up all night obsessed with this puzzle. Although I trust I'm correct, I also understand you, AND even see a third possibility given different perspectives. I'm not finished analyzing it as it appears to be another problem of using zero and/or infinity in the math.

My solution is from the perspective of the Guest where his perception is one of being indeterminate only until he opens the box and thus requires that he accepts all possibilities as given in my diagram. I'll read the rest of the posts to catch up before adding more to this.

No, i was really speaking about P(option1 ligne 1) = P(option1 line 3) not P(option1 ligne 1) = P(option1 line 2)
I mean by it, do you think the probability to be in the first case, equal the probability to be in the 3th case in you diagram.

I will let you going further before giving you more to read, i just wanted to confirm it wasn’t a mistake. Good night