The scams of Statistics...

[Scott Mayers]

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

6+3=9 no ?

The 12 comes from respecting the in-between act of picking an EVEN number, and then counting the two events after that. In the program experiment it is only intended to actually count the results without concerning ourselves with those 12 odds as you intended it to be constructed simply to see if a computer should actually guess close to your odds in multiple events regardless.

[Scott Mayers]

I understand your reference that some things are not equiprobable

Show me a simple example where the possibilities are not equiprobable, for me to confirm that you understand what it mean.

I clearly understand the distinction but you seem to be dragging me away from the actual problem into your own world.

The fact that regardless of interpreting your given puzzle where they are even as 1/2 o 1/6 = 1/12 is irrelevant to the Monty Hall Problem as in that case, they ARE equiprobable as = 0.

Please go to this to argue as this is what matters here. You won't opt to go simpler as I suggested but try to go to a more complex scenario.

What is the your proposed equiprobability in these cases?

Monty_Hall_Original.png

and

Fixed Solution.png

[dionisos]

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

6+3=9 no ?

The 12 comes from respecting the in-between act of picking an EVEN number, and then counting the two events after that. In the program experiment it is only intended to actually count the results without concerning ourselves with those 12 odds as you intended it to be constructed simply to see if a computer should actually guess close to your odds in multiple events regardless.

I am sorry but this is nonsense.
"12 odds" mean nothing, "in-between act of picking an EVEN number" mean nothing.
If there was 12 possibilities, you would be able to enumerate them, like i did with the 9 possibilities. (d1, d2-head, d2-tail, …)
You can’t enumerate the 12 possibilities, because there are only 9, this is why, you totally confuse yourself with english sentences.

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

If i add 3 possibilities, for each even number as you propose, to the 6 options of the dice rolls, i get 6+3+3+3=15
Outside the fact this calculus is totally meaningless, you don’t even get 12.

I am sorry but i stop this discussion, because either you don’t try to understand me, or you are just trolling me, or you are way too much delusional.

[Scott Mayers]

6+3=9 no ?

The 12 comes from respecting the in-between act of picking an EVEN number, and then counting the two events after that. In the program experiment it is only intended to actually count the results without concerning ourselves with those 12 odds as you intended it to be constructed simply to see if a computer should actually guess close to your odds in multiple events regardless.

I am sorry but this is nonsense.
"12 odds" mean nothing, "in-between act of picking an EVEN number" mean nothing.
If there was 12 possibilities, you would be able to enumerate them, like i did with the 9 possibilities. (d1, d2-head, d2-tail, …)
You can’t enumerate the 12 possibilities, because there are only 9, this is why, you totally confuse yourself with english sentences.

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

If i add 3 possibilities, for each even number as you propose, to the 6 options of the dice rolls, i get 6+3+3+3=15
Outside the fact this calculus is totally meaningless, you don’t even get 12.

I am sorry but i stop this discussion, because either you don’t try to understand me, or you are just trolling me, or you are way too much delusional.

It is you who is concerned about the 12, not me. The first program was set up to replicate your own but just in an assembly form pseudocode. As is, it does act as respecting 9 possibilities. But why are you focusing on this? If you don't want to go simple to begin with, then let's go to the actual puzzle.

I only want you to at least SEE where the actual problem lies in the actual game even if you disagree with the particular optional views. It is NOT about mistaking probabilities there, it is about interpreting whether there are two possibilities or one in the cases where one picks a car on as the first decision in the game. The probabilities of either one OR two possibilities are 0.

At least if you could recognize this we could attempt to see which one is correct.

[dionisos]

I will try to explain what i mean with a more simple problem, tell me if you understand, and if you see the link with our actual problem:

Imagine i roll a dice, i will have 6 possible cases: d1(i get 1 on the dice), d2, d3, d4, d5, d6.
A important property here, is that these cases are equiprobable, that mean, they have equal probability to happen: P(d1)=P(d2)=P(d3)=…
Now imagine i win the game if i got a 2 or a 3.
To calculate my probability to win, i will (please keep it mind here the goal is to show the general method):
1) See that P(d1)=P(d2)=P(d3)=… and that P(d1)+P(d2)+P(d3)+… = 1
2) Deduce that P(d1) = 1/6
3) Sum the probabilities of each cases where i win: P(d2)+P(d3) = 2/6

Note that what most people will do, is to count the number of cases where they win, and divide it by the total number of cases, and get 2/6, the good thing is that it is more simple, it include the 3 steps i just give, the bad thing, is that it hide the real reasoning.

Now let me show you a problem where cases aren’t equiprobable:

I roll a dice, if i got a even number i toss a coin, i will have 9 possible cases: d1, d2-head, d2-tail, d3, d4-head, d4-tail, d5, d6-head, d6-tail
Now imagine i win the game if i got 2 on the dice and a head on the coin, or if i got a 3 on the dice.
The naive way to calculate my probability to win would be to count the cases where i win, and divide it by the total number of cases: P(win) = 2/9
But this is false, because here, the cases are not equiprobable.
The real way to get the good result is to do:
1) P(d1)=P(d2)=P(d3)…=P(d6) and P(d1)+P(d2)+…=1
2) P(d1)=1/6
3) P(d2-head) = P(d2-tail) and P(d2-head) + P(d2-tail) = P(d2) = 1/6
4) P(d2-head)=(1/6)/2 = 1/12
5) Sum of the probabilities of each cases where i win: P(win) = P(d2-head)+P(3) = 1/12+1/6 = 3/12 = 1/4

You never said that P(option1 ligne 1) = P(option1 line 3), but the calculation you did, implied it, because without it, it was wrong.

This was my more clear and constructed answer, i want to end by it, because i should have stopped when, even in this more simple problem, and with my answer, you was thinking the probability to win here was 2/9.
And worst, that after that, you started to speak about changing the probability depending of the "perceptive".

[dionisos]

It is you who is concerned about the 12, not me.

You said there was 12 possibilities, when there was only 9, then yes, i was concerned about the 12.

[Scott Mayers]

Then prove that the case in the Monty Hall Problem to which you agree or disagree rather than your own.


OR attempt this:
In the simple two-box case, if I am the host of as simple game to which I said,

"There are two boxes in front of you. I have one ball that fits into it any one of them. I MAY place a ball in one of them."

In this simple game, there are three 'types' of possibilities (arrangements).

Box A....Box B
...Ball.......0
....0........Ball
....0.........0

The question here is without knowing anything, when you pick any box without looking into it, can you determine if it has a ball or not?
Select one: Determinate or Indeterminate

Is the probability 1/2 for each box or 1/3, or 1/4, considering a fair game?

If the host restarts the game and decides to initiate it by showing you an empty box leaving you the other to have to pick, what is the probability that it will have a ball? 1/2 or 1/3 or 1/4,... OR 2/3? OR...indeterminately both or all?

[dionisos]

Then prove that the case in the Monty Hall Problem to which you agree or disagree rather than your own.


OR attempt this:
In the simple two-box case, if I am the host of as simple game to which I said,

"There are two boxes in front of you. I have one ball that fits into it any one of them. I MAY place a ball in one of them."

In this simple game, there are three 'types' of possibilities (arrangements).

Box A....Box B
...Ball.......0
....0........Ball
....0.........0

The question here is without knowing anything, when you pick any box without looking into it, can you determine if it has a ball or not?
Select one: Determinate or Indeterminate

Is the probability 1/2 for each box or 1/3, or 1/4, considering a fair game?

If the host restarts the game and decides to initiate the it by showing you an empty box leaving you the other to have to pick, what is the probability that it will have a ball? 1/2 or 1/3 or 1/4? OR...indeterminately both?

It depend of the probabilities of the host to place a ball in one of the box, i suppose "I MAY place a ball in one of them." mean P(i place a ball in one of them) = 1/2. And that this is what you consider to be a fair game.

The question here is without knowing anything, when you pick any box without looking into it, can you determine if it has a ball or not?
Select one: Determinate or Indeterminate

The response is "no i can’t".
Or "if there is a ball or not is indeterminate".
The response to "can you X", is "yes or no", not "Determinate or Indeterminate".

If P(i place a ball in one of them) = 1/2, then
The probabilities to get a ball is 1/4.
If the host remove a empty box, the probability to get a ball is 1/2.

If i don’t know P(i place a ball in one of them), i don’t have enough information to know the probability to get the ball.
If P(i place a ball in one of them)≠1/2, them my last responses are wrong.

[Scott Mayers]

Is this not similar in part to the second half of the Monty Hall game when trying to determine the case or cases where the remaining options are empty?

When the host reveals the empty box in the first two arrangements above, he is 'forced' to choose in a determinate way; in the last arrangement, since both are empty, he has real options to select either Box A OR Box B to reveal.

Since both have no ball in them, whether you think of the the probability as a Result (what is not in the boxes in both collectively) OR if you consider them independent separate cases as the Act of him choosing Box A or B, exclusively, are they not equiprobable as '0' as one possibility OR two?

So is it:

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0 ? [Respecting the Result value of '0' only]

or

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0? [Respecting the Host Actions to reveal one of these two options independently?]
....0.........0.... <--- Probability = 0?

[dionisos]

Is this not similar in part to the second half of the Monty Hall game when trying to determine the case or cases where the remaining options are empty?

No it is not.

When the host reveals the empty box in the first two arrangements above, he is 'forced' to choose in a determinate way; in the last arrangement, since both are empty, he has real options to select either Box A OR Box B to reveal.

Yes.

Since both have no ball in them, whether you think of the the probability as a Result (what is not in the boxes in both collectively) OR if you consider them independent separate cases as the Act of him choosing Box A or B, exclusively, are they not equiprobable as '0' as one possibility OR two?

Your sentence have no meaning, we never consider probability "as a result", probability is about what we don’t know yet, and what we know about likeliness/frequencies.

So is it:

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0 ? [Respecting the Result value of '0' only]

or

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0? [Respecting the Host Actions to reveal one of these two options independently?]
....0.........0.... <--- Probability = 0?

You "probability=0" have no meaning, you should specify what is the probability you are speaking about, if you speak about P(having no balls in the box, and the host remove the first box), the probability is not equal to 0 at all.
P(having no balls in the box, and the host remove the first box) = (1-P(i place a ball in one of them)) / 2

[Scott Mayers]

Is this not similar in part to the second half of the Monty Hall game when trying to determine the case or cases where the remaining options are empty?

No it is not.

Look, if you don't think this has any representative meaning, then deal with the two diagrams I showed above for the actual puzzle and PLEASE tell me what you presume is at fault? Do you think there is NO relevant distinctions there? Do you not wonder how or why other mathematicians even with full degrees question the original presumptions of the puzzle? Or are they just imagining things? I'm asking you to pinpoint the particular problem in the diagram that demonstrates the cases of Switch options (in the dark grey boxes) I added. I'm at a loss here with you. I already understand the traditional position and how it is derived (as per the way they treat rows one and two in that diagram as merely one where I see it as distinctly separate events. ) But you don't seem to even be trying to interpret this from my perspective.

?? Here is the diagram I want you to scrutinize from above again:

Fixed Solution.png

[dionisos]

Do you not wonder how or why other mathematicians even with full degrees question the original presumptions of the puzzle? Or are they just imagining things?

I don’t wonder it, because it is false, about no mathematicians question the puzzle.
Some mathematicians do a mistake when first faced with the puzzle, because there is a trap, and anybody could fall in a trap, even very simple trap.
But i never see a mathematicians question the answer to the problem, after we show them the trap.

?? Here is the diagram I want you to scrutinize from above again:

Fixed Solution.png

Here in you diagram, in 2 cases you win by switching, and in 2 cases you lose by switching (win by staying).
Then, what ?
You can’t do anything with that, if you don’t know the probabilities of each 4 cases.
You can’t conclude that P(win when switching) = P(win when staying).

[dionisos]

I already understand the traditional position and how it is derived (as per the way they treat rows one and two in that diagram as merely one where I see it as distinctly separate events. )

It is two distinctly separate events, i said it from the start. (you could also consider it to be one event, it change nothing to the final result)
What you don’t seem to understand, is that probability is not just about counting the number of events.

[Scott Mayers]

Do you not wonder how or why other mathematicians even with full degrees question the original presumptions of the puzzle? Or are they just imagining things?

I don’t wonder it, because it is false, about no mathematicians question the puzzle.
Some mathematicians do a mistake when first faced with the puzzle, because there is a trap, and anybody could fall in a trap, even very simple trap.
But i never see a mathematicians question the answer to the problem, after we show them the trap.

?? Here is the diagram I want you to scrutinize from above again:

Fixed Solution.png

Here in you diagram, in 2 cases you win by switching, and in 2 cases you lose by switching (win by staying).
Then, what ?
You can’t do anything with that, if you don’t know the probabilities of each 4 cases.
You can’t conclude that P(win when switching) = P(win when staying).

Well, I looked up and found a good link to a New York magazine article on this HERE
I learned that I was correct on the perception thing. Read this article to see how you would interpret it based on the perception, that all that matters is the result and so you choose to switch 2/3 of the time. But this is contradictory if you think that this is actually increasing your odds. They are still even but this is perceived by recognizing the actual acts that lead you to the results independently.

That is my 1/2 understanding respects what I showed in the diagram above. For another illustration, (even recognizing the different probabilities using the math):

2015-09-05_070314.png

Add these up.

And:
1. Krauss, Stefen and Wang, X.T. (2003). "The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser," Journal of Experimental Psychology: General 132(1). Retrieved from http://en.wikipedia.org/wiki/Monty_Hall_problem.

2. Let A, B, and C be the events that the car is behind door 1, 2, and 3, respectively. We want to show what our intuition tells us: given the car isn't behind door 3, the probability it's behind door 1 is 50%. For those with some background in probability, this is a straightforward exercise in conditional probability. Using the given notation, we have

P(A | not C) = P(A and not C)/P(not C).

But P(A and not C) = P(A), since if the car is behind door 1, clearly it is behind door 1 and not behind door 3, and vice versa. Thus

P(A | not C) = P(A)/P(not C) = (1/3)/(2/3) = 1/2, as desired.

[Scott Mayers]

I already understand the traditional position and how it is derived (as per the way they treat rows one and two in that diagram as merely one where I see it as distinctly separate events. )

It is two distinctly separate events, i said it from the start. (you could also consider it to be one event, it change nothing to the final result)
What you don’t seem to understand, is that probability is not just about counting the number of events.

Perception is the problem. They are both true depending on how you look at it. But when you see it my way instead, it resolves it as being NON-Contradictory precisely for what I was saying.

My explanation is even better as I illustrate this as those diagrams to show how you actually have to allow each as distinct rows. And NO, this is NOT what the 2/3 solution uses. they use the two first rows as ONE row treating the variable option there for the Host to pick from as one out of three rather than four. ...again, precisely what I was explaining.