The scams of Statistics...

[Obvious Leo]

I already know your position and you aren't adding anything new.

This is not MY position and there is nothing new to add. Do you think this is a matter of opinion? Why have you failed to address the facts?

[Scott Mayers]

I already know your position and you aren't adding anything new.

This is not MY position and there is nothing new to add. Do you think this is a matter of opinion? Why have you failed to address the facts?

You are begging me to shut up! Address my arguments, not your fixed beliefs. If I am not allowed to disprove what you whole-heartedly believe without question, so be it.

[Obvious Leo]

Scott. This has got nothing to do with what I believe.

Please answer this question. The Monty Hall game has been played and simulated on computers probably millions of times in the last 20 years. On EVERY SINGLE OCCASION it has been shown that switching one's guess DOUBLES one's chance of winning.

How do you account for this?

[Scott Mayers]

Scott. This has got nothing to do with what I believe.

Please answer this question. The Monty Hall game has been played and simulated on computers probably millions of times in the last 20 years. On EVERY SINGLE OCCASION it has been shown that switching one's guess DOUBLES one's chance of winning.

How do you account for this?

READ WHAT I WROTE FIRST! I proposed the problem here and you are demanding me to appeal to your position without attending to anything I could possibly say. It is I who posited the problem and clearly proved it at fault with definitive closure. I also used to agree to the puzzle without actually paying close enough attention to it.

I already pointed out to you that you'd have to defer to the actual program being used to determine how they set any problem out. I don't believe you're concern to refer to such simulations unless I am allowed to see this code (the actual program being used).

The originators of this puzzle is either playing a game on us, or they are possibly sincerely blind, or they invested in it too much to believe they cannot back down, or....and even as weird as this may be, are testing us as a social experiment. (This last point relates to the fact of the interests are of the dominant interests of psychologists over the mathematicians in this subject!) The logic is flawed.

As a very, very simple example, imagine two closed boxes I present to which you can only choose once. If one box is 'A' and the other is 'B', is it not true that there are two possible choices you can make, not one? Imagine that both actually have nothing in them both OR both have something in each. [Empty-Empty or Coin-Coin] Where they both have the same value in their content, this may not matter as you either get the identical reward/non-reward in both. You conclude that since the value of the outcome is the same, you treat this as merely one case.

But it is NOT one simple case. If you pick box A, it is distinctly different than picking box B. This is why in this argument you have to split the option of the case where the host can choose either door when both are goats for the Monty Hall problem to reveal as distinct events.

If you accept the traditional solution as sound, then you MUST believe in miracles. If you include the game from the start as 1/3 using the denominator as a factor in the second play, this puzzle is suggesting that you can get 1/2 - 1/3 = 2/3 gain advantage for nothing. You cannot include the three choices once the host alters it to 1/2 because when both remaining doors are left with goats, this allows the host to RANDOMLY pick one of those doors to reveal, unlike the other two cases. This makes that case become two. The other two FORCE the host to pick the only remaining goat. Thus there are FOUR, NOT THREE event possibilities. So the 2/3 should be 2/4 = 1/2!!

[Scott Mayers]

Another point on the possible experiment using a simulation: If the vast majority doing the experiment are doing so in BELIEF that one should CHOOSE to Switch, this alone is enough to encourage that majority to opt to switch more often. As such, while the odds are still 1/2, if they measure the actual wins based on the majority placing faith in the switch only assures that when or where they DO win half the time, they are assured to check back and notice that it was due to a 'Switch' more often. Thus the very BELIEF is enough to throw people off. Such an experiment to be fair would have to require only people who never even heard of the problem, like testing with children, for instance.

[vegetariantaxidermy]

Scott. This has got nothing to do with what I believe.

Please answer this question. The Monty Hall game has been played and simulated on computers probably millions of times in the last 20 years. On EVERY SINGLE OCCASION it has been shown that switching one's guess DOUBLES one's chance of winning.

How do you account for this?

READ WHAT I WROTE FIRST! I proposed the problem here and you are demanding me to appeal to your position without attending to anything I could possibly say. It is I who posited the problem and clearly proved it at fault with definitive closure. I also used to agree to the puzzle without actually paying close enough attention to it.

I already pointed out to you that you'd have to defer to the actual program being used to determine how they set any problem out. I don't believe you're concern to refer to such simulations unless I am allowed to see this code (the actual program being used).

The originators of this puzzle is either playing a game on us, or they are possibly sincerely blind, or they invested in it too much to believe they cannot back down, or....and even as weird as this may be, are testing us as a social experiment. (This last point relates to the fact of the interests are of the dominant interests of psychologists over the mathematicians in this subject!) The logic is flawed.

As a very, very simple example, imagine two closed boxes I present to which you can only choose once. If one box is 'A' and the other is 'B', is it not true that there are two possible choices you can make, not one? Imagine that both actually have nothing in them both OR both have something in each. [Empty-Empty or Coin-Coin] Where they both have the same value in their content, this may not matter as you either get the identical reward/non-reward in both. You conclude that since the value of the outcome is the same, you treat this as merely one case.

But it is NOT one simple case. If you pick box A, it is distinctly different than picking box B. This is why in this argument you have to split the option of the case where the host can choose either door when both are goats for the Monty Hall problem to reveal as distinct events.

If you accept the traditional solution as sound, then you MUST believe in miracles. If you include the game from the start as 1/3 using the denominator as a factor in the second play, this puzzle is suggesting that you can get 1/2 - 1/3 = 2/3 gain advantage for nothing. You cannot include the three choices once the host alters it to 1/2 because when both remaining doors are left with goats, this allows the host to RANDOMLY pick one of those doors to reveal, unlike the other two cases. This makes that case become two. The other two FORCE the host to pick the only remaining goat. Thus there are FOUR, NOT THREE event possibilities. So the 2/3 should be 2/4 = 1/2!!

I can just see leo with smoke coming from his ears after reading this one.

[Scott Mayers]

I can just see leo with smoke coming from his ears after reading this one.

....IF, he reads it...

But I'm not trying to insult anyone here, especially Leo, as I value his input in fairness.

EDIT: My concern here is NOT to DEFEAT people here, only the improper LOGIC involved that leads to the faith in these Probability/Statistic errors. Yet, when people hold strong personal views, people can't help but internalize this as an attack against them as humans. Thus it is hard to actually do so without harming others sometimes regardless. And I'm no exception. But I'm already in the minority on many issues and though it isn't easy to get used to, I can't unlearn what I now know. Otherwise I'd be accepting defeat by whatever is already most popular. Should I give up just to please others? They certainly won't be rewarding me for anything anyways.

[Obvious Leo]

Scott. I do read all of your posts, believe me, but you are breaking my heart.

Look at this once more.

As I suspected there might be there is a Monty Hall game running on the internet.

As of now 278,324 contestants decided to stay with their first choice and won a total of 93,286 pretend cars, a success rate of just below 34%

168,244 contestants decided to change their choice and won a total of 111,535 pretend cars, a total of 66%.

Scott. Would you care to name the parties responsible for staging such a massive global conspiracy or would you prefer to apologise to the members of this forum for treating them like morons.

This is a sample size of 446,568 contestants. You're better at sums than me so YOU work it out but I'm willing to bet that for such a distribution of scores to occur when the true distribution should be 50/50, (according to you), is utterly IMPOSSIBLE to a million sigma order of probability.

This is not a maths problem, Scott, and your mathematical treatment of it is meaningless.

How do you account for the fact that switching guesses has doubled the contestants chances of winning EVERY SINGLE TIME THIS GAME HAS EVER BEEN PLAYED.

[mickthinks]

Leo, I don't know whether it is because you are deliberately behaving like an internet troll or you genuinely believe that you are helping Scott to understand where he is going wrong, but you are missing a very important point, which is that Scott believes he has a proof that your solution is false, and hence that any proof you offer must be faulty, just as you believe his proof must be faulty. When you understand the symmetry of the situation in which both of you are equally convinced by your own arguments and thus content to dismiss the arguments of the other as faulty by definition instead of addressing the detail of the other's argument and offering insights into where and how the mistakes in it have arisen, you should see that endlessly repeating your favourite arguments (and spicing them up with insults) is at best unproductive, and at worst, belligerent.

Scott has repeatedly asked you to address his arguments and I am beginning to think you might be incapable of doing so, perhaps because you don't really understand the mathematics involved and are merely parroting the ideas you've picked up from others.

I don't know how the experiments [with animals and children] were performed but this is a fact I've come across several times ...

I think VT was surprised to hear you suggest that there had been any experiments investigating the way children and animals play the Monty Hall guessing game, and I am too. I am inclined to think you've just made that up in order to add weight to your insults to Scott's intelligence. That you can't remember any details or cite any references or provide any links adds to those doubts.

How many times do you need to be told that this is not a maths problem?

wtf? It is self-evidently a probability problem and probability is a branch of applied mathematics. I suggest you stop trying to tell Scott it isn't. I think it makes you look rather foolish.

How do you account for this [that computer simulations of The Monty Hall game show that switching one's guess doubles one's chance of winning]?

Scott has already accounted for that when he said "if you initially use the very math that leads to the problem in a program set up that doesn't accept all possibilities by default, any tests based on such a program will only CONFIRM the expected result".

[Obvious Leo]

Mick. At no stage has Scott addressed the central point of the Monty Hall problem, which is the prior knowledge and actions of the game's host which alter the probabilities in this scenario. Neither have you.

[Scott Mayers]

Scott. I do read all of your posts, believe me, but you are breaking my heart.

Look at this once more.

As I suspected there might be there is a Monty Hall game running on the internet.

As of now 278,324 contestants decided to stay with their first choice and won a total of 93,286 pretend cars, a success rate of just below 34%

168,244 contestants decided to change their choice and won a total of 111,535 pretend cars, a total of 66%.

Scott. Would you care to name the parties responsible for staging such a massive global conspiracy or would you prefer to apologise to the members of this forum for treating them like morons.

This is a sample size of 446,568 contestants. You're better at sums than me so YOU work it out but I'm willing to bet that for such a distribution of scores to occur when the true distribution should be 50/50, (according to you), is utterly IMPOSSIBLE to a million sigma order of probability.

This is not a maths problem, Scott, and your mathematical treatment of it is meaningless.

How do you account for the fact that switching guesses has doubled the contestants chances of winning EVERY SINGLE TIME THIS GAME HAS EVER BEEN PLAYED.

Please reference your last statement here. I answered your request above regarding the bias of experiments. You'd have to show that they are immune to error by people's predisposed beliefs doing such. If it is an online experiment, it is exposed to the error of bias and we cannot determine its truth one way or the other. The people in the experiment have to be limited to those who have never even heard of the problem. Otherwise it is contaminated.

If you don't believe me, try publishing a set of definitive proofs for why any beliefs are irrational online. Then poll them to see if they agree. Even guessing the poll would result in just what the general population already believes in by averages, some will even attend to skew the results if it favors them. And while this may occur by both sides of the issue, the average of those in the crowd who would attend to it would still be more on the side of the popular view. Thus the apparent stats would show an even greater than normal belief because it is not controlled.

My question to you:
I gave you an example of two boxes. If both boxes have nothing in them, then do you believe that you should treat them as one equal unique act should you open one box OR the other? Or should you treat them as distinctly two events?

[Scott Mayers]

Mick. At no stage has Scott addressed the central point of the Monty Hall problem, which is the prior knowledge and actions of the game's host which alter the probabilities in this scenario. Neither have you.

The knowledge of the Host asserts that he has two KINDS of options.

In the cases where you initially do NOT pick the car, the host, knowing this is FORCED to choose the only remaining one with the goat to reveal.

In the one case where you initially DO pick the car, the host, knowing the remaining ones are both goats, is NOT FORCED to merely pick one. As such, he can randomly choose either one. This is two options by his ACT, not one option just because the RESULT is the same.

By contrast, it doesn't matter even from the perspective of the player. When the host reveals a door with a goat, to you, as a player, you are INDETERMINATE to be certain what is in the other alternative. If both remaining doors are goats, the player can realistically distinguish that he only has two options: STAY or SWITCH [that's two options]. If he chooses to STAY AND the car is there, he also has 2/3 of a chance to FAIL if he switches!!

Thus even using the same reasoning as the problem, if one has a 2/3 chance to win by switching, you have to recognize that you also have a 2/3 chance to lose by staying!! How is this possible?

[mickthinks]

Thus even using the same reasoning as the problem, if one has a 2/3 chance to win by switching, you have to recognize that you also have a 2/3 chance to lose by staying!! How is this possible?

Scott, I confess I don't understand your question. That you have a 2/3 chance of losing if you don't take the option which gives you a 2/3 chance of winning is not just possible, but necessarily true from the law that (independent exhaustive) probabilities sum to 1.

Code: Select all

option   |  A  |  B  | Total
P of win | 1/3 | 2/3 |  1
P of lose| 2/3 | 1/3 |  1
Total    |  1  |  1  |

[mickthinks]

Leo: At no stage has Scott addressed the central point of the Monty Hall problem, which is the prior knowledge and actions of the game's host which alter the probabilities in this scenario.

Apart from the fact that he has, this has nothing to do with anything I said to you about starting from where Scott is rather than trying to get him to concede a priori that your view of the problem is superior to his. It doesn't matter what you think the central point of the Monty Hall problem is. What matters is what Scott thinks about the problem. In other words, the problem here is not the Monty Hall problem, it is a Scott Mayers problem.

Leo: Neither have you.

Don't be a twit, Leo! I am not arguing with you about the Monty Hall problem.

[Scott Mayers]

Thus even using the same reasoning as the problem, if one has a 2/3 chance to win by switching, you have to recognize that you also have a 2/3 chance to lose by staying!! How is this possible?

Scott, I confess I don't understand your question. That you have a 2/3 chance of losing if you don't take the option which gives you a 2/3 chance of winning is not just possible, but necessarily true from the law that probabilities sum to 1.

Code: Select all

option   |  A  |  B  | Total
P of win | 1/3 | 2/3 |  1
P of lose| 2/3 | 1/3 |  1
Total    |  1  |  1  |

Your "P of win" is intended to say that 2/3 of the time, you win and 2/3 of the time you lose. This is just what I'm saying in part but that the Wins relate to what you consider favorable. Thus the "P of win" you speak of must relate to the odds of SWITCHING. Yet your "P of lose" has to refer to the odds of STAYING. To be accurate, you have to clarify all of these probabilities:

..................................A....B...Total
P to SWITCH AND WIN........ = 1/3 + 2/3 = 1
P to STAY AND NOT WIN..... = 2/3 + 1/3 = 1
P to SWITCH AND NOT WIN.. = 2/3 + 1/3 = 1
P to STAY AND NOT WIN.. = 1/3 + 2/3 = 1

IMPORTANT EDIT: I swear that I had written this correctly before but found this different than I originally wrote it. So I've repaired it.