The scams of Statistics...

[Scott Mayers]

Another clarification example:

Suppose we compare two different scenarios. In the first case, using the regular Monty Hall game, if you were to ALWAYS opt to switch, then the case reduces to imagining a person choosing any one of three doors and then simply deciding to always SWITCH, because we know that the host will always at least leave the 'known' car unrevealed by him.

Thus, in all cases, this is like having 1/3 chance to select and then deciding to change your mind and SWITCH WITHOUT YOU KNOWING whether your first choice is correct or not. This reduces to the odds being 1/3 no matter what!

The solution of this problem suggests that you always Switch. So you may as well opt to have someone put on headphones to play and ignore anything the Host might say but to always act like an automaton and simply SWITCH to the second door. This is indistinguishable logically to picking two doors out of three purposely aiming to have a better chance on the first round, then having one of them be shown to be a dud and then the host and asking you if you want to SWITCH to the remaining one door that hasn't been touched or stay. You certainly have better odds to STAY on the first round no matter what. But this reduces your odds to 1/2 since you get to choose two and gain even though you discover at least one of them as a dud! If you continue playing, this reduces your odds to 1/3 because instead of two doors if you choose to stay, you only have one out of the original three to play from.

[dionisos]

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

I am definitely not equating them as they are illustrated as being completely different events. Why you would think that is odd to me.

Maybe you are not reading this correctly. See my last post so that we may have a chance to reconstruct the problem step by step as given in the puzzle.

Thanks to ask, it show me you didn’t understand what i said, but i will try again.

What you do by counting the cases when you switch and you win, assume equiprobability of all the cases, even if you are not conscious of that. (in particular P(option1 ligne 1) = P(option1 line 3))

The real thing you should do, is sum the probabilities of the cases when you switch and you win.
If all cases have equal probability to happen, then summing the probabilities is the same thing as counting the cases when it happen, and divide it by the total number of cases, but without it, your operation have no meaning. (here is the subtle thing you don’t seem to get)

In 6 cases in the total of 12 cases, that mean 1/2 of the cases, you win when you switch, but it doesn’t give you the probability to win by switching, if these cases are not equiprobable.

[Scott Mayers]

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

I am definitely not equating them as they are illustrated as being completely different events. Why you would think that is odd to me.

Maybe you are not reading this correctly. See my last post so that we may have a chance to reconstruct the problem step by step as given in the puzzle.

Thanks to ask, it show me you didn’t understand what i said, but i will try again.

What you do by counting the cases when you switch and you win, assume equiprobability of all the cases, even if you are not conscious of that. (in particular P(option1 ligne 1) = P(option1 line 3))

The real thing you should do, is sum the probabilities of the cases when you switch and you win.
If all cases have equal probability to happen, then summing the probabilities is the same thing as counting the cases when it happen, and divide it by the total number of cases, but without it, your operation have no meaning. (here is the subtle thing you don’t seem to get)

In 6 cases in the total of 12 cases, that mean 1/2 of the cases, you win when you switch, but it doesn’t give you the probability to win by switching, if these cases are not equiprobable.

You disappeared for so long and need to catch up some. Please read what you missed to see if I answered your own questions there as you are digressing me back to where I've already provided much depth in responding here. The second-last post I've made above should suffice for your question HERE.

[Obvious Leo]

Scott. I'm hoisting the white flag and retiring from the battlefield. I did my best.

[dionisos]

I read the messages i missed, i saw a lot of mistakes, from both people thinking we have 1/2 chance to win by switching and people thinking we have 2/3.
I also see we got some new way to see and to solve of the problem, and i have also much to say to that.

I have just too much to say, this is why i took the decision to only answer your questions, to come back to our previous discussion.

On the link you give me, i agree with mickthinks (outside the little mistake), but not with you, but still, i think it would confuse the discussion if i answer to that.
I will answer if you explain me the link between that and my own previous answer, but i would not otherwise.

And i don’t see any other things that could answer to what i said to you.

Also i accept we deal with the problem with the "representation" of your choice, but not to change before fixing the disagreement with this "representation".
I didn’t say it before, then if you want we can change now.
I will only focus on what you say, and not on what other say.

Sorry to put these kind of rules, but it would be too hard for me otherwise.

[Scott Mayers]

I read the messages i missed, i saw a lot of mistakes, from both people thinking we have 1/2 chance to win by switching and people thinking we have 2/3.
I also see we got some new way to see and to solve of the problem, and i have also much to say to that.

I have just too much to say, this is why i took the decision to only answer your questions, to come back to our previous discussion.

On the link you give me, i agree with mickthinks (outside the little mistake), but not with you, but still, i think it would confuse the discussion if i answer to that.
I will answer if you explain me the link between that and my own previous answer, but i would not otherwise.

And i don’t see any other things that could answer to what i said to you.

Also i accept we deal with the problem with the "representation" of your choice, but not to change before fixing the disagreement with this "representation".
I didn’t say it before, then if you want we can change now.
I will only focus on what you say, and not on what other say.

Sorry to put these kind of rules, but it would be too hard for me otherwise.

I'm having a hard time following your broken English, dionisos.

You mentioned above a concern to assert that something is or is not "equiprobable" without defining what you mean by it. In your reference to my line 1 vs. line 3, in the diagram you insist I equate anything is NOT true at all. I equate line 1 and line 2 as options where the initial pick by the guest is the CAR. Lines 3 and 4 are where the guest does NOT pick the CAR. The first two differ significantly to the 3rd and 4th also because the Host has two 'equal' outcome values but require two different doors in which the Host can pick from which force the Switch option by the Guest to be distinctly two different doors.

The third and fourth lines are unique because in each case the Host is FORCED to reveal the goat and NOT the car. Thus, the Host has a variable option when you first select the car and a constant unique option in the other two cases.

You have to follow my lead on this because you've opted to come to an English language site to which makes me unaware if you're actually translating me fair. If you are using "Google Translate" or some similar software, you could be seeing something mistranslated that I didn't actually write. This is why I'm questioning why you'd find something I'm equating between lines 1 and 3 as they have nothing in common neither particularly nor generally.

[Scott Mayers]

Please look back at my example at viewtopic.php?f=26&t=16671&start=105#p220495. This is the post on page 8. I had to edit it as I looked at it and it wasn't what I'm certain that I didn't post before. This is the second time today that the HTML was altered on me here. [The last was the same link above that I certainly created and tested but then discovered it was altered on me!]

[dionisos]

I'm having a hard time following your broken English, dionisos.

You mentioned above a concern to assert that something is or is not "equiprobable" without defining what you mean by it. In your reference to my line 1 vs. line 3, in the diagram you insist I equate anything is NOT true at all. I equate line 1 and line 2 as options where the initial pick by the guest is the CAR. Lines 3 and 4 are where the guest does NOT pick the CAR. The first two differ significantly to the 3rd and 4th also because the Host has two 'equal' outcome values but require two different doors in which the Host can pick from which force the Switch option by the Guest to be distinctly two different doors.

The third and fourth lines are unique because in each case the Host is FORCED to reveal the goat and NOT the car. Thus, the Host has a variable option when you first select the car and a constant unique option in the other two cases.

You have to follow my lead on this because you've opted to come to an English language site to which makes me unaware if you're actually translating me fair. If you are using "Google Translate" or some similar software, you could be seeing something mistranslated that I didn't actually write. This is why I'm questioning why you'd find something I'm equating between lines 1 and 3 as they have nothing in common neither particularly nor generally.

I use google translate a lot when i don’t understand some words, but yes it is possible i misunderstand you because of english, even if i do my best on it.

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.

[Scott Mayers]

dionisos,

You err if you define a win as uniquely a (2 AND a HEAD) or a (3). You would have to include all 9 because it is NOT the case that a
(2 AND NOT HEAD) is equal to a (2 AND a HEAD) in fact as an Act NOR Result. I find this also confused the issue by adding more factors than the original argument. Note that in the Monty Hall Game, this rule is unique to only three original options ONLY. But I can't even relate to your claim here.

So limit this to simply two boxes (or doors). Each box, or door, we'll label as distinct things. So if you label the first with a Number '1' and the other Number '2', you can't rearrange them.

TwoBoxes.png

So in this initially simple idea, if I were to ask you to pick a box, how many possibilities do you have? Ignore initially whether one, the other, both, or neither have anything in them.

[dionisos]

You err if you define a win as uniquely a (2 AND a HEAD) or a (3).

I don’t see how it could be a error, i define my game like it.
If you get a 2 and a head, you win, if you get a 3, you win, otherwise you lose.

it is NOT the case that a (2 AND NOT HEAD) is equal to a (2 AND a HEAD) in fact as an Act NOR Result.

Yes, but i never compare "(2 AND NOT HEAD)" and "(2 AND a HEAD)" i compare "P(2 AND NOT HEAD)" and "P(2 AND a HEAD)"
Maybe you are not accustomed to this kind of notation we use in probability. P(X) mean, the probability to have X.
It mean you have as much chance to get "a 2 on the dice, and a tail on the coin", than to get "a 2 on the dice, and a head on the coin".
If you don’t agree, please tell me what are the probabilities of these two cases.

I find this also confused the issue by adding more factors than the original argument

.

No, i assure you it don’t confuse thing at all, on the contrary, if you think i made a error on my simple problem with the dice and the coin, we should really, really, fix it before anything.
And here it seem you misunderstand what P(X) mean, and it explain very well why you didn’t understand me from the start.
The problem with the dice is much less confusing that the Monty Hall Game, please we should fix it before.

So limit this to simply two boxes (or doors). Each box, or door, we'll label as distinct things. So if you label the first with a Number '1' and the other Number '2', you can't rearrange them.

TwoBoxes.png

So in this initially simple idea, if I were to ask you to pick a box, how many possibilities do you have? Ignore initially whether one, the other, both, or neither have anything in them.

I have two possibilities.
Note that we never disagree on the number of possibilities.

[Scott Mayers]

dionisos,

You may be using different terms of probability than we use in the West. A "possibility" is any type of thing that can be "posited" or "proposed" for some domain. These can then be used as inputs for some problem [note the root word of ""probability" here too from "to probe", meaning to seek]. Thus a probability is a comparison of some particular kind of possibility in comparison to all possibilities in some domain.

So, in your dice example, you have to first enumerate for all possibilities in the domain which is 9, in your given scenario. Then, since you are posing two unique possibilities, the probability of getting these is thus, 2/9, and is the correct number. When dealing with the rest of the probabilities, the possibilities add up to all possibilities but the others are the "complement" of the one you are seeking. Thus the complement of 2/9 = 7/9. So 7/9 represents the probability of NOT getting what you are seeking in the first case. This includes your (2 and not Head). Yet you seem to be trying to think of the dependency of the dice (= 2) as all that matters because one of the possibilities it leads to happens to be what you want. This can lead to trouble because a dice roll of Even numbers is what you defined as requiring a further step to get the final result. Thus it is indeterminate from simply getting an Even roll until you toss the coin.

I find your example though lacking clarity as well as simplicity and why I am trying to ask you about those boxes. So you appear to agree so far that those two boxes and the given instruction to 'pick one' leads to distinctly two possibilities.

Now notice that the ACT of "picking one unique box" represents a type of possibility as a dynamic process. I haven't introduced any intention to what we could seek for inside those boxes yet. But when we do, we consider the set of those things that could be placed in those boxes as a static (or non-dynamic) set of possibilities to which we could discover as a RESULT of opening one. So my next question is whether you understand this distinction and the definitions so far?

Note too that we have the boxes themselves that represent a set of possibilities in a static way. These differ from the RESULT types of possibilities in that where the RESULTs are "constant" when discovered, the boxes act as "variables". In this way, each box acts as an "indeterminate" kind of possibility until we "determine" it by opening the box to discover what is inside. The boxes themselves are relative RESULT-like constants in themselves and so we can also think of the ACT of selecting determines the 'constant' of a set of two unique boxes. Then, if we want to go further, the ACT of opening the box treats the box as an indeterminate variable that represents the set of possibilities of the contents of the box that reveal another relative RESULT.

Do you follow this too? It will help in how we can describe things with better clarity where we might be confusing the types of possibilities that we use to interpret our probabilities.

[dionisos]

No, i could make mistake in english, but know pretty well what is a probability.

Thus a probability is a comparison of some particular kind of possibility in comparison to all possibilities in some domain.

No, it is not, probability is the measure of the likeliness that an event will occur. (and this is not equivalent of what you just said)
If you play my game 1000 time, you will get something like it:
you will get d1 about 166 times
you will get d2 and head about 83 times
you will get d2 and tail about 83 times
you will get d3 about 166 times
you will get d4 and head about 83 times
you will get d4 and tail about 83 times
you will get d5 about 166 times
…
the likeliness to have d2 and head, that mean P(d2 and head) is near of 83/1000.
in fact P(d2 and head) = 1/4 = 0.8333333333…

If you don’t believe me, just play the game 1000 times, and count each time you get a 1, each time you get a (2 and a head), etc… (evidently, don’t do it manually, do a program).

So, in your dice example, you have to first enumerate for all possibilities in the domain which is 9, in your given scenario. Then, since you are posing two unique possibilities, the probability of getting these is thus, 2/9, and is the correct number.

I explained why it is not the case here, it is the case only if all possibilities are equiprobable.
You make exactly the mistake i thought you was doing.
Counting the possibilities you have to get somethings, and dividing it by the total number of possibilities, is meaningless without equiprobability.

When dealing with the rest of the probabilities, the possibilities add up to all possibilities but the others are the "complement" of the one you are seeking. Thus the complement of 2/9 = 7/9. So 7/9 represents the probability of NOT getting what you are seeking in the first case. This includes your (2 and not Head).

Yes, if P(X)=2/9, then P(not X)=7/9,
P(2 and head)=1/4, and then P(not(2 and head))=3/4.
not(2 and head) here is (1, or 2 and tail, or 3 or 4 and head, or 4 and tail or 5 or 6 and head, or 6 and tail)
not having 2 and head, is having one of the other 8 possibilities.

Now notice that the ACT of "picking one unique box" represents a type of possibility as a dynamic process. I haven't introduced any intention to what we could seek for inside those boxes yet. But when we do, we consider the set of those things that could be placed in those boxes as a static (or non-dynamic) set of possibilities to which we could discover as a RESULT of opening one. So my next question is whether you understand this distinction and the definitions so far?

The act is what we do, and the result what we get by doing it ?
By example the act is "opening box 1", and the result will be "i see the car in the box" ?

[Scott Mayers]

No, i could make mistake in english, but know pretty well what is a probability.

Thus a probability is a comparison of some particular kind of possibility in comparison to all possibilities in some domain.

No, it is not, probability is the measure of the likeliness that an event will occur. (and this is not equivalent of what you just said)
If you play my game 1000 time, you will get something like it:
you will get d1 about 166 times
you will get d2 and head about 83 times
you will get d2 and tail about 83 times
you will get d3 about 166 times
you will get d4 and head about 83 times
you will get d4 and tail about 83 times
you will get d5 about 166 times
…
the likeliness to have d2 and head, that mean P(d2 and head) is near of 83/1000.
in fact P(d2 and head) = 1/4 = 0.8333333333…

If you don’t believe me, just play the game 1000 times, and count each time you get a 1, each time you get a (2 and a head), etc… (evidently, don’t do it manually, do a program).

I'm not in total disagreement with you. But your answer is still wrong even here. I understand how you derived this as 1/4 but I had to introduce the concepts above to show how even your thinking can be skewed by interpreting probability depending on what counts or not. You predefine the ACT of rolling a dice but then create an illusion that when considering what is "even" matters by introducing a second ACT to toss a coin where even. Thus you have to alter your assumptions about possibilities here.

In this case, only 2/6 = 1/3 of the dice throws are permitted to 'matter' initially. But only on the toss of a '2', is where the coin toss 'matters'. And so this only adds one extra possibility to the whole. So we have actually 7 possibilities that matter in this game as you defined.

Thus, the odds to WIN have to be 2/7, not 1/4. You cannot include those events where you think that other even dice throws participate in this example. For instance, you already know before-hand the rules. If you first told someone that they had to play according to the rules before assigning the result, while the odds by the perspective of the individual playing can later infer 1/4, from one perspective, it is 2/7 in another perspective.

This is the reason I'm expanding on the particular (and appropriate) way to begin our investigation as it has to respect things with understanding possibilities and how each kind represent different types or sets to which we have to be specific about. You can do a computer simulation above to assure the result as 1/4 as much as you can to demonstrate it as 2/7 depending on the perspectives.....which depend on specifying the possibilities according to appropriate classes with clarity. In your example, the actual odds that matter to get the RESULT of winning with all rules in mind are 2/7. If you play it by unfolding the rules in play, while it appears as 1/4 because you place a pretense of significance on ALL even numbers at first including their coin tosses, this only creates an illusion that some of the ACTS matter. Can you begin to at least see the problem we have to contend with when arguing probabilities of the Monty Hall Problem?

[dionisos]

No, all possibilities 'matter' here, forget about the part of winning or not, it seem to confuse you.
If you roll a dice, and then toss a coin if the dice give a even number, then your probability to get a 2 and a head, is 1/12. P(d2 and head) = 1/12
And you probability to get a 3 is 1/6, P(d3)=1/6

as much as you can to demonstrate it as 2/7 depending on the perspectives

Ok, i wait to see your program then, here would be mine, i could translate it in python to get the result, but i want yours:

Code: Select all

results = {}

do it 100000 times
  dice_result = a random integer between 1 and 6
  if 2 divide dice_result then
    c = a random integer between 0 and 1.
    coin = "head" if c = 0, and "tail" if c=1
    results["d"+x+"-"+coin] = results["d"+x+"-"+coin] +1 // we initialize results["d"+x+"-"+coin] to 0 if it doesn’t exists.
  else
    results["d"+x] = results["d"+x] +1

print(results)

You could see i only use the game rules, i don’t do any reasoning about probability here.
Now i wait to see your program, i can also translate it in python if you want to see the result.

We agree there are 9 cases in my game, i gave you the probability of each 9 cases, i wait for you to do the same.

[Scott Mayers]

The act is what we do, and the result what we get by doing it ?
By example the act is "opening box 1", and the result will be "i see the car in the box" ?

Yes, on the first question. In your example, this depends on the perspective, but this is one perspective that is correct by recognizing one which one begins with as an act relative to what one ends on as a result, which is what is inside the box.

You can also say that the ACT of "selecting the box" and the end of each particular chosen box, is a local RESULT.

Then, the ACT of "opening the box" treats the last RESULT as the beginning of another sub-perspective (as a subject) to receive the final object, "what is inside the box" as the secondary RESULT.

The, there is the perspective of the whole set of acts and results collectively. So we can describe beginning with an ACT (or a subject being presumed to be some type of "result" simply assumed, like the Guest, whom we assume is a result of one opting to play a game) of "picking a box" that ends with a final RESULT, like winning a car, as the end goal to which one wants from beginning to end.