Game Show Problem

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Re: Game Show Problem

Postby Gofer » Thu Aug 31, 2017 4:37 pm

Robert, you are wasting your time trying give reasons for why we can deduce (as in "logically follows") the correct answer from the problem description when there isn't one, the reason being it isn't clear whether such a description refers to a happenstance or event or to actual rules of the game, and even if it were, such as the host always opening a goat-door, we couldn't rule out it applying to all goats, even those initially selected by the contestant, which would then instead imply the answer 1/2 to the problem.
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Re: Game Show Problem

Postby robert 46 » Fri Sep 01, 2017 7:43 am

Gofer wrote:
> Robert, you are wasting your time trying give reasons for why we can deduce (as in
> "logically follows") the correct answer from the problem description when there
> isn't one, the reason being it isn't clear whether such a description refers to
> a happenstance or event or to actual rules of the game,

Even if it is a one-time-occurrence, never to be repeated, we can work out a probability. We just assume the opportunity to switch was spontaneously given without bias.

> and even if it were [the rules of the game], such
> as the host always opening a goat-door, we couldn't rule out it applying to all
> goats, even those initially selected by the contestant, which would then instead
> imply the answer 1/2 to the problem.

We rule out the host revealing the player's door has a goat because it either does not offer the opportunity to switch, or winning the car is a certainty- not requiring a probability calculation [1].

We are tasked with finding an answer which is superior to all others. We use information provided and reasonable assumptions implied by the problem statement. For example:
The problem does not say the host deliberately reveals a goat, but the fact that the host knows where the prizes are implies that the host knows what is behind the door which will be opened before it is opened; so this lack of uncertainty as to what will be revealed can be interpreted as implying deliberate action.


[1] Consequent to the reduction of getting the higher value prize behind the remaining doors.
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Re: Game Show Problem

Postby Gofer » Wed Sep 06, 2017 8:38 am

Robert, I'm condensing your theory to a single statement:

Pr(HS|Pc) = x = Pr(HS|Pg), for some 0 < x < 1,

where HS is the host opening a goat-door other than the player's and offer a switch, and Pc is the player initially choosing the car-door, and Pg is the player initially choosing a goat-door,

which basically says that the host action, as far as revealing a goat and offer a switch, is independent of the player action.

But nothing in the problem description allows us to make such a deduction, which can only be made by appealing to reasonable assumptions.

The host knowing the location of the car does nothing for determining the distribution of the random variable associated with him.
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Re: Game Show Problem

Postby Gofer » Sat Nov 11, 2017 5:29 pm

Jeff has previously pointed out that one must use Bayes' law in order to solve the Game Show Problem (GSP). But error lingers around the corner in such a construction if the correct probability space associated with this view isn't properly defined.

However, there is a better way to solve the GSP, namely without Bayes', and expressing the solution as odds:

Pr(H and C) vs. Pr(H and not C)

where H is the event the host opened the door he just opened and displayed a goat, and C is the event the contestant chooses the car on the first try.

Notice how this is valid under any probability space associated with Pr. Of course, under the normal assumptions, we'd expect 'H and (not C)' to occur twice as often as 'H and C' because the host a choice between two doors in the latter but not in the former.
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Re: Game Show Problem

Postby JeffJo » Tue Nov 14, 2017 8:01 am

Gofer: Jeff has previously pointed out that one must use Bayes' law in order to solve the Game Show Problem (GSP).

This, of course, is fabrication and misrepresentation that Gofer uses to make himself appear superior. I have repeated, over and over, that "The simplest CORRECT explanation of the Game Show Problem is that, with the normal assumptions, it is twice as likely to have arrived at the current game state (one door open, revealing a goat) if the contestant picked the a goat, than if she picked the car."

Sometimes I have added that this must use conditional probability, which is where the "current game state" comes in. Since Bayes Law (Pr(A|B)=Pr(B|A)*Pr(A)/Pr(B)) is just a re-statement of the definition of conditional probability (Pr(A|B)=Pr(A&B)/Pr(B)), one can USE Bayes Law to do this. But I never said you have to.

And Gofer's solution to the more general problem, "If I were to go on this show and play this hypothetical game, what strategy should I use?" Reduces to Bayes Law the same way. But he makes use think it doesn't by not actually following through with the details, like I do. If he did, we would see that his Pr(H&C) and Pr(H&~C) are just simplified conditional probabilities, and that his general solution does not apply to the problem where we see what doors get chosen by the host and the contestant. I'm not saying they get the wrong answer, just that more details are needed to make them apply. Ignoring those details doesn't make his solution simpler, it makes it incomplete.

And finally, it doesn't explain, to the uninitiated, why the probability is not 1/2. Mine does.
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Re: Game Show Problem

Postby robert 46 » Tue Nov 14, 2017 9:16 am

JeffJo wrote:
> Gofer: Jeff has previously pointed out that one must use Bayes' law in order to solve
> the Game Show Problem (GSP).
>
> This, of course, is fabrication and misrepresentation that Gofer uses to make himself
> appear superior. I have repeated, over and over, that "The simplest CORRECT explanation
> of the Game Show Problem is that, with the normal assumptions, it is twice as likely
> to have arrived at the current game state (one door open, revealing a goat) if the
> contestant picked the a goat, than if she picked the car."

Whether or not one arrives at the game state of one door open revealing a goat is contingent on the circumstances whereby this happens. Whereas there is no indication of bias on the part of the host, one can assume that the game state happens either always or randomly- both situations preclude bias.

With this reasonable assumption, the problem can be solved analytically as follows:

If I choose a door with a goat, and the host reveals the other goat, then I must necessarily switch to the car.

If I choose the door with the car, and the host reveals either goat, then I must necessarily switch to the other goat.

Whereas I have no knowledge of bias in the placement of prizes, all doors are equally likely to have the car.

Whereas there are two goats and one car, my chances of choosing the door with the car are 1/3.

Thereby, derived from the above, my chances of switching to a goat are 1/3, and the chances of switching to the car are 2/3.

Conclusion: it is rationally beneficial to switch.

> And finally, it doesn't explain, to the uninitiated, why the probability is not 1/2.
> Mine does.

My analysis, above, also explains why the probability is not 1/2.

To apply Bayes' Law is to more or less mindlessly apply rote method, but to apply analysis is to come to a justifiable conclusion which is supported by the cogency of the analysis.
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Re: Game Show Problem

Postby Gofer » Tue Nov 14, 2017 2:16 pm

Robert > both situations preclude bias.

And this is where you go wrong. It is one thing to assume no bias; quite another to preclude it. Since the original problem says nothing about the host being biased or not, we cannot preclude either way. Necessarily being an assumption, it cannot be used in any kind of formal proof or derivation of the correct answer.

And by the way, if the host or car are biased, your answer is incorrect, but for the a-priori probabilities of an unbiased contestant - which you seem to be arguing.

----

The problem with Jeff's solution, specifically "Pr(Open=3)", is that it is not invariant under a change of probability space associated with the measure "Pr" - whereas mine is - meaning that his solution is incomplete without specifying - which Jeff didn't on page 133 - the correct said space, thus essentially providing the solution.

For example, is Pr(Open=3) to be evaluated in a space describing grandma watching game show after game show on T.V., thus implying the answer 1/3,
or something else? If we apply this probability space to my solution, grandma observes the event H&C approximately 1 in 6 and H&notC [1] 1 in 3, when the contestant chooses a particular door, implying the odds 1:2 in favor of the car being behind the other door.

[1] "notC" is here to be interpreted as the car not being behind the contestant's door, and not as the complement of some event in some sigma-algebra.
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Re: Game Show Problem

Postby JeffJo » Wed Nov 15, 2017 11:07 am

Gofer: The problem with Jeff's solution, specifically "Pr(Open=3)", is that it is not invariant under a change of probability space associated with the measure "Pr" - whereas mine is - meaning that his solution is incomplete without specifying - which Jeff didn't on page 133 - the correct said space, thus essentially providing the solution.

The point Gofer ignores, in his almost pathological need to make others believe he is right, is that on page 133 I clearly stated that I was solving a problem from a single contestant's point of view. At the time I said it, I had no idea how determined Gofer would be to misinterpret what I said, so I didn't make it so verbose as to be impossible to misinterpret - besides, Gofer has since demonstrated that he will ignore such attempts because they use too many words. And to ignore any attempt to re-state the intent.

The problem I established as the one I was addressing, is one where the contestant chose door #1, and the host opened door #3. While these numbers are examples, the fact that the problem is always presented **AFTER** these decisions are made is what makes Gofer's invariant solution incorrect, and necessitates the use of conditional probability. His solution is correct only **BEFORE** they are made. And that is never the problem that is presented.

Further, parroting a solution that gives a different answer, even if it is correct, does not explain what is wrong with the solution "There are two closed doors, so each had a 1/2 chance." If you can convince the person who says that, that your solution (A) addresses the same problem (as explained above, Gofer does not), and (B) is correct (Gofer won't convince anybody with his incorrect and confusing symbology), then all you have done is establish a paradox.

PRESENTING A SOLUTION THAT SEEMS TO BE CIRRECT DOES NOT SHOW WHAT IS WRONG WITH ANOTHER SOLUTION, THAT GIVES A DIFFERENT ANSWER, BUT ALSO SEEMS TO BE CORRECT. Do I need to reprat this with more emphasis?

My simple explanation can convince them of what they did wrong, since it points it directly, and shows how correcting their error leads to the correct answer.
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Re: Game Show Problem

Postby robert 46 » Wed Nov 15, 2017 6:32 pm

Gofer wrote:
> Robert > both situations preclude bias.
>
> And this is where you go wrong. It is one thing to assume no bias; quite another
> to preclude it.

It appears you are resorting to equivocation. If the host always or randomly gives the opportunity to switch it does in fact preclude circumstantial bias. The point is that we are tasked with solving the problem based on the information supplied by the problem statement- we are not attempting to divine some actual game show occurrence.

> Since the original problem says nothing about the host being biased
> or not, we cannot preclude either way.

You appear to be confusing an actual occurrence with a hypothetical. The problem statement begins "Suppose you are on a game show..."

> Necessarily being an assumption, it cannot
> be used in any kind of formal proof or derivation of the correct answer.

The "assumption" is implied by the problem statement. An actual indication of bias, and what kind of bias it is, is required to factor bias into the analysis. The only bias we are sure of is that the host reveals a goat, which gives the player the opportunity to switch. JeffJo calls this "a rule of the game".

> And by the way, if the host or car are biased, your answer is incorrect,

Clearly, you are treating the problem as an actual instance of the game show rather than as a hypothetical.

> but for the a-priori probabilities of an unbiased contestant

The "contestant" cannot be biased because it makes no difference which door is chosen. There is double randomization where only one randomization is required (car is placed randomly, player chooses randomly)

>- which you seem to be arguing.

You show little comprehension of what I am arguing.
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Re: Game Show Problem

Postby Gofer » Thu Nov 16, 2017 12:20 pm

> The point Gofer ignores, in his almost pathological need to make others believe he is right,

Ad hominem and obvious lie (AHOL)

> is that on page 133 I clearly stated that I was solving a problem from a single contestant's point of view.

Statement without supporting evidence.

>At the time I said it, I had no idea how determined Gofer would be to misinterpret what I said,

AHOL

> so I didn't make it so verbose as to be impossible to misinterpret - besides,

Here Jeff seems to be admitting that his solution isn't complete.

>Gofer has since demonstrated that he will ignore such attempts because they use too many words. And to ignore any attempt to re-state the intent.

AHOL

>The problem I established as the one I was addressing, is one where the contestant chose door #1, and the host opened door #3. While these numbers are examples, the fact that the problem is always presented **AFTER** these decisions are made is what makes Gofer's invariant solution
incorrect, and necessitates the use of conditional probability. His solution is correct only **BEFORE** they are made. And that is never the problem that is presented.

Another lie by Jeff. My solution takes into account the actions of the host, the player and the car, thus making it complete under any interpretation, hence the event H&C. Your have not demonstrated the opposite.

>Further, parroting a solution that gives a different answer, even if it is correct, does not explain what is wrong with the solution "There are two closed doors, so each had a 1/2 chance." If you can convince the person who says that, that your solution (A) addresses the same problem (as explained above, Gofer does not), and (B) is correct (Gofer won't convince anybody with his incorrect and confusing symbology), then all you have done is establish a paradox.

Incomprehensible gibberish.

>PRESENTING A SOLUTION THAT SEEMS TO BE CIRRECT DOES NOT SHOW WHAT IS WRONG WITH ANOTHER SOLUTION, THAT GIVES A DIFFERENT ANSWER, BUT ALSO SEEMS TO BE CORRECT. Do I need to reprat this with more emphasis?

Here Jeff seems to be arguing that I argue his solution wrong - although this being not the case - just incomplete.

> My simple explanation can convince them of what they did wrong, since it points it directly, and shows how correcting their error leads to the correct answer.

Your explanation, yes, but hardly "simple".

-----

My solution basically says:

How often does the host open the door he opened with the contestant choosing the car to when he doesn't. With the contestant choosing a particular door, we'd expect the former to occur, in a fair game, roughly 1/6 of the time, and the latter 1/3 of the time, making it twice as likely the car being behind the other door.
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Re: Game Show Problem

Postby Gofer » Thu Nov 16, 2017 12:37 pm

>>> Robert > both situations preclude bias.
>
>> And this is where you go wrong. It is one thing to assume no bias; quite another
>> to preclude it.

>It appears you are resorting to equivocation.

Nonsense! My writing here is clear.

>If the host always or randomly gives the opportunity to switch it does in fact preclude circumstantial bias.

The "always" is not part of the problem description, making your statement here a red herring.

>The point is that we are tasked with solving the problem based on the information supplied by the problem statement- we are not attempting to divine some actual game show occurrence.

Correct, and which you haven't done!

>> Since the original problem says nothing about the host being biased
>> or not, we cannot preclude either way.

>You appear to be confusing an actual occurrence with a hypothetical. The problem statement >begins "Suppose you are on a game show..."

You appear to be confusing that I don't.

>> Necessarily being an assumption, it cannot
>> be used in any kind of formal proof or derivation of the correct answer.

>The "assumption" is implied by the problem statement.

Nonsense! Nowhere in the problem statement does it say that an assumption is implied.

> An actual indication of bias, and what kind of bias it is, is required to factor bias into the analysis.

So is any lack of bias. Or are you biased in your bias?

>The only bias we are sure of is that the host reveals a goat,
which gives the player the opportunity to switch. JeffJo calls this "a rule of the game".

This is error! Nowhere in the description does it say this being a "rule of the game".

>> And by the way, if the host or car are biased, your answer is incorrect,

>Clearly, you are treating the problem as an actual instance of the game show rather than as a hypothetical.

Clearly I'm not, since it is entirely possible for a hypothetical game to include bias.

> You show little comprehension of what I am arguing.

I know exactly what you're arguing. You're trying to deduce, as in "logically follows", the answer from the problem description, where no such logical deduction can be made.
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Re: Game Show Problem

Postby JeffJo » Thu Nov 16, 2017 4:49 pm

The Game Show Problem seems to be two problems in one. The first problem - the intended one - is how to make the best choice in a game of chance. The majority of people who try to solve it see only this problem, and they see it as a choice between two equivalent options. So their answer is that each option has a 50% chance.

The problem has two predessors where the intended problem is identical: Martin Gardner's Three Prisoners and Steve Selvin's Monty Hall Problem. But each of those made the rules of the game more explicit than the one that was asked of Marilyn vos Savant. There is also a more modern version, the Three Pancakes Problem, that is also identical. The issue in these versions is also that people think there are two equivalent options, making the answer 50%.

The second problem in the GSP, is determining what the first problem is. This is really only an issue for those who like to appear superior to others. This is demontrated by the fact that 99% of the time it is discussed, the issue people have is the same issue they have with those other three versions where the problem is not ambiguous. Yet the wanna-be-superiors will endlessly debate how and why the details of the game should be resolved, often making up (and not defining) terms like "happenstance,", "only examples," "one time occurrence," or "the problem is *about* XXX because it mentions XXX"[1], that have absolutely no significance. None of these terms are used in the mathematics of probability, and raising them serves no purpose other than demonstrating that the originator thinks he is superior.

The intended problem, as all but these wanna-be-superiors understand it, is that the host is required to open a door that does not have the car, and is not the one that the contestant chose. This is understood, and reasonable, because it is perceived as typical behavior on an American game show. No other behavior is reasonable, and few allow a definitive conclusion to be reached.

Once the first problem is defined this way (and also in those other three problems), the error in the 50% solution can only be explained by pointing out why the two options are not equivalent.[2] The reason is that one of the two options could have been different, but not the other.

In the GSP, if the choice actually is "my door or door X," it could also have been "my door or door Y." But never "door X or door Y." Since "door X" could have been different, but "my door" could not, the assumption that the options are equivalent is wrong. This difference is measured, probabilisticly speaking, by the probability that the other set of options would have occured. This is ostensibly split 50%:50%, and despite what others claim, I have never said otherwise. Only that a split is required for a correct solution, and that different splits give different answers (which demonstrates its significance).

+++++

[1] This is used in the logically, but not numerically, equivalent Two Child Problem, where the wanna-be's insist a problem that mentions boys must be "about boys" and could not have included a similar option about girls.

[2] Yes, there are other ways to arrive at the correct answer, but they do not explain what is wrong with the 50% solution. So all they do is create a paradox.
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Re: Game Show Problem

Postby robert 46 » Thu Nov 16, 2017 7:19 pm

Gofer wrote:
> >>> Robert > both situations preclude bias.
> >
> >> And this is where you go wrong. It is one thing to assume no bias; quite another
> >> to preclude it.
>
> >It appears you are resorting to equivocation.
>
> Nonsense! My writing here is clear.

I invite interested readers to peruse all of Gofer's posts to discover that his writing style is characteristically squirrely.

> >If the host always or randomly gives the opportunity to switch it does in fact preclude
> circumstantial bias.
>
> The "always" is not part of the problem description, making your statement here a
> red herring.

I have given an analysis which identifies the conditions where circumstantial bias is precluded. The problem statement only identifies a one-time-occurrence. This could be consequent to the host always giving the opportunity to switch (a rule-of-the-game) or randomly from time-to-time. What the problem statement does NOT DO is imply that giving the opportunity to switch is not a rule-of-the-game, or is consequent to some favoritism.

> >The point is that we are tasked with solving the problem based on the information
> supplied by the problem statement- we are not attempting to divine some actual game
> show occurrence.
>
> Correct, and which you haven't done!

This kind of blanket denial is merely a ploy.

> >> Since the original problem says nothing about the host being biased
> >> or not, we cannot preclude either way.
>
> >You appear to be confusing an actual occurrence with a hypothetical. The problem
> statement >begins "Suppose you are on a game show..."
>
> You appear to be confusing that I don't.

A typical squirrely statement.

> >> Necessarily being an assumption, it cannot
> >> be used in any kind of formal proof or derivation of the correct answer.
>
> >The "assumption" is implied by the problem statement.
>
> Nonsense! Nowhere in the problem statement does it say that an assumption is implied.

If a problem statement says something indicative then it is a condition, it cannot say something which is an assumption: assumptions follow from what is not explicitly said, but is reasonably implied.

> > An actual indication of bias, and what kind of bias it is, is required to factor
> bias into the analysis.
>
> So is any lack of bias. Or are you biased in your bias?

The necessary default is that there is no bias unless it is somehow indicated and identifiable.

> >The only bias we are sure of is that the host reveals a goat,
> which gives the player the opportunity to switch. JeffJo calls this "a rule of the
> game".
>
> This is error! Nowhere in the description does it say this being a "rule of the game".

We are sure the host reveals a goat just because we are told the host reveals a goat in the one-time-occurrence. This would qualify as a rule-of-the-game where the game is "give the player the opportunity to switch possibly to the car". The latter is described by the one-time-occurrence.

> >> And by the way, if the host or car are biased, your answer is incorrect,
>
> >Clearly, you are treating the problem as an actual instance of the game show rather
> than as a hypothetical.
>
> Clearly I'm not, since it is entirely possible for a hypothetical game to include
> bias.

The hypothetical must follow from the problem statement. Otherwise it is merely extraneous speculation.

> > You show little comprehension of what I am arguing.
>
> I know exactly what you're arguing. You're trying to deduce, as in "logically follows",
> the answer from the problem description, where no such logical deduction can be
> made.

Gofer believes that all uncertainty must be resolved by a problem statement in order to make a logical deduction. The task otherwise is to make a reasonable analytical deduction. The Game Show Problem is not a logic problem, but an analytical problem. It would be nice if Gofer was not so confused about what kind of problem it is we are dealing with.
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Re: Game Show Problem

Postby Edward Marcus » Thu Nov 16, 2017 7:51 pm

I interpret GSP rules as providing the opportunity to switch the outcome of the original choice.

P(Car, switch offered) = P(Goat, no switch offered)=2/3

Further reading of the GSP's intent is "weaving the thread of wisdom too finely" -Ben Franklin
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Re: Game Show Problem

Postby robert 46 » Sat Nov 18, 2017 8:13 am

JeffJo wrote:
> The Game Show Problem seems to be two problems in one. The first problem - the intended
> one - is how to make the best choice in a game of chance.

Whereas the player has no reason believe that any door is more likely to have the car than any other door, the chances of choosing the door with the car is 1/n where n is the number of doors. In the problem we are given n=3. Therefore it is impossible for the player to make a "best choice" among the doors. The second part of the problem is whether it is preferable to switch one's choice of door given the circumstance that a goat is revealed behind a door not the players. This is a matter of analysis: what does the circumstance do to the likelihood of winning the car?

> The majority of people
> who try to solve it see only this problem, and they see it as a choice between two
> equivalent options. So their answer is that each option has a 50% chance.

Perhaps this only means that the majority of people are general-purpose dolts characterized by a lack of analytical ability consequent to insufficient programming. My father told me that it wasn't until he got to graduate school that any of his teachers actually tried to teach him how to think. All his prior schooling was rote learning of facts. The reason for this obviously is that the vast majority of lower-school teachers are as lacking in thinking ability as the general population, and thus are incapable of teaching anyone how to think. (An equally ignorant school board may impose a curriculum which stifles teachers with any ability to think from imparting such understanding to one's students.)

> The problem has two predecessors where the intended problem is identical: Martin Gardner's
> Three Prisoners and Steve Selvin's Monty Hall Problem. But each of those made the
> rules of the game more explicit than the one that was asked of Marilyn vos Savant.

The Three Prisoners Problem is subtly different from the Game Show Problem.

> There is also a more modern version, the Three Pancakes Problem, that is also identical.

Which has been discussed here, here:
viewtopic.php?f=1&t=2014

> The issue in these versions is also that people think there are two equivalent options,
> making the answer 50%.

There are two disparate meanings of "think": to get a notion which pop's up in one's mind; to analyze in a purposeful way.

> The second problem in the GSP, is determining what the first problem is.

It is primarily an analytical problem; in minor degree a probability problem; and not a formal logic problem.

> This is
> really only an issue for those who like to appear superior to others. This is demontrated
> by the fact that 99% of the time it is discussed, the issue people have is the same
> issue they have with those other three versions where the problem is not ambiguous.
> Yet the wanna-be-superiors will endlessly debate how and why the details of the
> game should be resolved, often making up (and not defining) terms like "happenstance,",
> "only examples," "one time occurrence," or "the problem is *about* XXX because it
> mentions XXX"[1], that have absolutely no significance.

JeffJo's epithet "wanna-be-superiors" can be nicely contrasted with "intellectual-zombie-inferiors".

> None of these terms are
> used in the mathematics of probability, and raising them serves no purpose other
> than demonstrating that the originator thinks he is superior.

Old-stick-in-the-muds characterized by only believing what they have been taught, and subject to the not-invented-here bias, are the self-appointed "superiors".

> The intended problem, as all but these wanna-be-superiors understand it, is that
> the host is required to open a door that does not have the car, and is not the one
> that the contestant chose. This is understood, and reasonable, because it is perceived
> as typical behavior on an American game show. No other behavior is reasonable, and
> few allow a definitive conclusion to be reached.

Not because it is typical behavior: after-all, you state that Monty Hall claimed never to give an opportunity to switch in the Let's Make a Deal game show in the manner described by the problem statement. It makes no difference if it is typical or atypical behavior: what matters is merely that it is the problem we are given, and which needs to be analyzed in its own context.

> Once the first problem is defined this way (and also in those other three problems),
> the error in the 50% solution can only be explained by pointing out why the two
> options are not equivalent.[2] The reason is that one of the two options could have
> been different, but not the other.

I have explained the matter succinctly, which the three follow-on posters have mostly ignored; so I will repeat it:

robert 46 wrote:

Whether or not one arrives at the game state of one door open revealing a goat is contingent on the circumstances whereby this happens. Whereas there is no indication of bias on the part of the host, one can assume that the game state happens either always or randomly- both situations preclude bias.

With this reasonable assumption, the problem can be solved analytically as follows:

If I choose a door with a goat, and the host reveals the other goat, then I must necessarily switch to the car.

If I choose the door with the car, and the host reveals either goat, then I must necessarily switch to the other goat.

Whereas I have no knowledge of bias in the placement of prizes, all doors are equally likely to have the car.

Whereas there are two goats and one car, my chances of choosing the door with the car are 1/3.

Thereby, derived from the above, my chances of switching to a goat are 1/3, and the chances of switching to the car are 2/3.

Conclusion: it is rationally beneficial to switch.

End quote.

> In the GSP, if the choice actually is "my door or door X," it could also have been
> "my door or door Y." But never "door X or door Y." Since "door X" could have been
> different, but "my door" could not, the assumption that the options are equivalent
> is wrong.

What JeffJo is trying to say here (unclearly) is that the remaining door might have been different. Who cares? All that is relevant is that there is a goat behind whichever door the host does open, and that he does not open the player's door. The remaining door retains the same significance: it is as likely to have the car as the player's door is likely to have a goat.

> This difference is measured, probabilisticly speaking, by the probability
> that the other set of options would have occured. This is ostensibly split 50%:50%,
> and despite what others claim, I have never said otherwise. Only that a split is
> required for a correct solution, and that different splits give different answers
> (which demonstrates its significance).

This is muddy and irrelevant.

> +++++
>
> [1] This is used in the logically, but not numerically, equivalent Two Child Problem,
> where the wanna-be's insist a problem that mentions boys must be "about boys" and
> could not have included a similar option about girls.

Of course it might have been about girls, but because the probability of the birth of a boy or girl are considered equal as a simplification the answer is the same: the woman would have a 1/3 chance of two girls when the problem is about girls, and a 1/3 chance of two boys when the problem is about boys. What the problem is NOT about is the probability of reporting ALOB or ALOG; nor the probability of the problem being about boys or about girls.

> [2] Yes, there are other ways to arrive at the correct answer, but they do not explain
> what is wrong with the 50% solution. So all they do is create a paradox.

Not my analysis quoted above.
robert 46
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