Game Show Problem

The Game Show problem, logical fallacies, Marilyn's daily diet, multi-tasking and achieving your potential.

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Re: Game Show Problem

Postby JeffJo » Thu May 11, 2017 6:46 am

Gofer stolidly refuses to look up the definition of an event. Try here: en.wikipedia.org/wiki/Event_(probability_theory). But no, Gofer thinks that the definition is "anything that can be put inside the parentheses of the Pr() function. Maybe he should look here: en.wikipedia.org/wiki/Probability_space#Conditional_probability.

In the post that Gofer finally referenced after almost a year of arguing, I clearly defined a problem. Part of that clear, unambiguous problem definition was "Say you play 300 games, and pick Door #1 in all of them." From that problem definition, a probability space can be easily inferred, if you are such a pedantic **** that you feel the need to have one. And anybody but that pedantic **** would easily see that the door chosen by the contestant is not subject to randomness, so "Pr(Open=3|P=1)" would have been incorrect.
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Re: Game Show Problem

Postby Gofer » Sat May 13, 2017 3:52 pm

Jeff apparently has the short-term memory of a goldfish. Just a couple of posts back, I clearly explained events. They are the points of the sigma algebra.

> Part of that clear, unambiguous problem definition was "Say you play 300 games, and pick Door #1 in all of them."

Jeff conveniently forgets to inform the readers that that quote wasn't part of his formal derivation on page 133, but a separate section of the same posting, delimited by "++++++".

Here's what really happened. Jeff probably had the experiment, and thereby the implied probability space, in the back of his mind, but forgot to include it in the derivation, hence his statement "we know", on page 133, and which references prior knowledge or derivation, hence his usage of "=" rather than ":=".

> so "Pr(Open=3|P=1)" would have been incorrect.

Not if Pr is associated with a probability space describing an experiment where all the three actions of the host, player and the car are taken into consideration.
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Re: Game Show Problem

Postby JeffJo » Sun May 14, 2017 6:11 am

Gofer apparently has the short-term memory of a goldfish. Just a couple of posts back, I clearly explained that, while the elements of the sigma algebra specific to the probability space, that relationship defines neither a sigma algebra, nor an event.

But he thinks "A|B" is denotes an event, he is free to show us the sigma algebra it belongs to.


> Part of that clear, unambiguous problem definition was "Say you play 300 games, and
> pick Door #1 in all of them."

What Gofer conveniently forgets to inform the readers, is that the formal derivation of a solution has to address the problem it is solving, not what Gofer whats it to have been to he can claim there was a mistake where there was none. So Gofer is free to show us what part of the problem that was addresses requires a random variable addressing the contestant's choice.

>> so "Pr(Open=3|P=1)" would have been incorrect.
>
> Not if Pr is associated with a probability space describing an experiment
> where all the three actions of the host, player and the car are taken into
> consideration.

Gofer's favorite method of lying is to preface his assertions with "if XXX," and then not justifying why we should accept XXX. Or, as in this case, deliberately avoiding why we can't.

Here's what really happened. Gofer doesn't like to be called wrong, so he made up an error, and waited a year to explain what he meant.
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Re: Game Show Problem

Postby IQprofessor » Mon May 22, 2017 8:23 am

Actually almost professors are fool.
Because they are just good at memorizing.
But unfortunately they are good at IQ test.

So we must divide IQ into 3 kinds of IQs like computer.(CPU IQ,RAM IQ, HDD IQ)
I think CPU IQ is important to solve this problem. I could solve this problem easily evenif my IQ is 95.

For example
Me(normal IQ 95)
CPU IQ 126(I can solve the problem)
RAM IQ 70
HDD IQ 90

Normal professor(normal IQ 130)
CPU IQ 110(they can not solve the problem)
RAM IQ 120
HDD IQ 160
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Re: Game Show Problem

Postby robert 46 » Fri Jul 21, 2017 8:07 am

At Parade:
"Marilyn's Intuition
July 21, 2017 – 5:00 AM
By Marilyn vos Savant

Just wondering: When you first read the game show problem, did you intuitively understand that switching would have a 2/3 chance of winning? Or did you initially assume that the chance of winning was 1/2, regardless of whether you switched?

—Ivan in Hong Kong

Marilyn responds:

Actually, neither! I have a long-standing habit of approaching probability problems (and many other kinds of problems, for that matter) with no assumptions."


Other than Marilyn's characteristic hyperbolic use of the exclamation point, this is interesting in that the only way to answer the Game Show problem conclusively is to make assumptions. The main assumptions required are:
1. The host either always or randomly gives the opportunity to switch: i.e there is no situational bias in allowing the player the option to switch.
2. The host deliberately chooses a door with a goat to open.
3. The host randomly chooses which door with a goat to open. [1]

The justifications for these assumptions are:
1. The one-time-occurrence described has no indication of bias in giving this (or an other) opportunity to switch.
2. The host knows where the prizes are, so the host has the capability to deliberately open a door with a goat.
3. The one-time-occurrence described has no indication of bias in the host's choice of which goat-door to open.

With these assumptions, one can derive the conclusive answer that there is a 2/3 probability of winning the car by switching, so it is advantageous to switch. Without them, there is no definite answer.


[1] It is necessary to recognize that the door #s given are examples. This is conveyed by the word "say".
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Re: Game Show Problem

Postby Gofer » Fri Jul 21, 2017 4:14 pm

>> With these assumptions, one can derive the conclusive answer that there is a 2/3 probability of winning the car by switching,

No, one cannot, because the host knowing the location of the car doesn't necessarily change the probabilities.

>> Without them, there is no definite answer.

Yes, there is, namely a formula having the necessary probabilities.

The formula is

Pr(O&X) vs Pr(O&S) in favor of switching, where O, X and S are events describing, respectively, the host opening the door he just opened and offer a switch; the car being behind the switch-door; and the car being behind the player's initial door.

to add: under the normal assumptions of the host always opening a goat-door, and always choosing uniformly random if given a choice, and the car being placed uniformly random, and accounting for the player's initial choice, we'd expect the event O&X to occur 1 time in 3, S to occur 1 time in 3, and O&S to occur 1 time in 6 (because of the host having a choice between two doors); the odds come out as 2 to 1 in favor of switching.
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Re: Game Show Problem

Postby JeffJo » Sat Jul 29, 2017 1:41 pm

robert 46> Other than Marilyn's characteristic hyperbolic use of the exclamation point, this is
> interesting in that the only way to answer the Game Show problem conclusively is to make
> assumptions.

And other than robert’s typical misinterpretation of the question and the response, there is nothing new in his reply. Or particularly interesting, except that he is slowly morphing his rhetoric into what I have been saying for years, and he keeps calling wrong even though he agrees with it.

The question in this particular column was about whether Marilyn **ASSUMES AN ANSWER**, and then tries to find logic to support it. Robert should be familiar with this, since it is what he usually does. And while I disagree about whether Marilyn does make such assumptions, she has admitted to some of the assumptions robert talks abut that are needed to arrive at the solution:

Marilyn> [In the original answer, ] we published no significant reason to view the host as anything
> more than an agent of chance who always opens a losing door and offers the contestant the
> opportunity to switch … Nearly all of my critics understood the intended scenario, and few
> raised questions of ambiguity.

robert 46> The main assumptions required are:
> 1. The host either always or randomly gives the opportunity to switch: i.e there is no situational
> bias in allowing the player the option to switch.

Whether or not robert thinks it gives the same answer, there is no justification for assuming the offer is optional.

> 3. The host randomly chooses which door with a goat to open. [1]

#3 is the assumption that drives the correct solution, and that few realize is important. It is what I’ve tried to get robert to recognize for years, and he keeps arguing with it while (previously) trying to avoid stating it.

> The justifications for these assumptions are:
> 1. The one-time-occurrence described has no indication of bias in giving this
> (or an other) opportunity to switch.

There is no indication that he will not offer a switch, and that isn’t the point of the problem. So that part is just more rhetoric. Robert refuses to define how we can tell what is, or is not, a “one-time occurrence” in a problem like this. In the entire history of probability solutions, he is the only one who has used this term as he does, and he is the only one who knows what he thinks it means. Or why he can use it differently, based on how he wants the solution to apply. Here, it means it always happens. But...

> 3. The one-time-occurrence described has no indication of bias in the host's choice of which
> goat-door to open.

... here, it means it is a random selection among the set of possibilities.

Probably every detail in every probability problem that Marilyn has ever published is described as something that happens once. Yet robert chooses how to handle them, based on the result he wants to get. Here, robert uses it to justify, without cause or explanation, why he dismisses any representation of the (non-)bias he finally described. He refuses to grasp, because he would have to admit he was wrong in other problems where he cites a “one-time occurrence,” that NOBODY ARGUES WITH THIS CONCLUSION.

The point of representing the (non-)bias in the solution, even if it is a 50:50 choice, is this: some solutions, like his, are logically incorrect even though they get the right answer. Such solutions typically say that the contestant’s original 1/3 chance can’t be affected by the revelation of a goat, because a goat can always be revealed.

The fact that robert refuses to address, is that the chances are affected by the (non-)bias, but in a value-neutral way.

> With these assumptions, one can derive the conclusive answer that there is a 2/3 probability
> of winning the car by switching, so it is advantageous to switch. Without them, there is no
> definite answer.
>
> [1]It is necessary to recognize that the door #s given are examples. This is conveyed by the word "say".

More empty rhetoric. With those assumptions, it is completely unnecessary to recognize that the numbers are just examples. The answer is the same whether they are not mentioned at all, whether they are considered to be examples (which is logically the same thing), or we are only supposed to consider the 1/6 of all games where these are the doors involved. This insistence on “say means examples” is just robert’s way to avoid using his same logic in another problem, where he gets the wrong answer by not using it.

But …
Gofer> >With these assumptions, one can derive the conclusive answer that there is a 2/3
> > probability of winning the car by switching,
>
> No, one cannot, …

Yes, one can.

> … because the host knowing the location of the car doesn't necessarily change the probabilities.

And "host knows the location" was in the problem statement, so it isn't even an assumption. This argument is not only a complete non-sequitur, it doesn;t seem to have a rational point.

Probabilities don’t “change.” The expression Pr(Car=1)=1/3 is always true under the assumptions, because it represents only what is called a prior probability. Prior probabilities are updated, not changed. That means that we compute a different kind of probability, called a posterior or conditional probability and written Pr(Event1|Event2).

You can tell that this is a different kind of probability my the notation, which you don't seem to understand. In a prior probability, the argument is a single event. You can name them, as you tried to do, but you can also describe them by values of random variables. In a posterior probability, two events are used; the could have been separated by a comma, but the difference in the purpose of the two events is emphasized by the "|" symbol.

In other words, in Pr(Event1|Event2), “Event1|Event2” is not an event as you have claimed.

In this problem, and using the explicit door numbers since IT DOESN’T MATTER WHAT YOU CALL THEM AS LONG AS YOU REMAIN CONSISTENT (that’s for robert), the prior probability Pr(Car=1)=1/3 is updated to the posterior probability:

Pr(Car=1|Open=3)
= Pr(Open=3|Car=1)*Pr(Car=1) / Pr(Open=3)
= Pr(Open=3|Car=1)*Pr(Car=1) / [Pr(Open=3|Car=1)*Pr(Car=1) + Pr(Open=3|Car=2)*Pr(Car=2)]
= [B*(1/3)] / [B*(1/3) + (1)*(1/3)]
= B/(B+1)
= 1/3 if B=1/2.

Here, B is the (non-)bias robert finally recognized. As you can see by this formula (and unlike what you said in your reply, it actually is a formula), the prior probability Pr(Car=1)=1/3 is just one term used to derive Pr(Car=1|Open=3) which has the same value, but could have a different one under different assumptions.

> > Without them, there is no definite answer.
>
> Yes, there is, namely a formula having the necessary probabilities.

An actual formula that contains unknowns is not a definite answer, so without the assumptions that allow use to assign values, there is no definitive answer. However, …

> The formula is
>
> Pr(O&X) vs Pr(O&S) in favor of switching,

That is not a formula.

What it seems you meant, but can’t seem to express in an understandable manner, is that “vs” means relative odds. As in “the odds are 1 to 5 that you will rolls a 6 on a six-sided die.” So you should have said “to” instead of “vs,” or even used a colon. That still isn’t a formula, but at least it has understandable meaning in the context.

Interpreting your poorly-worded definitions of the events, and using Marilyn’s examples, Pr(O&X) is the prior probability that, in a random game, the host would open Door #3 and the car would be behind Door #2. That’s 1/6, because you didn’t include the condition that you once assumed me of leaving out (even though I had included it in the problem set up).

Pr(O&S) is the probability that, in a random game, the host would open Door #3 and the car would be behind Door #1. That’s also 1/6, so your non-formula is wrong.
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Re: Game Show Problem

Postby Gofer » Sat Jul 29, 2017 4:08 pm

Oh, boy, Jeff surely managed to dish out a spunky stew of misrepresentation, non-sequiturs, outright lies and errors. Let's have a look at them; and remember, if I don't address something Jeff wrote, it means it is irrelevant.

> And "host knows the location" was in the problem statement, so it isn't even an assumption. This argument is not only a complete non-sequitur, it doesn;t seem to have a rational point.

The point is obvious. We cannot infer the host always opening a goat-door, from the host knowing the location of the car, and therefore cannot INFER the correct answer. But we can make reasonable assumptions, hence my doing exactly that.

> An actual formula that contains unknowns is not a definite answer, so without the assumptions that allow use to assign values, there is no definitive answer. However, …

Of course it can be seen as a definite answer. This is just another of Jeff's ploys of trying to sound authoritative.

> > Pr(O&X) vs Pr(O&S) in favor of switching,

> That is not a formula.

Of course it's a formula, a formula being defined as "a general fact, rule, or principle expressed in usually mathematical symbols" (merriam-webster).

Jeff pats himself on the back how I should have used ":" or "to" instead of "vs" to express odds. While that may be true, it doesn't change what I wrote was correct. Obviously yet another sounding-authoritative-and-clever ploy.

> Pr(O&X) is the prior probability that, in a random game, the host would open Door #3 and the car would be behind Door #2.

No, O is the event the host opened the door he just opened.

>That’s 1/6, because you didn’t include the condition that you once assumed me of leaving out (even though I had included it in the problem set up).

No, it's 1 in 3, just like I wrote, because I clearly stated "accounting for the player's initial choice", which is "the condition [... left out]". And this "accounting" is exactly what "chooses" the correct probability space, making any use of "Pr" refer to this.

> Pr(O&S) is the probability that, in a random game, the host would open Door #3 and the car would be behind Door #1. That’s also 1/6, so your non-formula is wrong.

And Jeff obviously can't read. I explicitly wrote that O&S occurs 1 in 6.
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Re: Game Show Problem

Postby JeffJo » Fri Aug 04, 2017 8:53 am

Gofer> Oh, boy, Jeff surely managed to dish out a spunky stew of misrepresentation,
> non-sequiturs, outright lies and errors.

Oh, boy, Gofer sure thinks his faulty impressions from superficial reading must be correct, since he is so sure of this completely wrong accusation.

>>>> With these assumptions, one can derive the conclusive answer …
>>> No, one cannot, because the host knowing the location of the car doesn't necessarily
>>> change the probabilities.
>> And "host knows the location" was in the problem statement, so it isn't even an
>> assumption. This argument is not only a complete non-sequitur, it doesn’t seem
>> to have a rational point.
> The point is obvious. We cannot infer the host always opening a goat-door, from the
> host knowing the location of the car, …

We didn’t infer it from the host knowing the location of the car, we inferred it from "these assumptions" that were mentioned. Which include (1) The host always offers a switch after opening another door that (2) does not have a goat.

> … and therefore cannot INFER the correct answer. But we can make reasonable
> assumptions ...

“We” did. “Gofer” refuses to recognize them when other people make them.

>> An actual formula that contains unknowns is not a definite answer …
> Of course it can be seen as a definite answer.

If I say the answer is (14X+Y)/cos(Z), how is that definitive?

>> That is not a formula.
> Of course it's a formula, a formula being defined as "a general fact, rule, or principle
> expressed in usually mathematical symbols" (merriam-webster).

Including one symbol Gofer didn’t define, so it was not a formula. But you can think of a formula being the equivalent of an English statement, using these symbols instead of words. Gofer's non-formula lacked the equivalent of a verb, so it wasn't an expression of any fact. It was a collection of symbols that he used incorrectly.

And even if we take what seemed to be his intention, that "vs." meant to define odds, he called it a probability. So it was a non-formula for something other than the probability he said it was a formula for.

>> Pr(O&X) is the prior probability that, in a random game, the host would open Door #3 and
>> the car would be behind Door #2.
> No, O is the event the host opened the door he just opened. [sic]

In my example, that is door #3. Please note that it was possible for the host to open Door #2 as well.

And as Gofer defined X, it is the event [where] the car is behind what Gofer called the “switch door.” In my example, that is #2. So “O&X” is the event where the car is behind #2, and the host opens #3. Like I said.

Finally, from what Gofer skipped in whatever probability class he claims to have taken, Pr(EVENT) is the notation for the prior probability that the outcome is in the set EVENT, whether you use its value before, or after, you know that what happened is in what Gofer called O.

> > That’s 1/6,…
> No, it's 1 in 3, …

If Gofer would only you set up a truth table for ***ALL*** the prior probabilities in his system, he could see that there is a 1/3 chance that the car is behind #2. Then, there are equal chances that the host will open either #1 (if the contestant chose #3) or #3 (if the contestant chose #3). That makes the prior probability Pr(O&X) equal to 1/6.

Of course, I know that isn't what Gofer meant. But he needs to use proper terminology.

> …just like I wrote, because I clearly stated "accounting for the player's initial choice",
> which is "the condition [... left out]".

But a condition does not affect the prior probabilities. It affects the overly-complicated way Gofer defined his events, which I took into account by simplifying them.

But let’s assume Gofer is just ignorant about actual notation, and failed to read the lesson I provided that explained notation. That is, that Gofer really meant to define the event C to be where the contestant choose the door she did, and the meant to write “a formula for the odds is Pr(O&X|C) to Pr(O&S|C).”

This is the condition he accused me of leaving out. And spent a year arguing about the results before explaining how that was the basis of his complete misunderstanding.

Note that, given C (and X and S defined from it - that is, "taking it into account"), then we need Pr(O&X|C) + Pr(O&S|C) = Pr(O|C). Which is 1, by the assumptions. So they can't be 1/3 and 1/6, as Gofer claims. Note only that, they represent the actual answers we want to get from a formula, so claiming they are the terms that go into such a formula is circular logic.

>> Pr(O&S) is the probability that, in a random game, the host would open Door #3 and the car would be behind Door #1. That’s also 1/6, so your non-formula is wrong.
> And Jeff obviously can't read. I explicitly wrote that O&S occurs 1 in 6.

No, Gofer obviously can’t tread, since what I was saying is that the “formula” he said evaluated to “1/3 vs. 1/6” was actually “1/6 vs. 1/6”. Since Gofer wants it to be “2Q to Q,” for some Q, it was wrong.
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Re: Game Show Problem

Postby JeffJo » Fri Aug 04, 2017 9:22 am

The MHP is actually pretty trivial to model, but you do have to be careful to include everything.

1) The problem statement doesn't say the doors are numbered. Even if it did, we can use whatever labels we want to, as long as we clearly distinguish the doors. So, say that the game identifies them by the names Peter, Paul, and Mary. But before deciding whether to switch, we run on stage and write numbers over the labels with these names. We but a "1" on the door we chose, a "3" on the door the host opened, and a "2" on the remaining door.

2) Being careful means that everything that was originally possible has to be considered, even if it is contradicted by what happens later. So we can represent the three possible locations of the car by the events C=1, C=2, and C=3. We can represent the door the host opened by the events H=2 and H=3 (note: we assume he can't open #1).

3) Basic probability theory now says:

Pr(C=2|H=3) = Pr(H=3|C=2) / [Pr(H=3|C=1) + Pr(H=3|C=2) + Pr(H=3|C=3)]

(Note: I left out the mulpliers Pr(C=1)=Pr(C=2)=(Pr(C=3)=1/3 since they divide out).

4) Our assumptions provide all the values we need:

Pr(H=3|C=1) = 1/2.
Pr(H=3|C=2) = 1.
Pr(H=3|C=3) = 0.

Pr(C=2|H=3) = (1) / [(1/2) + (1) + (0)] = 2/3.

+++++

Where people go wrong:

Most first-timers: "The two doors closed are the same, so each is equally likely." This is an intuition-based response that ignores that the host could have opened #2. It is equivalent to saying Pr(H=3|C=1)=Pr(H=3|C=2)=1, or that the MHP "is about opening the door we saw the host open." It is also equivalent to saying that the Two Child Problem "is about boys" because it says we know about a boy.

Naive appeal to intuition: "If there were a million doors, and the host opened all but one, you'd switch pretty quick." This merely changes how you apply intuition, it is not an actual solution.

Educated, but ill-considered: "Your choice initially had a 1/3 chance, and this can't change because the host can always open a door." This is based on the incorrect statement "this can't change." It can, it just doesn't in this problem. So an actual solution needs to explain both how it can, and why it doesn't.
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Re: Game Show Problem

Postby Gofer » Fri Aug 04, 2017 10:49 am

> “We” did. “Gofer” refuses to recognize them when other people make them.

My original response was directed to Robert; and he maintains that one can "infer" the correct answer, merely from the condition of the host knowing the location of the car, which is error in logic.

> If I say the answer is (14X+Y)/cos(Z), how is that definitive?

If you define X, Y and Z, it could be definitive.

> Including one symbol Gofer didn’t define, so it was not a formula.

What didn't I define? But you're right, it wasn't exhaustive, but not for the reasons you presented, and therefore waived others.

> But you can think of a formula being the equivalent of an English statement, using these symbols instead of words.

Not according to the definition of "formula" I quoted from merriam-webster.

> Gofer's non-formula lacked the equivalent of a verb, so it wasn't an expression of any fact. It was a collection of symbols that he used incorrectly.

A formula doesn't need a verb to be correct.

> And even if we take what seemed to be his intention, that "vs." meant to define odds, he called it a probability.

I never called it a probability. The use of "vs", as in "versus", was implied to mean odds. The fact that you're even arguing this is pathetic.

> So it was a non-formula for something other than the probability he said it was a formula for.

Only Jeff knows what he speaks of 50% of the time.


> Finally, from what Gofer skipped in whatever probability class he claims to have taken, Pr(EVENT) is the notation for the prior probability that the outcome is in the set EVENT,

And where did I ever said it wasn't?

> If Gofer would only you set up a truth table for ***ALL*** the prior probabilities in his system, he could see that there is a 1/3 chance that the car is behind #2. Then, there are equal chances that the host will open either #1 (if the contestant chose #3) or #3 (if the contestant chose #3). That makes the prior probability Pr(O&X) equal to 1/6.

If Jeff would only ever read what I actually write, instead of jumping the gun, misrepresent, and then proclaim victory by rebutting a statement never advanced by me, a s.c. straw man. The condition "accounting for the player's initial choice" is specifically the condition that bars half of the cases Jeff speaks of, namely the player choosing "door 3", and is thusely what is "choosing" the correct probability space.

> But a condition does not affect the prior probabilities.

There are no prior probabilities in my solution.

> But let’s assume Gofer is just ignorant about actual notation, and failed to read the lesson I provided that explained notation. That is, that Gofer really meant to define the event C to be where the contestant choose the door she did, and the meant to write “a formula for the odds is Pr(O&X|C) to Pr(O&S|C).”

Let's assume not, but instead assume that Jeff likes to read things he wishes his opponent state so as to be wrong.

> This is the condition he accused me of leaving out. And spent a year arguing about the results before explaining how that was the basis of his complete misunderstanding.

What condition? You mean Pr(...|C)? As I've stated before, the use of Pr refers to a particular probability space describing an experiment, and if using the wrong experiment, Jeff's answer is incorrect. More on this later.

> Note that, given C (and X and S defined from it - that is, "taking it into account"), then we need Pr(O&X|C) + Pr(O&S|C) = Pr(O|C). Which is 1, by the assumptions.

Wrong! O is the event the host opened THAT PARTICULAR door, which, after the player having chosen the door he chose, occurs 1/2, all things being equal.

+++++++++++++++++++++
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Re: Game Show Problem

Postby Gofer » Fri Aug 04, 2017 11:15 am

The best way to solve this problem is the use the tools handed to us by mathematics and probability theory, namely the use of probability spaces to describe experiments.

Let's imagine an experiment starting AFTER the player made his initial choice, and ending with the revelation of the car. Let H be the set of all host actions, and C be the set of all car actions, which we assume are finite.

Now, we can generate the probability space P from HxC, where "x" is the Cartesian Product, and from the standard assumptions of the game, such as the host always opening a goat-door other than the player's and offers a switch, and the host always choosing uniformly random if given a choice, and the car being placed uniformly random. These assumptions are codified in the probability measure Pr. The triple, Pr, HxC, and the sigma algebra generated from HxC, constitute P, and is a complete description of our experiment.

The solution can now simply be stated as:

Pr(O&X) to Pr(O&S), in P, in favor of switching, where O, X and S are events describing, respectively, the host opening the door he just opened, the car being behind the switch-door, and the car being behind the player's door.
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Re: Game Show Problem

Postby JeffJo » Fri Aug 04, 2017 12:52 pm

Gofer> My original response was directed to Robert; and he maintains that one can "infer"
> the correct answer, merely from the condition of the host knowing the location of
> the car, which is error in logic.

As much as I think robert is a blowhard who makes things up in order to achieve the answer he likes, he never said that. I even repeated what he actually said for you. Maybe you should take a course in remedial reading.

> If you define X, Y and Z, it could be definitive.

If you had defined "vs.", or what you meant by Pr(event) (which is not the standard mathematical definition, by the correct meaning of your statements which you do not understand), or how you got the probabilities (correctly), your answer would have been closer to being a definitive formula.

But you don't.

> The use of "vs", as in "versus", was implied to mean odds.

Yes, I got that. I said that I got that. But you didn't say it, yet think you did, demonstrating that you do not understand what it is you are saying.

And since what you were replying to "the conclusive answer that there is a 2/3 probability of winning the car by switching," and you were saying your non-formula provided that answer, you were indeed calling it a probability.

The fact that you're even arguing this is pathetic.

>> Finally, from what Gofer skipped in whatever probability class he claims to have
>> taken,Pr(EVENT) is the notation for the prior probability that the outcome is in the set EVENT,
>
> And where did I ever said it wasn't?

When you claim that Pr(O&X) was a probability conditioned on what door was chosen originally, and did not have the value I said it should have as a prior probability. This isn't difficult.

> If Jeff would only ever read what I actually write, instead of jumping the gun, misrepresent,
> and then proclaim victory by rebutting a statement never advanced by me, ...

If Gofer would only realize that most of what he writes is incomplete and incorrect...

> The condition "accounting for the player's initial choice" is specifically the condition that
> bars half of the cases Jeff speaks of, ...

And based on this statement, what you meant is properly written Pr(O&X|C), not Pr(O&X). Based on others, where you purport to draw these events from a sample space where C is not a condition but "before the experiment," it is not a condition. But how to determined Pr(O&X) is undefined.

Maybe if you try to understanding what "O&X|C" means, which you don't. It does not represent an event. You have said, repeatedly, that it does.

> There are no prior probabilities in my solution.

Then you haven't "conditioned on what door was chosen originally," a phrase which represents taking prior probabilities and updating them with a condition. As I keep telling you, the notation you used was entirely in prior probabilities. And if they weren't meant to be, they were already the answers you needed, and didn't say how to get them. WHICH REQUIRES UPDATING FROM PRIOR PROBABILITIES.

> As I've stated before, the use of Pr refers to a particular probability space describing an
> experiment, ...

Describing the experiment, yes. Not an instance where you can limit the events to a subset of the event space in a properly-formed probability space. That's called conditioning, and requires forming the entire probability space before you apply the condition Like "taking the chosen door into account" (is it one now? or not?).

Now, it is true that you can call the results, of trimming a deterministic probability space down by limiting it to a condition, another probability space. The problem with trying to claim an answer based on that "conditional probability space" is that you have no way to determine the probability values unless you go back to the "prior probability space."

What part of this do you not understand?

>> Note that, given C (and X and S defined from it - that is, "taking it into account"), then we
>> need Pr(O&X|C) + Pr(O&S|C) = Pr(O|C). Which is 1, by the assumptions.
>
> Wrong! O is the event the host opened THAT PARTICULAR door, which, after the player
> having chosen the door he chose, occurs 1/2, all things being equal.

The probability that the host will open door 1 is 1/3. The probability that the host will open door 2 is 1/3. The probability that the host will open door 3 is 1/3. Any of these doors can be what you want to call "the chosen door," so you have not eliminated any of them by "taking it into account."

+++++

> Let's imagine an experiment starting AFTER the player made his initial choice, ..

This is what I did over a year ago, that Gofer insisted I didn't do and needed to. But it is also equivalent to how I handled the door numbers, so it is nothing new.

It is not what is meant by "taking the choice into account," or using it as a condition, but it isn't clear that Gofer understands these differences. Which is why I wrote that simple solution, one that you can actually use to get an answer. Not just non-deterministic formula.

> Let H be the set of all host actions, and C be the set of all car actions, which we assume
> are finite.

How are the elements of these sets to be described? All you provide, are abstractions.

> Now, we can generate the probability space P from HxC, ...

Only if you describe how H and C are formed. Without that, your "formula" is just as informative as saying "count the number SW of times you win by switching, and the number of times ST you win by staying, and the odds are SW:ST in favor of switching."

YOU HAVEN'T SAID ANYTHING OF USE.
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Re: Game Show Problem

Postby Gofer » Fri Aug 04, 2017 2:27 pm

>> If you define X, Y and Z, it could be definitive.

> f you had defined "vs.", or what you meant by Pr(event)

Not having defined "vs" is obviously miles away from you not defining X,Y and Z in your formula, and could almost be considered a red herring.

> And since what you were replying to "the conclusive answer that there is a 2/3 probability of winning the car by switching," and you were saying your non-formula provided that answer, you were indeed calling it a probability.

No, I was calling it odds, just like I wrote, "the odds come out as 2 to 1 in favor of switching". And yet again Jeff fails to read what I write.


> When you claim that Pr(O&X) was a probability conditioned on what door was chosen originally, and did not have the value I said it should have as a prior probability. This isn't difficult.

I did nothing of the sort. I said "accounting for", meaning that our probability space does not contain player action.

>> The condition "accounting for the player's initial choice" is specifically the condition that
>> bars half of the cases Jeff speaks of, ...

> And based on this statement, what you meant is properly written Pr(O&X|C), not Pr(O&X).

This is error. See previous comment.

> Based on others, where you purport to draw these events from a sample space where C is not a condition but "before the experiment," it is not a condition.

Eh, what? Please try to make sense!

> But how to determined Pr(O&X) is undefined.

Only because I didn't explicitly write out all the elements of the sample space, or write out the measure Pr, hence my statement that the probability space is GENERATED from those conditions.

> Maybe if you try to understanding what "O&X|C" means, which you don't. It does not represent an event. You have said, repeatedly, that it does.

I've said that it can be seen as an event in a subspace of the space associated with Pr. Why are you trying to misrepresent me?

> Then you haven't "conditioned on what door was chosen originally," a phrase which represents taking prior probabilities and updating them with a condition.

Because it isn't necessary, because we only need to consider the proper events in the proper probability space. This isn't hard!

> As I keep telling you, the notation you used was entirely in prior probabilities.

Try to make sense!

> And if they weren't meant to be, they were already the answers you needed, and didn't say how to get them. WHICH REQUIRES UPDATING FROM PRIOR PROBABILITIES.

Completely incomprehensible!

> Describing the experiment, yes. Not an instance where you can limit the events to a subset of the event space in a properly-formed probability space. That's called conditioning, and requires forming the entire probability space before you apply the condition Like "taking the chosen door into account" (is it one now? or not?).

Wrong! The experiments starts AFTER the player's choice, hence the probability space is determined then.

> Now, it is true that you can call the results, of trimming a deterministic probability space down by limiting it to a condition, another probability space. The problem with trying to claim an answer based on that "conditional probability space" is that you have no way to determine the probability values unless you go back to the "prior probability space."

Duh! To form a subspace, one needs the original space. Tautologous much?

> The probability that the host will open door 1 is 1/3.
The probability that the host will open door 2 is 1/3. The probability that the host will open door 3 is 1/3. Any of these doors can be what you want to call "the chosen door," so you have not eliminated any of them by "taking it into account."

... depending on the probability space the action is being evaluated in. This isn't hard!



>> Let's imagine an experiment starting AFTER the player made his initial choice, ..

>This is what I did over a year ago, that Gofer insisted I didn't do and needed to. But it is also equivalent to how I handled the door numbers, so it is nothing new.

You never used those words, and certainly didn't refer to probability spaces as describing experiments.

> It is not what is meant by "taking the choice into account," or using it as a condition, but it isn't clear that Gofer understands these differences.

It is never clear what Jeff writes.

> Which is why I wrote that simple solution, one that you can actually use to get an answer. Not just non-deterministic formula.

And yours was correct, had you properly defined the probability space; otherwise it could be wrong.

>> Let H be the set of all host actions, and C be the set of all car actions, which we assume
>> are finite.

> How are the elements of these sets to be described? All you provide, are abstractions.

Because it's trivial to define them, but isn't worth the effort, hence my using the word "generated".

> YOU HAVEN'T SAID ANYTHING OF USE.

Of course I have, namely that a formula such as Pr(O) may be correct or incorrect depending on the probability space, and hence the experiment, associated with Pr.
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Re: Game Show Problem

Postby klaralles » Mon Aug 28, 2017 12:22 am

Where does logic end and reasoning begin?
The problem is couched in a quasi-real world situation involving a 'game show host' who knows what is behind each door. The correct answer hinges on the behaviour of the host.
SCENARIO ONE: MALEVOLENT HOST
As you are not the one who pays the game show host's salary, you can assume that the host is playing against you, offering choices and opening doors which minimise your chance of winning.
You have picked Door #1, but the host has declined to open that door, asking instead if you might choose a different door. Why?
Obviously, the car is behind the door you picked. If there had been a goat behind it, the host would have simply opened your choice of door making you the instant loser. If you stick with your initial choice, you win every time.
SCENARIO TWO: IMPARTIAL HOST, GAME RULES RULE.
You assume the aim of the game host is to maximize the drama of the moment for the audience, and will always offer a chance to switch, opening either Door #2 or Door #3, no matter what was behind Door #1.
In this scenario, there are two chances in three that the car is not behind Door #1, hence two chances in three that the car is behind the door (choice of #2 or #3) which the host does not open. If you switch, you win two times in three.
Last edited by klaralles on Mon Aug 28, 2017 9:22 pm, edited 1 time in total.
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