Game Show Problem

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Re: Game Show Problem

Postby Gofer » Fri Feb 09, 2018 4:28 pm

> Gofer wrote:
> Robert,
>
> take 3 cups, a sugar cube, and one standard die:
>
> roll the die and accordingly place the cube under one of the cups. Roll the die again,
> choosing one cup. Again roll the die and accordingly flip one of the remaining two
> cups. Discard events where the cube was shown. Flip the third cup and note whether
> it has the cube.

>Invalid analogy. The host does not randomly choose among the remaining two doors, but chooses to >reveal a goat in all circumstances.

But you previously said that all that mattered was the host revealing a goat, yet now states the probability ("in all circumstances") in doing so also affects the answer. Guess we are all on the same page then ...
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Re: Game Show Problem

Postby robert 46 » Fri Feb 09, 2018 4:54 pm

Gofer wrote:
> > robert 46 wrote: Invalid analogy. The host does not randomly choose among the remaining
> > two doors, but chooses to reveal a goat in all circumstances.
>
> But you previously said that all that mattered was the host revealing a goat, yet
> now states the probability ("in all circumstances") in doing so also affects the
> answer.

Would you care to quote whatever it is you are referring to? Guardian stated that the probability of switching to a goat cannot be higher than the probability of switching to the car, so it is beneficial to switch. True enough, yet there is nothing in the problem statement which changes the expected probability of winning the car by switching from 2/3. What guardian is saying is that factoring in irrelevant speculation still does not reverse the potential benefit of switching.

> Guess we are all on the same page then ...

I find no purpose in factoring in un-implied speculation. The particular door opened has no significance per se. What is significant is that a goat has been revealed, not behind the player's door.
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Re: Game Show Problem

Postby guardian » Fri Feb 09, 2018 4:56 pm

robert 46 wrote:
> guardian wrote:
> > robert 46 wrote:
> > > Assume the game configurations are equally likely (the car is equally likely
> to
> > be
> > > behind any door). Then in configuration:
> > > Pc Hg Hg the player switches to a goat (probability 1/3)=
> > > Pc Sg Hg and
> > > Pc Hg Sg;
> > > Pg Sc Hg the player switches to the car (probability 1/3);
> > > Pg Hg Sc the player switches to the car (probability 1/3).
> > >
> > > The probability of the player switching to the car is 1/3+1/3=2/3=1-Pr(Pc).
> > > [note: clarifications added]
> >
> > I completely agree with everything you have here. But in the game, the player gets
> > to see which door the host opens. So at the time the player has to make their
> final
> > decision, the possibilities are either:
> >
> > Pc Sg Hg vs. Pg Sc Hg
> > or
> > Pc Hg Sg vs. Pg Hg Sc
> >
> > As you pointed out, in both pairs the latter term has probability 1/3. But the
> two
> > former terms have a probability that TOTAL to 1/3. So in either case, the latter
> > term must be at least as likely as the former term, and thus the player should
> switch.
>
> This is reasonable.
>
> However, that JeffJo admits the only implied value is Q=1/2 is a tacit agreement
> that there is no significance to the actual door which is observed to be opened.

Actually, this is very significant. Without this observation, the player wouldn't know which door to switch to. So there is information being conveyed here.

> It is merely the contents, a goat, and that it is not the player's door, which is
> significant because this means the player will switch to the opposite prize from
> that initially chosen. Thus the probability of the player having chosen a goat is
> the controlling factor. The host has no influence on this at any time because this
> is consequent to the fact that there are two goats and three doors and the player
> must choose a door. There are two chances in three that the door chosen has a goat,
> and if it does then the player necessarily switches to the car.

This is all fine, and a great starting point. The flaw in your argument is that it seems to imply that there are only two strategies - always switch or never switch. You make a very convincing case that the former is superior to the latter. But what if someone asked, "Well, if I have a 2/3 chance of winning by switching, maybe it would be better if I rolled a die and only switched when it came up 1-4"? Or how about, "I will pick an odd door and only switch if the host reveals the other odd door"? Obviously these are inferior strategies, but your solution does not explain why. It only explains that there is a way to win 2/3rds of the time on average; it doesn't explain why this is the best that can be done.

Suppose you gave your solution to someone and they said, "Oh, I see, by always switching I win with a probability that will be equal to the combined probability of the doors I didn't pick. And this is even before I get the extra information of which door the host reveals. Can I use that information to do even better?" The paragraph you wrote above is longer than your original answer, and touches on why the answer is no. But to provide a complete explanation, and to answer the question of why is 2/3 the best the player can do in general, you solution requires some form of additional exposition. To me, this is not parsimonious.

If you look at my solution above, it simply states that when it is time to make a final choice of doors, there will only be two possible game states. The game state where the player wins by not switching cannot be more likely, and sometimes (if not always) will be less likely, than the game state where the player wins by switching. So the player should switch. No further explanation is needed because this holds for all possible values of car, player, and host door. Very parsimonious, no?
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Re: Game Show Problem

Postby robert 46 » Fri Feb 09, 2018 5:59 pm

guardian wrote:
> If you look at my solution above, it simply states that when it is time to make a
> final choice of doors, there will only be two possible game states. The game state
> where the player wins by not switching cannot be more likely, and sometimes (if
> not always) will be less likely, than the game state where the player wins by switching.
> So the player should switch. No further explanation is needed because this holds
> for all possible values of car, player, and host door. Very parsimonious, no?

The two games states are either: the player has chosen a goat and will switch to the car; or has chosen the car and will switch to a goat. The former is twice as likely as the latter.
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Re: Game Show Problem

Postby guardian » Sat Feb 10, 2018 12:06 am

robert 46 wrote:
> guardian wrote:
> > If you look at my solution above, it simply states that when it is time to make
> a
> > final choice of doors, there will only be two possible game states. The game state
> > where the player wins by not switching cannot be more likely, and sometimes (if
> > not always) will be less likely, than the game state where the player wins by
> switching.
> > So the player should switch. No further explanation is needed because this holds
> > for all possible values of car, player, and host door. Very parsimonious, no?
>
> The two games states are either: the player has chosen a goat and will switch to
> the car; or has chosen the car and will switch to a goat. The former is twice as
> likely as the latter.

No, you silly goose. The two game states are either: the player has chosen a goat AND the host HAD to reveal the door that they did; or the player has chosen the car AND the host CHOSE to reveal the door that they did. You can ASSUME that the former is twice as likely as the latter, but that is introducing unnecessary speculation into the solution (tsk, tsk). For those of us who prefer to be parsimonious, all that is required is that the latter cannot be more likely than the former.

And just so we are clear on what this means, let's review. By revealing a door, all of the possibilities that include the car being behind that door get removed. The possibility that the car is behind the player's door and the other door gets revealed is also removed. So all of the probability of the revealed door, some of the probability of the player's door, and none of the probability of the other door get removed. When we look at what is left, we have all of the probability of the other door and some of the probability of the player's door. Assuming these two doors started with the same probability, it has to be more likely that the car is behind the other door (unless there was absolutely no chance of the host revealing the other door when the car is behind the player's door, in which case they are equally likely).

As I said (and you ignored), all your solution says is that there is a way to win 2/3rds of the time. It doesn't say if this is the best you can do, and if so, why. It only says that it is superior to using the opposite strategy, which is obvious. The better answer is to show that the player's odds are never worse by switching no matter what happens in the game. My answer does, yours does not.

I see you are still accusing me of irrelevant speculation. I would love to know what that entails. All I say is that when the player chooses the door with the car, the host has a choice of doors to open. That is all I say. Nothing about bias, nothing about how the host makes the choice, nothing about the significance of the choice. The ONLY thing that I use is the fact that host will only open one of the doors, so by definition this means the host must make a choice. And by the way, you use this too, because if the host didn't always open one door, "always switch" would not be well-defined. So by all means, please tell me what irrelevant speculation I am introducing here...
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Re: Game Show Problem

Postby robert 46 » Sat Feb 10, 2018 5:45 am

guardian wrote:
> The two game states are either: the player has chosen a goat
> AND the host HAD to reveal the door that they did; or the player has chosen the
> car AND the host CHOSE to reveal the door that they did.

The two games states are either: the player has chosen a goat
(AND the host HAD to reveal the door that he did); or the player has chosen the car
(AND the host CHOSE to reveal the door that he did).

The first parenthetical remark is consequent to "the player has chosen a goat".
The second parenthetical remark is consequent to "the player has chosen the car".
From the player's point of view, there is no insight into whether the host HAD or CHOSE to open the door he did, so the parenthetical remarks are irrelevant to solving the problem. We can say the same thing about goat color: there is no insight into whether the host HAD or CHOSE to reveal the colored goat he did. The fact that the player sees a particular door opened, or sees a goat of a particular color, IS IRRELEVANT TO SOLVING THE PROBLEM BECAUSE IT PROVIDES NO USEFUL INFORMATION (other than trivially designating the remaining door, or color of goat that one might win if the goat colors are known in advance).

That "the player has chosen a goat" is twice as likely as "the player has chosen the car" is NOT an assumption: it is a calculation based on there being two goats and three doors. Nothing else can be calculated based on speculation about whether the host HAD or CHOSE to open the door he did, or HAD or CHOSE to reveal the colored goat he did.

If the player chose a goat THEN the host had to open the door he did.
If the player chose the car THEN the host chose to open the door he did.

The consequent is deduced from the antecedent, and thereby provides no additional useful information to change the probability- IT IS REDUNDANT AND SUPERFLUOUS.
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Re: Game Show Problem

Postby guardian » Sat Feb 10, 2018 7:26 am

robert -

I would rather discuss numbers than semantics. So consider the following game:

I have 20 red marbles, 10 blue marbles, and 2 bags. I let you see that I put 10 red marbles in each bag. Then, out of your sight, I distribute the 10 blue marbles between the two bags. Next I choose a bag to give you. The probability that I choose a bag is proportional to the number of marbles in it, so, for example, if both bags have 15 marbles there is a 50/50 chance of you getting either bag; if one bag has 18 marbles and the other has 12, there is a 60% chance I hand you the 18 marble bag and 40% chance I hand you the 12 marble bag. You don't know how many marbles are in the bag I give you.

You will then blindly select a marble out of the bag I hand you. Before you do, I ask you to guess which color you will get. If your prediction is correct, you win. Which color do you guess and why?
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Re: Game Show Problem

Postby robert 46 » Sat Feb 10, 2018 7:36 am

guardian wrote: ...

Evasion through redirection.

The consequent of a deduction provides no more information than was contained in the antecedent from the start.
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Re: Game Show Problem

Postby guardian » Sat Feb 10, 2018 7:56 am

robert 46 wrote:
> guardian wrote: ...
>
> Evasion through redirection.
>
> The consequent of a deduction provides no more information than was contained in
> the antecedent from the start.

Nope, redirection in the hope of getting somewhere useful. I don't want to be stuck here for years like jeffjo. You can either answer my question or I can stop talking to you again. The choice is yours.
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Re: Game Show Problem

Postby guardian » Sat Feb 10, 2018 8:21 am

robert 46 wrote:
> guardian wrote: ...
>
> Evasion through redirection.
>
> The consequent of a deduction provides no more information than was contained in
> the antecedent from the start.

In case it gives you any more incentive to answer my question, I will explain why I didn't pursue your line of argument: my game states were described generically but were meant to be applied to specific configurations. So if the player chose door 1 and the host revealed door 3, the two possible games states would be "a goat is behind door 1 and the host had to reveal door 3" and "the car is behind door 1 and the host chose to reveal door 3". The probability of the first state is 1/3, the probability of the second state is no more than 1/3. I wanted a statement that applied to all possibilities (not just 1 and 3), but the generic statement allowed you to construe it in a way that let you go back to your original argument, which doesn't apply to the example I was trying to frame. So we could argue about the meaning of my words or we could try something different. I believe the answer you give to the problem I posed will be very useful in trying to illustrate my point.
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Re: Game Show Problem

Postby JeffJo » Sat Feb 10, 2018 10:20 am

robert can only make points by misrepresenting what others say:

> ... JeffJo admits the only implied value is Q=1/2 ...

Wrong. No value is implied, and none can be inferred. But you if you must assume a value[1], the only value you can assume is 1/2. Robert just can't get it into his limited capability to analyze a problem that inference and assumption are different things.

I've been saying this for years, so why is robert just noticing it now?

Robert continues in his misrepresentation:

> ... is a tacit agreement that there is no significance to the actual door which is
> observed to be opened.

Wrong. The significance is that the door is chosen. Since only that door can be chosen if the contestant picked a goat, but both could have been chosen if she picked the car, it is the choice that makes the circumstances for the remaining possibilities different.

This establishes both what the answer is - whether of not you assume (not infer) a value for Q - and why the "don't switch" argument is wrong. Robert's solution, which tacitly infers (not assumes) a value for Q while denying that he does, in incorrect because of the inference and because it answers the wrong question. And, it doesn't explain why the other solution attempt is wrong.

> Thus the probability of the player having chosen a goat is the controlling factor.

It is an irrelevant factor. The only "controlling factor" is the probability Q. Whther or not you assume Q=1/2.

+++++

[1] The only time you need to, is if you are asked for the probability of winning. Since we aren't (another fact robert hasn't seemed to notice), we don't need to.
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Re: Game Show Problem

Postby robert 46 » Sun Feb 11, 2018 9:06 am

If the player chose a goat and the host opens another door to reveal a goat THEN the player will switch to the car (and, by-the-way, the host had to open the door he did).

If the player chose the car and the host opens another door to reveal a goat THEN the player will switch to a goat (and, by-the-way, the host chose to open the door he did.)

The player will not know which parenthetical consequent is the case until the contents of the remaining door are revealed. Thus a parenthetical consequent cannot be used to change the probability of winning the car by switching.

Opponents believe that a vacuous parenthetical consequent may be used as an antecedent to change the probability of winning the car by switching. This, of course, is absurd; but it doesn't stop them from trying to argue that a vacuous parenthetical consequent may be used as an antecedent.

If the host favors a door to open AND this favoritism is known to the player AND there are numeric values also known to the player THEN favoritism CAN be factored in as an antecedent. However, if the host favors a door to open BUT there are no numeric values known to the player THEN favoritism CANNOT be factored in as an antecedent. Nothing in the problem statement implies that the host has any favoritism for a door to open, nor implies that the player is aware that the host has favoritism, nor provides any numeric values which would make favoritism relevant. Thus it is parsimonious to leave out of an analysis all considerations of host favoritism for a door to open.
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Re: Game Show Problem

Postby guardian » Sun Feb 11, 2018 1:47 pm

I see robert has respectfully declined to answer my question about the marbles. That is sort of a shame, since there was potentially some insight to be gained.

In honor of parsimonious robert, I will now present "Parsimonious GSP Problem with Parsimonious Solution". Enjoy:

A game show has 3 doors. One door has a car behind it, the other 2 goats. Each door is equally likely to have the car placed behind it. Before the player guesses a door, the host tells them the following: "If the car is behind door 3, I will open door 2. If it is behind door 2, I will open door 3. If it is behind door 1, I will open either door 2 or door 3." In a particular game, the host opens door 3. Which door should the player choose?

Answer: The player should choose door 2 because it cannot be more likely that the host opens door 3 when the car is behind door 1 than when it is behind door 2.

Thank you, thank you very much.

P.S. We could have also babbled on about how a player who never picks door 1 will win 2/3rds of the time on average. But that wouldn't have been parsimonious. So as everyone can now see, "parsimonious robert" is actually an ironic nickname.
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Re: Game Show Problem

Postby robert 46 » Sun Feb 11, 2018 2:21 pm

Whatever point guardian is trying to make is obscure because the player chooses a door BEFORE the host opens another door to reveal a goat.

Thus the host is not going to make his speech until after the player has chosen a door, and then adapts his speech to the circumstances:
Host: "I see that you have chosen door #1. If your door has the car I will open door #2 or door #3, If door #2 has the car, I will open door #3. If door #3 has the car, I will open door #2. Is this clear?
Player: "It is clear that you intend to open a door having a goat."
Host: "I am opening door #3, and, as you can see, door #3 contains a white goat. Is this clear?"
Player: "It is clear that the car is behind either door #1 or door #2. What is the color of the other goat?"
Host: "The other goat is a black goat. Would you like to stay with door #1 or switch to door #2?"
Player: "Do you prefer to open a door with the white goat, the black goat, or have no preference?"
Host: "I decline to say."
Player: Do you prefer to open door #3 if you can, door #2 if you can, or have no preference?"
Host: "I decline to say."
Player: "In that case, I choose to switch to door #2."

Note that since the host declines to say whether he has a preference for the white or black goat, or door #2 or door #3, the goat color and particular door opened are irrelevant to making the decision to switch. The player recognizes that he had a 2/3 chance of having chosen a goat, and will switch to the car if he has.
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Re: Game Show Problem

Postby guardian » Sun Feb 11, 2018 3:31 pm

robert 46 wrote:
> Whatever point guardian is trying to make is obscure because the player chooses a
> door BEFORE the host opens another door to reveal a goat.
>
> Thus the host is not going to make his speech until after the player has chosen a
> door, and then adapts his speech to the circumstances:
> Host: "I see that you have chosen door #1. If your door has the car I will open door
> #2 or door #3, If door #2 has the car, I will open door #3. If door #3 has the car,
> I will open door #2. Is this clear?
> Player: "It is clear that you intend to open a door having a goat."
> Host: "I am opening door #3, and, as you can see, door #3 contains a white goat.
> Is this clear?"
> Player: "It is clear that the car is behind either door #1 or door #2. What is the
> color of the other goat?"
> Host: "The other goat is a black goat. Would you like to stay with door #1 or switch
> to door #2?"
> Player: "Do you prefer to open a door with the white goat, the black goat, or have
> no preference?"
> Host: "I decline to say."

> Player: Do you prefer to open door #3 if you can, door #2 if you can, or have no
> preference?"

This question is unnecessary and thus not parsimonious. If the host opens door #2 when they can, then the car is behind door 2 and the player should switch. If the host has no preference, the player will win on average 2/3rds of the time by switching, so they should switch. If the host opens door 3 when they can, then there is a 50/50 chance of the car behind door 1 or 2, and it doesn't matter if the player switches. So the only reason to ask is if the player didn't want to switch unless it was necessary to improve their odds, in which case they should have picked a different door to begin with. Simply put, there is no answer to this question that will make not switching a better choice than switching. So why ask?

> Host: "I decline to say."
> Player: "In that case, I choose to switch to door #2."
>
> Note that since the host declines to say whether he has a preference for the white
> or black goat, or door #2 or door #3, the goat color and particular door opened
> are irrelevant to making the decision to switch. The player recognizes that he had
> a 2/3 chance of having chosen a goat, and will switch to the car if he has.

Yes, he HAD a 2/3 chance of choosing a goat, but the player recognizes that, after a door is revealed, that chance can change. Your answer is "the good news is that he started off with 2/3 chance of winning by always switching, so he can ignore this intermediate step and continue to believe his chance is 2/3 even though at the moment it may actually be higher or lower." My answer is "the good news is that no matter which door the host reveals, the player will still have at least a 50% chance of winning by switching." Which better describes the player's situation at the time of the decision? If you don't think it is mine, go back and answer my marble question.

Anyway, your excessive, agenda-based exposition is totally NOT parsimonious. But since you are too dense to see that the player initially choosing a door was extraneous in my original formulation, I now present "Parsimonious GSP Problem with Parsimonious Solution 2.0":

A game show has 3 doors. One door has a car behind it, the other 2 have goats. Each door is equally likely to have the car placed behind it. The host first asks the player to pick a "choice door". In this particular game, the player chooses door 1. The host then tells them the following: "If the car is behind door 3, I will open door 2. If it is behind door 2, I will open door 3. Since door 1 is my choice door, if the car is behind door 1, I will open either door 2 or door 3." In this particular game, the host opens door 3. If the player wants the car, which door should the player choose?

Answer: The player should choose door 2 because it cannot be more likely that the host opens door 3 when the car is behind door 1 than when it is behind door 2. No questioning of the host is needed to come to this decision.

You're welcome.
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