Game Show Problem

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Re: Game Show Problem

Postby guardian » Tue Jan 09, 2018 3:38 pm

***New and exciting content included between the asterisks below...***

Finally...

robert 46 wrote:
> guardian wrote:
> > When the contestant chooses the player's door and the host reveals the host's door,
> > the host reveals the host's door for one of two reasons:
> >
> > (1) the car was placed behind the remaining door
>
> So the host chose to reveal the other door because it was the only door available
> having a goat. Thus:
>
> (1) the car was placed behind the remaining door AND the host chose to reveal the
> host's door because it is a rule of the game to reveal a goat.
>
> > (2) the car was placed behind the player's door AND the host *the* chose to reveal
> > the host's door.
>
> Five times you have failed to correct this grammatical error. Do you ever reread
> your posts?
>
> (2) the car was placed behind the player's door AND the host chose to reveal the
> host's door because it is a rule of the game to reveal a goat.
>
> > We have two explanations that are not necessarily equally likely - the second condition
> > needs 2 things to happen, and the first part of the second condition has the same
> > chance as the entire first condition. The 2nd part of the 2nd condition cannot
> make
> > the condition more likely to be true, it can only make it less likely. In other
> > words, if the probability of (1) is X, the probability of (2) is XY, where 0<=Y<=1.
> > So the first condition must be at least as likely as the second.
>
> When it is stated correctly- that it is a rule of the game to reveal a goat behind
> another door- your argument disintegrates.

Why? The host was not bound by the rules of the game to pick a particular door. Your phrasing is misleading - it makes it sound like the rules of the game forced the host to choose the door that they did. It should be written like this:

(2) the car was placed behind the player's door AND, because it is a rule of the game to reveal a goat, the host had to reveal one of the remaining doors, and the host chose the host's door instead of the remaining door (they could have chosen the remaining door, but didn't).


*******************************************
Just to make clear what I am trying to capture in (1) and (2), suppose the host has a bag with 10 red, 10 white, and 10 blue marbles. Before each game, the host chooses a marble from the bag at random. The car is put behind door 1 if it is red, door 2 if it is white, and door 3 if it is blue. The host then sticks the marble in their pocket for the duration of the game.

Now suppose the player chooses door 1 and is shown door 3. What can the player say about the color of the marble? The player knows it cannot be blue, but cannot rule out the marble being red or white. Since there were originally 10 red and 10 white marbles, it would seem that the probability that the marble in the pocket is red is 1/2. But this would mean the probability of winning by switching would be 1/2. What is the error in this reasoning? The answer to this question is equivalent to the point I am trying to make with (1) and (2).

*******************************************

>
> > I know that if you had to bet on one of these outcomes, you would bet on the first,
> > which is analogous to always switching. So why is this not the solution to the
> GSP?
> > If it is the solution, how is your solution equivalent?
>
> The only occasion where the host has an alternate for goat-doors to open is when
> the player chose the car.
> The only occasion where the host does not have an alternate for goat-doors to open
> is when the player chose a goat.
> The former is half as likely to occur as the latter.
> When the former occurs the player switches to a goat.
> When the latter occurs the player switches to the car.
> Conclusion: it is half as likely that the player switches to a goat than that the
> player switches to the car. That the host has an alternate or not is subsidiary
> to the prize the player has chosen. So the controlling factor is the prize the player
> has chosen- not whether the host has an alternate or not.

This is all true. Now tell me how it implies "always switch". Yes, the average probability of winning by switching is 2/3 over the two possible doors that could be revealed. But how do I know there isn't a "bad" door that I shouldn't switch to? So far, your answer has been "how could there be?" or "that is not indicated by the problem". Neither of these are sufficient explanations. The problem does not indicate that the probability of winning by switching is 2/3, but we don't just say "how could it not be?" We show why it is. Even if it is obvious, it is still part of a complete solution. My solution says "the average is 2/3 and each individual door can't be below 1/2". Yours seems to be saying either "the average is 2/3 and it is 2/3 for each possible door that can be revealed", or it is saying "the average is 2/3 and if the doors are unequal the player can't know to what extent or which is which". A complete solution would clarify and fully explain this.
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Re: Game Show Problem

Postby guardian » Thu Jan 11, 2018 12:24 pm

C'mon, robert, the spectators are anxiously waiting for you to resolve my paradox:

Suppose the host has a bag with 10 red, 10 white, and 10 blue marbles. Before each episode of the GSP, the host chooses a marble from the bag at random. The car is put behind door 1 if it is red, door 2 if it is white, and door 3 if it is blue. The host then sticks the marble in their pocket for the duration of the game. Suppose the player knows everything about this process except for the color of the marble selected, and thus which door has the car.

Suppose the player chooses door 1 and is shown door 3. The player knows the marble cannot be blue, but cannot rule out the marble being red or white. Since there were originally 10 red and 10 white marbles, there is a 50/50 chance that the marble in the host's pocket is red, and thus a 50/50 chance of winning by switching.

Similarly, if the player chooses door 1 and is shown door 2, the player knows the marble cannot be white, but cannot rule out the marble being red or blue. Since there were originally 10 red and 10 blue marbles, there is a 50/50 chance that the marble in the host's pocket is red, and thus a 50/50 chance of winning by switching.

Thus no matter which door is shown, there is a 50/50 chance of winning by switching, contradicting robert's assertion that the player will win on average 2 out of 3 times by switching.

Do you see any problems here, robert?
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Re: Game Show Problem

Postby robert 46 » Thu Jan 11, 2018 2:45 pm

guardian wrote:
> ***New and exciting content included between the asterisks below...***

Note: No notification of *edits* to posts is conveyed to those receiving notifications of *new* posts.

> > robert 46 wrote: ...
> This is all true. Now tell me...

*****
> guardian wrote:
> C'mon, robert, the spectators are anxiously waiting for you to resolve my paradox:

They are invited to come down from the viewing stands and say what they like.

> ....
> Do you see any problems here, robert?

> guardian wrote, Aug 27, 2016 12:18 am, p 155:
> I am not sure what the original argument is about, but I figure if people are still arguing about > it, I would provide a rigorous but intuitive explanation as to what MVS got right and wrong
> about this problem.
> ....
> Hopefully this clears everything up and this decades-long argument can come to an end...

It is clear that guardian has yet to complete the task of reading the entire topic to understand what the arguing is all about. I suggest he take a break from posting to do so.

Guardian first came to this topic as the savior who dispels all confusion about the Game Show Problem: was responded to, but did not respond back.

Guardian returned to the topic over a year later:

> guardian wrote, Nov 19, 2017 2:32 am, p 194: Your [robert 46's] answer is not wrong, but it is
> incomplete. Since you have assumed no bias on the part of the host, you have assumed away
> the part of the problem that would force you to see in what sense your answer is incomplete.

Guardian still thinks my answer is incomplete even when he agrees with what I say; which comes to a conclusive answer to the problem based on the facts and reasonable assumptions which one gets from the problem statement.

Principle: Where there is no implication to the contrary, we should assume that an element randomly takes on a value. Host choice of a goat-door to open is an element where there is no implication to the contrary that the choice is random. One can invoke a probability-agent for this purpose. A "probability-agent" is an idealized device which randomly produces values in a domain, where the values are equally likely (and, I might add: have a greater than zero probability). All such random elements can be handled in the GSP by using an honest die to select one of three doors, or one of two doors [1]. The host's random choice of where to place the car is made before the game begins. The host's random choice of which relative goat-door to open if the need arises, can also be made before the game begins.

Thus, there being no implication to the contrary, host favoritism for a goat-door to open is a speculative, irrelevant, non-issue. In guardian's case, his insistence on answers to irrelevancies is indicative of an attention-seeking personality trait. That he repeated one question five times (without correcting a minor grammatical error), and another question twice (so far) reinforces the observation. He appears to manifest the personality of one who thinks himself a savior, craves attention, and seeks adulation from the masses.

If spectators would like to comment, I wholeheartedly invite them to do so. Be sure to lavish praise on guardian if you feel it is apropos.


[1] For the former: opposite faces count as the same: i.e 1-6, 2-5, 3-4.
For the latter: odd, even.
Last edited by robert 46 on Thu Jan 11, 2018 4:25 pm, edited 1 time in total.
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Re: Game Show Problem

Postby Edward Marcus » Thu Jan 11, 2018 3:07 pm

Robert's solution handles this symmetric decision maker

[1] For the former: opposite faces count as the same: i.e 1-6, 2-5, 3-4.
For the latter: odd or even.


Guardian's handles this asymmetric one
[1] For the former: opposite faces count as the same: i.e 1-6, 2-5, 3-4.
For the latter: (1-6 or 2-5) or 3-4.
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Re: Game Show Problem

Postby guardian » Thu Jan 11, 2018 4:24 pm

Hey robert -

What was the name of that fallacy of when you just launch a bunch of personal attacks instead of responding to the question? I seem to remember you complaining about that...

My older posts employed sarcasm. I am guessing the majority of the people realized it. Your analysis about me, like your analysis of competitive cheerleading, is based on complete ignorance. But more importantly, it is irrelevant to the GSP. Your little game of playing lawyer trying to score points with a jury is intellectually dishonest. This is a math problem - there is no need to take a vote. There are standards as to what constitutes a valid solution. Yours falls short. I have explained this, and even offered two possible interpretations that result from your incomplete solution. You wrote a lot, but didn't address this very fundamental point. Instead you keep babbling about choice of goat door, which I don't even use in my solution (other than to say there is one), much less care about for yours. I have told you this, but, using your "when all you have is a hammer, everything looks like a nail" principle, you keep hammering away on this irrelevant point. This is either academic dishonesty or total cluelessness, and, frankly, neither is attractive on someone who is already very unpleasant.

I knew you wouldn't resolve the paradox, and I am pretty sure you don't know how. But I will give you another chance using robert-speak. Instead of drawing a marble, the host rolls one of your robert-approved dice, and writes the corresponding door number down on a slip of paper and puts it in their pocket. Proceed in the obvious fashion.
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Re: Game Show Problem

Postby robert 46 » Thu Jan 11, 2018 5:23 pm

The viewpoint that after a goat is revealed there are two closed doors, one must have a goat and the other must have the car, so the odds are 50:50, is the naïve opinion of the ignorant masses (and, at one time, apparently a not insignificant number of mathematical professionals).

This matter was resolved by considering the case of the Little Green Woman (LGW). After the host opens a door to reveal a goat, the LGW lands on the stage in her flying saucer, and steps out. She is told that behind one of the unopened doors there is a car, and behind the other a goat. Then asked to choose which door she believes is most likely to have the car. She says that they are equally likely, and she is right; but only from her perspective. The crucial element is that the LGW does not know which of the two doors is the one the player chose.

If the LGW was to choose the player's door, the probability of winning the car is the same as if the player stays. If the LGW was to choose the remaining door, the probability of winning the car is the same as if the player switches. However, the LGW doesn't know which is which; so has 1/2 probability of picking the stay door, 1/2 probability of picking the switch door. So the overall probability for her is (1/2)*(1/3)+(1/2)*(2/3)=1/2: a 1/2 chance of picking the stay-door with a 1/3 chance of having the car and a 1/2 chance of picking the switch-door with a 2/3 chance of having the car. Yet if the LGW actually does pick a particular door, it has the same probability of having the car as the player would determine for that door; but the LGW doesn't know what that numeric probability actually is, although she can work out that it is either 1/3 or 2/3 if she is told that one of the unopened doors was chosen by the player before a door was deliberately opened to reveal a goat.
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Re: Game Show Problem

Postby Gofer » Thu Jan 11, 2018 5:39 pm

> The viewpoint that after a goat is revealed there are two closed doors, one must have a goat and the other must have the car, so the odds are 50:50, is the naïve opinion of the ignorant masses (and, at one time, apparently a not insignificant number of mathematical professionals).

Which exactly are the odds when assuming the host favors no door to open - meaning he could open the player's. It has nothing to do with being naïve, but with recognizing how the host chooses. Since there's nothing in the problem description that the host is bound to always open a goat-door, I think we can forgive them, or even praise them.

----

The gist of Robert's theory is that all that matters is that the host revealed a goat other than the player's door. But as I showed, this is blatantly false under the assumption of the host always opening a goat-door, and in fact equals the probability of the prize being behind a particular door, being 1/3 given the normal assumptions.
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Re: Game Show Problem

Postby guardian » Thu Jan 11, 2018 5:50 pm

Edward Marcus wrote:
> Robert's solution handles this symmetric decision maker
>
> [1] For the former: opposite faces count as the same: i.e 1-6, 2-5, 3-4.
> For the latter: odd or even.
>
>
> Guardian's handles this asymmetric one
> [1] For the former: opposite faces count as the same: i.e 1-6, 2-5, 3-4.
> For the latter: (1-6 or 2-5) or 3-4.

robert's solution actually could apply to either the symmetric or asymmetric decision maker. The fact that this is not clear is my primary point of contention with his solution. robert's solution only shows that on average there is a 2/3rds chance of winning by switching. Because the question asks "should the player switch?", to answer "yes" we need to show that either there is always at least a 50/50 chance of winning by switching, or we need to show than on average we come out ahead and there is no way to know when the odds are worse when we switch.

So really all robert needs to do is to add one of the following 2 explanations to his solution:

(1) Since the problem gives no indication of how the host chooses a door to reveal when the player chooses the door with the car, we will assume that the host is equally likely to show either unchosen door. Since we would then have both doors having the same chance of initially having the car and the same chance of being revealed when there is a choice, they must have the same probability of winning by switching. Since we know the average is 2/3, this means they both must be 2/3. Hence the player should always switch.

(2) The average probability of winning by switching is 2/3. Since both doors not chosen by the player have the same initial probability of having the prize, the only way they could have different probabilities of winning by switching is if the host is not equally likely to reveal either door when the player chooses the door with the car. Since we are given no information about how the host makes this choice, we have no way of determining if one door is inferior, and if so, which door that is. Thus the best the player can do is to use the fact that on average they will be better off by switching and always switch.

The drawback of (1) is that it makes an extra assumption - that the host is equally likely to choose either door when the player chooses the door with the car. The drawback of (2) is that it doesn't make the strongest statement possible - even if one door is inferior, it is impossible for the inferior door to ever have a probability of winning by switching below 50%. This is a stronger endorsement of the strategy of "always switch", because on average the player cannot make their odds worse by doing so.

Even though (1) has an extra assumption that is not really necessary to solve the problem and (2) doesn't tell us everything there is to know about the properties of the game, I still consider them valid solutions. But I believe the explanations are necessary to qualify the statement "on average the player wins 2/3rds of the time by switching." 2/3 can be a straight average of 1/3 and 1, or even a weighted average of 0 and 1 (and we wouldn't want to switch if we knew our odds of winning by switching were 1/3 or 0), so it is important to put this number in the proper context - specifically, to show that "always switch" will never include an avoidable situation of the player switching to worse odds.
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Re: Game Show Problem

Postby guardian » Thu Jan 11, 2018 6:10 pm

robert 46 wrote:
> The viewpoint that after a goat is revealed there are two closed doors, one must
> have a goat and the other must have the car, so the odds are 50:50, is the naïve
> opinion of the ignorant masses (and, at one time, apparently a not insignificant
> number of mathematical professionals).

Um, yeah, no kidding, genius. We all know it isn't true, that was sort of a given. I was asking for you to find the specific flaw in my reasoning, as in which part of the argument is false and why it is false. So here it is again in its new incarnation:

Suppose the host has a fair six sided die with 1 on two of the faces, 2 on two of the faces, and 3 on two of the faces. Before each episode of the GSP, the host rolls the die and puts the car behind the door corresponding to the number on the die. The host also writes down this number on a slip of paper and sticks the paper in their pocket. Suppose the player knows everything about this process except for the outcome of the die roll, and thus which door has the car.

Suppose the player chooses door 1 and is shown door 3. The player knows the paper does not have a 3, but cannot rule out it having a 1 or 2. Since both these numbers are equally likely to be rolled, there is a 50/50 chance that the paper has a 1, and thus a 50/50 chance of winning by switching.

Similarly, if the player chooses door 1 and is shown door 2, the player knows the paper does not have a 2, but cannot rule out it having a 1 or 3. Since both these numbers are equally likely to be rolled, there is a 50/50 chance that the paper has a 1, and thus a 50/50 chance of winning by switching.

Thus no matter which door is shown, there is a 50/50 chance of winning by switching, contradicting robert's assertion that the player will win on average 2 out of 3 times by switching.

P.S. This has nothing to do with perspective. The probability for the paper is exactly the same as the probability for the car, since the number on the paper is the same as the number of the door with the car.
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Re: Game Show Problem

Postby robert 46 » Thu Jan 11, 2018 6:42 pm

If guardian wants to ignore the LGW issue because he can't understand it, I suggest he read the entire topic to see how the issue unfolded. It was initially introduced in the first post, Marilyn's, so it is germane to the overall GSP.
Last edited by robert 46 on Thu Jan 11, 2018 8:02 pm, edited 2 times in total.
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Re: Game Show Problem

Postby guardian » Thu Jan 11, 2018 7:44 pm

robert 46 wrote:
> If guardian wants to ignore the LGW issue because he can't understand it, I suggest
> he read the entire topic to see how the issue unfolded. It was initially introduced
> in the first post, Marilyn's, so it is germane to the overall GSP.

I presented an argument. I asked you to find the flaw in what I wrote. There are many things germane to the overall GSP that are not relevant to this particular task. I want to ignore the LGW because it is not relevant.

The probability of the paper is the same as the probability of the door. Saying that a player who chooses door 1 should switch is the same as saying the player should bet against the slip of paper in the host's pocket having a 1 on it. I am asking how these two events can be reconciled. You cannot answer because you have hidden for years behind your explanation of winning the opposite of what was behind the door initially picked instead of trying to truly grasp how the probability mechanism actually worked. Once a door is revealed, one of the numbers is eliminated from being on the paper and one of the remaining numbers can become less likely, and on average will be less likely, to be on the paper than the other remaining number. I am asking you why.
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Re: Game Show Problem

Postby robert 46 » Thu Jan 11, 2018 8:02 pm

> guardian wrote: robert's solution only shows that on average there is a 2/3rds chance of winning by switching.

Nonsense. My solution shows that for the one-time-occurrence there is a 2/3 chance of winning the car by switching. From this, one can project the expected statistics: that 2/3s of players who switch will, on average, win the car.

> Because the question asks "should the player switch?", to answer "yes" we need to show that either there is always at least a 50/50 chance of winning by switching,

There is a very easy way of establishing this using the LGW issue: The probability of the player switching to the car cannot be less than the probability of the LGW choosing the door with the car- not knowing which door is the player's. One door must have the car and the other must have the other goat: one door must be the player's and the other must be the remaining; one door must be the stay door and the other the switch door. Which is which is unknown.

The LGW perspective reduces to randomly choosing to stay or switch: the projected statistics is that half the LGWs who actually choose a door will win the car on average. If the players randomly choose to stay or switch, they can't do any worse than the LGWs on average. If choosing to stay, player then has a 1/3 chance of winning the car; so the chances of winning the car by switching must be such that the average of it and 1/3 equals 1/2.

> or...

Preempted.
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Re: Game Show Problem

Postby Gofer » Fri Jan 12, 2018 4:47 pm

> Nonsense. My solution shows that for the one-time-occurrence there is a 2/3 chance of winning the car by switching.

Obviously a lie, since I just showed that your solution actually produces the value 1/3 given the normal assumptions.
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Re: Game Show Problem

Postby robert 46 » Fri Jan 12, 2018 6:24 pm

Gofer wrote:
> > robert 46 wrote: Nonsense. My solution shows that for the one-time-occurrence there is a
> 2/3 chance of winning the car by switching.
>
> Obviously a lie, since I just showed that your solution actually produces the value
> 1/3 given the normal assumptions.

All you have shown is that the rote method cannot handle the formalized expression of the problem statement; which is the fault of the rote method- not of the problem. It can be solved analytically, using brain-power, where the programmed-automatons fail.

> > robert 46 wrote: Conclusion: ... That the host has an alternate or not is subsidiary to the
> > prize the player has chosen. So the controlling factor is the prize the player has chosen-
> > not whether the host has an alternate or not.
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Re: Game Show Problem

Postby guardian » Sat Jan 13, 2018 7:29 am

robert 46 wrote:
> > guardian wrote: robert's solution only shows that on average there is a 2/3rds
> chance of winning by switching.
>
> Nonsense. My solution shows that for the one-time-occurrence there is a 2/3 chance
> of winning the car by switching. From this, one can project the expected statistics:
> that 2/3s of players who switch will, on average, win the car.

Weren't you the guy that said we can't do statistical analysis based on one-time occurrences? At the conclusion of this game, the only "statistic" we can project is that the car is always behind whatever door it wound up being behind. We will have no real-life evidence that it can be behind any other door. We will have no real-life evidence that there is a 1/3rd chance of the car being behind each door. We will just know that it was behind one particular door 1 out of 1 time. The only thing we can "project" based on this information is that the car will always be behind this particular door. Saying that on average 2/3rds of the people who switch will win the car is a probability based on your assumptions, not a statistic based on realized evidence.
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