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 Gofer » Mon Feb 12, 2018 7:08 am

Robert46> 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).

The useful information is that the host chose that particular door over the one he also could have chosen. And if the host must always reveal a goat, then the mere goat-revelation alone likewise provides no useful information - which I proved by evaluating Pr(car behind third door|host reveals a goat) to 1/3.
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Re: Game Show Problem

Postby guardian » Tue Feb 13, 2018 6:59 am

To paraphrase robert 46:

That robert 46 appears to have ceased posting, indicates to me that he recognizes he is on the losing side of the debate.

I asked two questions, and you seem to be using evasion through evasion.
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Re: Game Show Problem

Postby robert 46 » Tue Feb 13, 2018 8:38 am

If players decide to stay with their initial choice of door, 1/3 on average are expected to win the car. If players randomly choose either of the two closed doors, 1/2 on average are expected to win the car. Randomly choosing a door combines staying with switching, and means that some choose to stay and others to switch. With a binary random choice, 1/2 on average are expected to make one choice, and 1/2 the other choice.

Given 2*n players (n a large increasing number), on average n are expected to stay, and n are expected to switch. Thus of the n who stay 1/3*n=n/3 are expected to win the car. The number expected to win the car by switching is n-n/3=2*n/3. If all players choose to stay then m/3 are expected to win the car, but if all players choose to switch then 2*m/3 are expected to win the car.

So we see that randomly choosing to stay or switch gives greater chances of winning the car than staying, and that switching gives greater chances of winning the car than randomly choosing to stay or switch. Whereas C>B>A, necessarily C>A; so switching is the preferable choice. Which other door is opened by the host to reveal a goat, and whether the host had to, or chose to, open the door he did, is entirely irrelevant to this analysis.



P.S. It is forum etiquette to post a response in five days or sooner. My previous post was two days ago, so guardian appears to be "chomping at the bit", as-it-were.
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Re: Game Show Problem

Postby Edward Marcus » Tue Feb 13, 2018 12:03 pm

robert 46 wrote:
> Which other door is opened by the host to reveal a goat, and whether the host had
> to, or chose to, open the door he did, is entirely irrelevant to this analysis.


It is irrelevant to your analysis-because you are basing your analysis on the assumption the constraint the goat is to be revealed is the only factor of the probability, whereas the other camp is basing their probability on the specificity of the door which is revealed when the player chooses the car.
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Re: Game Show Problem

Postby robert 46 » Tue Feb 13, 2018 12:50 pm

Edward Marcus wrote:
> robert 46 wrote:
> > Which other door is opened by the host to reveal a goat, and whether the host had
> > to, or chose to, open the door he did, is entirely irrelevant to this analysis.
>
>
> It is irrelevant to your analysis-because you are basing your analysis on the assumption
> the constraint the goat is to be revealed is the only factor of the probability,
> whereas the other camp is basing their probability on the specificity of the door
> which is revealed when the player chooses the car.

As I have explained: There is nothing in the problem statement which implies the host has any favoritism for a goat door to open; there is no indication that the player is aware of any host favoritism; there are no values related to host favoritism provided which would be relevant to a consideration of host favoritism. When there is no implication to the contrary, a variable is reasonably assumed to be random: this includes placement of the prizes, player choice of a door, host choice of a goat-door to open. There is no implication to the contrary for any of these three variables, so they are reasonably assumed random in the context of the problem statement. The analysis of the problem is from the perspective of what the player knows and can reasonably infer: the player can only reasonably infer that these three variables are random, so the analysis is constrained to ignore speculation related to host choice of a goat-door to open.

My previous analysis shows that host choice of a goat-door to open is irrelevant to the conclusion that switching has a higher probability of winning the car than randomly choosing a closed door, which has a higher probability of winning the car than staying.
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Re: Game Show Problem

Postby guardian » Tue Feb 13, 2018 1:44 pm

robert 46 wrote:
> If players decide to stay with their initial choice of door, 1/3 on average are expected
> to win the car. If players randomly choose either of the two closed doors, 1/2 on
> average are expected to win the car. Randomly choosing a door combines staying with
> switching, and means that some choose to stay and others to switch. With a binary
> random choice, 1/2 on average are expected to make one choice, and 1/2 the other
> choice.
>
> Given 2*n players (n a large increasing number), on average n are expected to stay,
> and n are expected to switch. Thus of the n who stay 1/3*n=n/3 are expected to win
> the car. The number expected to win the car by switching is n-n/3=2*n/3. If all
> players choose to stay then m/3 are expected to win the car, but if all players
> choose to switch then 2*m/3 are expected to win the car.
>
> So we see that randomly choosing to stay or switch gives greater chances of winning
> the car than staying, and that switching gives greater chances of winning the car
> than randomly choosing to stay or switch. Whereas C>B>A, necessarily C>A; so switching
> is the preferable choice. Which other door is opened by the host to reveal a goat,
> and whether the host had to, or chose to, open the door he did, is entirely irrelevant
> to this analysis.

I gave an answer is two sentences. This rambling, pointless exercise is NOT parsimonious.

In fact, any explanation that uses the value of 2/3 is not parsimonious. This is because:

1) The probability of winning by switching may not be 2/3 at the moment the player makes their final decision.
2) The question can be answered conclusively without this value.

A particular door is revealed every time the car is behind the other door not chosen by the player, not necessarily every time the car is behind the door chosen by the player, and never when the car is behind that particular door. Since we assume the car is equally likely to be placed behind each of the 3 doors, it cannot be more likely that particular door is revealed when the car is behind the door chosen by the player than when it is behind the other door not chosen by the player. So there must be at least a 50% chance of winning by switching. Any discussion of 2/3 is unnecessary and thus not parsimonious.
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Re: Game Show Problem

Postby Gofer » Thu Feb 15, 2018 4:28 am

guardian> So there must be at least a 50% chance of winning by switching.

While possibly true, the problem asks if this chance is larger ("advantage"); so any solution must address this.

------

GSP : A prize (Z) is placed behind any of three closed doors D:={p,x,h}. The player must then choose exactly one door in D, whereupon the host (H) must open a door in D except for the player's and the prize's, and then give the player the option to switch to the third door.

scenario 1: the player has chosen p, and the host opened h.
scenario 2: scenario 1 + host favors no door.
scenario 3: scenario 1 + host favors doors in order h,x,p.

q: is Pr(Z=x|H=h) larger than Pr(Z=p|H=h)?

---------

To my understanding, Robert's model equates scenario 1 and 2 - which is error - and equates the answer to q for 2 and 3 because "all that matters is that the host revealed a [no-prize door]" - which again is error.
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Re: Game Show Problem

Postby guardian » Thu Feb 15, 2018 5:53 am

Gofer wrote:
> guardian> So there must be at least a 50% chance of winning by switching.
>
> While possibly true, the problem asks if this chance is larger ("advantage"); so
> any solution must address this.

Then I will be more explicit: The phrase "not necessarily every time the car is behind the door chosen by the player" implies that we (and the player) do not know how the host chooses a goat door when the player initially chooses the door with the car. At worst, there is a 50% chance of winning by switching, but we are not assuming anything that would allow us to distinguish the "at worst" case from all of the other cases. Since it can't result in worse odds and will often (if not always) result in better odds, the player should always switch.

If you want to be even more explicit, we could say "switching will always result in better odds unless the host always reveals door 3 when the player chooses door 1 and the car is behind door 1, in which case it is 50/50 either way." Personally, I feel this is overkill.
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Re: Game Show Problem

Postby Gofer » Thu Feb 15, 2018 6:33 am

What guardian is trying to say, albeit oratorically, is that Pr(Z=x|H=h) can't be less than Pr(Z=p|H=h) under the normal assumptions, meaning we are no worse off switching, and we should therefore switch, given the chance.
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Re: Game Show Problem

Postby robert 46 » Thu Feb 15, 2018 9:09 am

guardian wrote:
> 1) The probability of winning by switching may not be 2/3 at the moment the player
> makes their final decision.

This shows that guardian does not understand the concept that the probability is calculated from a particular perspective, and that there can be many different perspectives. In the problem we are given, the probability is to be calculated from the player's perspective. Guardian seems to believe that there is some absolute probability which is in effect immediately prior to the player's decision to switch. I can assure guardian that this problem is not quantum mechanical and that there is no collapse of the probability wave function.

> 2) The question can be answered conclusively without this value.

But from the player's perspective, 2/3 chance of winning the car by switching is exact based on the information available and reasonable assumptions.

> A particular door is revealed every time the car is behind the other door not chosen
> by the player, not necessarily every time the car is behind the door chosen by the
> player, and never when the car is behind that particular door. Since we assume the
> car is equally likely to be placed behind each of the 3 doors, it cannot be more
> likely that particular door is revealed when the car is behind the door chosen by
> the player than when it is behind the other door not chosen by the player. So there
> must be at least a 50% chance of winning by switching. Any discussion of 2/3 is
> unnecessary and thus not parsimonious.

It is necessarily parsimonious because it takes into consideration all that is known and reasonably assumed by the player, and does not take into consideration any speculation which cannot be resolved to change the expected 2/3 chance of winning the car by switching.
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Re: Game Show Problem

Postby Edward Marcus » Thu Feb 15, 2018 1:29 pm

robert 46 wrote:
> In the problem we are given, the probability is to be calculated from the player's
> perspective


Robert- it has been shown the probability does not need to be calculated for the question of the optimal GSP strategy to be effectively addressed.
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Re: Game Show Problem

Postby robert 46 » Thu Feb 15, 2018 2:47 pm

Edward Marcus wrote:
> robert 46 wrote:
> > In the problem we are given, the probability is to be calculated from the player's
> > perspective
>
>
> Robert- it has been shown the probability does not need to be calculated for the
> question of the optimal GSP strategy to be effectively addressed.

The probability of choosing the door with the car is fundamental and "up front". It is deduced that the player will switch to the opposite prize from the one initially chosen. So the probability of the result of switching is consequent to the prize initially chosen. Nothing the host does changes this fundamental relationship.

The player is more likely to choose a goat, so the host is more likely to have had to open a particular goat-door; so the player is more likely to switch to the car. This shows that the intermediate statement is irrelevant because it is derived from the first as an auxiliary fact.
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Re: Game Show Problem

Postby guardian » Thu Feb 15, 2018 6:28 pm

robert 46 wrote:
> guardian wrote:
> > 1) The probability of winning by switching may not be 2/3 at the moment the player
> > makes their final decision.
>
> This shows that guardian does not understand the concept that the probability is
> calculated from a particular perspective, and that there can be many different perspectives.

Someone who believes in magical "probability swaps" should probably not be commenting on what I do and don't understand.

> In the problem we are given, the probability is to be calculated from the player's
> perspective.

Correct. And the player now knows more than "2/3rds of the time I pick a door with a goat."

>Guardian seems to believe that there is some absolute probability which
> is in effect immediately prior to the player's decision to switch.

There is a new probability mechanism in place at this time. The player knows the mechanism, but does not know all of the values needed to calculate an exact probability.

> I can assure
> guardian that this problem is not quantum mechanical and that there is no collapse
> of the probability wave function.

Attempt at humor?

> > 2) The question can be answered conclusively without this value.
>
> But from the player's perspective, 2/3 chance of winning the car by switching is
> exact based on the information available and reasonable assumptions.

It is not exact and it is not necessary. The player has sufficient information to utilize the new probability mechanism.

>
> > A particular door is revealed every time the car is behind the other door not chosen
> > by the player, not necessarily every time the car is behind the door chosen by
> the
> > player, and never when the car is behind that particular door. Since we assume
> the
> > car is equally likely to be placed behind each of the 3 doors, it cannot be more
> > likely that particular door is revealed when the car is behind the door chosen
> by
> > the player than when it is behind the other door not chosen by the player. So
> there
> > must be at least a 50% chance of winning by switching. Any discussion of 2/3 is
> > unnecessary and thus not parsimonious.
>
> It is necessarily parsimonious because it takes into consideration all that is known
> and reasonably assumed by the player, and does not take into consideration any speculation
> which cannot be resolved to change the expected 2/3 chance of winning the car by
> switching.

It does NOT take into consideration "all that is known and reasonably assumed by the player." Once the player chooses door 1, the player knows there are 4 possibilities: the car is behind door 3 and the host reveals door 2 (probability=1/3); the car is behind door 2 and the host reveals door 3 (probability=1/3); the car is behind door 1 and the host reveals door 2 (probability=[1/3]*X, 0<=X<=1); the car is behind door 1 and the host reveals door 3 (probability=[1/3]*[1-X], 0<=[1-X]<=1). When a door is revealed, two of these possibilities are eliminated, leaving a possibility that had probability 1/3 and a possibility that had probability of the form [1/3]*Y where 0<=Y<=1. The possibility that had probability of 1/3 must be at least as likely as the possibility with probability [1/3]*Y. ALL of this is known to the player. The only "speculation" here is that the car was initially equally likely to be behind each door and that the host would not reveal the car or the player's door, all of which are used in your answer as well. Everything else follows directly, is always true given the assumptions, and is knowable to the player at that point in the game. The only thing the player cannot do is calculate the exact value of (1/3)/[(1/3)+(1/3)*Y], but since this simplifies to 1/(1+Y), and 0<=Y<=1, then the probability of winning by switching must now be between 1/2 and 1. And that is all the player needs to make the correct decision. It could still be 2/3, but the player doesn't know that, and the player doesn't need to know it.

Think back to my marble game - the host has 2 bags and puts 10 red marbles in each bag. The host then takes 10 blue marbles, divides them between the bags, and then has the player guess which color marble will be selected out of a randomly chosen bag. You could say the player should bet on red because 2/3rds of the marbles are red. But the better, and more relevant answer is that the player should bet on red because no matter which bag the marble is being selected from, there will always be at least as many red marbles in the bag as blue marbles. This is the difference between your solution and mine.

Using a better version of your large number analysis, let's say 30,000 players choose door 1. We would expect the car to be behind door 1, 2, and 3 10,000 times each. So we know at least 10,000 players will be shown door 2, and at least 10,000 will be shown door 3. The total number shown door 2 will be 10,000+A and the total number shown door 3 will be 20,000-A, where A is the number of 10,000 players that play a game with the car behind door 1 who are shown door 2. The percentage of players who are shown door 2 that would win by switching is 10,000/(10,000+A), which must be between 1/2 and 1. The percentage who are shown door 3 that would win by switching is 10,000/[20,000-A], which must also be between 1/2 and 1. Playing with the math, we see one number will always be between 1/2 and 2/3 and the other will always be between 2/3 and 1, but we have no way of knowing which is which. Since 1/2 is the worst case scenario, it is clear that always switching is the best policy.
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Re: Game Show Problem

Postby robert 46 » Thu Feb 15, 2018 9:03 pm

There is no factor X which is knowable to the player. Focusing on speculative hidden motives of the host is an odd fixation.
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Re: Game Show Problem

Postby guardian » Fri Feb 16, 2018 5:22 am

robert 46 wrote:
> There is no factor X which is knowable to the player. Focusing on speculative hidden
> motives of the host is an odd fixation.

Refusing to include all of the information and structure of the problem is even odder. If I hand you a bag and tell you it has 10 red marbles and no more than 10 blue marbles, which color do you bet on being drawn? Unless your answer is something other than "red, because there are at least as many red marbles as blue", you can stop with the whole "speculative" nonsense. The motives of the host are irrelevant, because it can be easily shown that there is no way to reveal a door where the "win by not switching" probability is greater than the "win by switching" probability.

robert 46 wrote:
> The probability of choosing the door with the car is fundamental and "up front".
> It is deduced that the player will switch to the opposite prize from the one initially
> chosen. So the probability of the result of switching is consequent to the prize
> initially chosen. Nothing the host does changes this fundamental relationship.

No, only the RESULT of switching is consequent to the prize initially chosen. The player starts with a 1/3rd chance of initially selecting the car, and ends with a 0% or 100% chance. The probability changes as information is learned. The host cannot change the prize behind the player's door, but the host can take actions that change the player's assessment about how likely their door is to have the car. For example, if the host honks a horn 95% of the time the player chooses the door with the car and 5% of the time the player chooses a door with a goat, and blows a whistle 5% of the time the player chooses the door with the car and 95% of the time the player chooses a door with a goat, a player hearing a horn knows they probably chose the correct door. They would assess a new probability of winning by switching that was significantly less than 2/3. Where the initial 2/3rds probability comes in is that the horn is only about half as likely to be sounded as the whistle. But once a sound is heard, the initial probabilities are useless.

Likewise, when the host opens a door, information is learned and the initial probabilities can change. A solution, like mine, can say what they change to; or a solution, like yours, can try to explain why the new information is no better than the original information. My explanation is superior because it definitively shows what the new probability mechanism is and how it can be used to answer the question, and it is more concise. Yours requires ancillary explanations that justify using an average probability, which seems to stem from some oddball belief that we must have a value for the probability. As with my bag of marbles example, the question doesn't ask for the probability value, and we don't need it to answer the question.
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