## Saturday, 28 February 2015

### The Objections of chrysics - Part 2

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong (meaning that chrysics and Mathematician and irishsultan and ChalkboardCowboy were all right from the start and I should have listened to them rather than arguing with them).  Fortunately, I didn't because, for me at least, this little intellectual journey has been far more interesting than it would have otherwise been.

The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).

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The discussion goes on between chrysics and me.  Please note that this has been written with one reader in mind, so I haven't provided much in the way of context.  There are, however, links below to other relevant posts which might be of assistance to other readers.  Note that the comments below that I am responding to are from /r/math over at reddit.

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First my preceding comment as context:

(quoted) chrysics: And if we say that there's a Red Mary game where the contestant has picked Red and White: your argument is again correct, there are two equally likely possibilities (MAC and MCA). The contestant has a 50% chance of winning by switching.

me (as wotpolitan): That's been my point right from The Reverse Monty Hall Problem, although it might be more clearly stated in Marilyn Gets My Goat.
As I said on my blog, everything else has been a diversion, likely due to me not making myself absolutely crystal clear. I agree that the argument is not valid when we have a Red Mary game in which Red Mary is not revealed, but this would then an isotropic White Ava or Green Ava game.

I appreciate your analysis of why, overall, the likelihood of winning from switching across multiple iterations, is 1/3 (or 2/3 in the classic Monty Hall Problem). However, ChalkboardCowboy argues that this is in contravention of the Law of Large Numbers. Would you have a response to that? (Note, my suspicion is that ChalkboardCowboy is misapplying the Law of Large Numbers.)

PS: I know that we have been at this for a long time, but could you please take another look at the original Reverse Monty Hall Problem article and see if I somehow failed to describe what I meant to describe.

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And then chrysics’ post that I am responding to:

Perhaps my last comment was unclear. I am not saying that the whole of your argument is valid, and the contestant will deduce that they have a 50% chance of winning by switching. It is emphatically not the case that the contestant has a 50% chance of winning by switching, if there are no further qualifiers applied to that statement. Your argument about equally likely options existing, and thus providing a 50% chance of winning, is valid only in specific circumstances (the contestant selected Red and Green; the contestant selected Red and White), and only to somebody with knowledge that is unavailable to the contestant (the location of Mary, even if she is not revealed).

There is also an additional scenario in which the argument does not apply. Your definition of a Red Mary game as being any game in which the Red Door holds Mary puts no constraints on which doors are selected by the contestant. As such, it requires that you consider the additional scenario (the contestant does not pick the Red Door at all. They can never win by switching, as that gets them the Red Door which by definition holds Mary) as equally likely to each of the others. The probability of winning by switching, given that Mary is behind the Red Door, is thus reduced from 50% to 1/3.

What I'm saying is that if there exists an outside observer who knows:

1. The contestant's choice of doors, and

2. The location of Mary (assumed to be red, for the sake of simplicity)
then that observer will, in some but not all scenarios, deduce that the contestant has a 50% chance of winning. In another scenario, that observer will deduce that the contestant has a 0% chance of winning by switching.

The contestant themselves can never (correctly) deduce that switching provides them with a 50% chance of winning. They do not have the same information available to them. The contestant is asked, once the door is opened and they find themselves in a Revealed Red Mary (or a Revealed Green/White Ava, as the case may be), to evaluate the probability that they win by switching. To do that, they must weigh up all of the possible ways in which they could potentially arrive at the scenario they now find themselves in, and must also consider how likely each of those ways is to actually produce the scenario they find themselves in. In doing so, they follow a procedure equivalent to that I outlined in my earlier post.

Your argument is valid only in the scenario that the Red Door is one of the contestant's two chosen doors. Your argument does not show that the contestant has a 50% chance of winning a Red Mary game. It shows that the contestant has a 50% chance of winning a Red Mary game if and only if they have selected the Red Door as one of their two. Which is true only 2/3 of the time.

The analysis I'm providing - let me make it very clear, once again - is entirely independent of how many times you play the game. It is not an analysis exclusively of the probabilities you get when playing repeatedly. Nor is it an analysis exclusively of the probabilities when playing a single iteration, but it can be treated as such if that is what interests you. Because the probabilities deduced by the contestant are in no way dependent upon how many times the game is played.

My response to ChalkboardCowboy's statement would be that ChalkboardCowboy is exactly right. I'd be interested to know why you think the law of large numbers cannot be applied here (or why you think it applies but in a different way than ChalkboardCowboy says, if that's the case).

(quoted) wotpolitan: could you please take another look at the original Reverse Monty Hall Problem article and see if I somehow failed to describe what I meant to describe.

I believe you've described exactly the same process throughout, with the exception of the one time you said that the host is committed to open the Red Door if it holds Mary. I'll accept this was an error as you've otherwise been consistent both before and after in saying the host will choose randomly if both selected doors hold goats. Your initial blog post seems perfectly clear to me. You ask for the likelihood, as deduced by the contestant, that switching will win the car. There's not really any room for ambiguity here. We can narrow it down to, for example, only considering the possibilities in which Mary is behind the Red Door, and perform the analysis that way, and this does not affect the result as it is one of many equally likely scenarios which each produce the same set of probabilities.

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First off, I have to reiterate my rather strange sounding claim that, overall, with The Reverse Monty Hall Problem, the contestant will benefit from a policy of staying, winning 2/3 of the time, if the game is played repeatedly.  I’ve not denied that.

Secondly, I’ll quote myself from The Reverse Monty Hall Problem (I keep linking to it because my original words are there and I think that many people are responding to what they think I said and not to what I actually said, I don’t feel particularly responsible for arguing for what they think I said):

One day you decide to go out to buy a new puzzle book at the massive Honty Mall.  When you enter, however, you are confronted by three doors and a rather dishevelled amateur philosopher who swiftly talks you into trying out a variation of an old game show puzzle (you obviously like puzzles, so it was an easy task).

The puzzle is put to you as briefly yet comprehensively as possible:

·         There are three doors, there is a goat behind two of the doors and behind the third is a car.

·         If, at the end of the game, you open the door with the car behind it, you win the car.

·         First, you select two doors (not the one door of the Classic Monty Hall Problem).

·         The philosopher will then open one of the doors you selected, revealing a goat.

·         You then have the option to switch from your remaining selected door or stay.

·         Before being allowed to open a door, you must provide the likelihood that a switch will win you the car (even if you choose to stay).

·         The placement of the goats and car is randomised.

Do you switch or stay, and what is the likelihood of winning from a switch?

In the exact situation in which the contestant finds herself, two doors have been selected (later described in Marilyn Gets My Goat as Red-White, Red-Green or White-Green, and specified as Red-Green) and one door has been opened (specified as the Red Door in Marilyn Gets My Goat).  While I do discuss the specific example of Mary revealed behind the Red Door (referred to as “Red Mary”) after the Red and Green Doors were selected, if I have correctly understood the term this situation is isomorphic with:

Red Mary & Red-White

White Mary & Red-White

White Mary & White-Green

Green Mary & Red-Green

Green Mary & White-Green

Red Ava & Red-White

Red Ava & Red-Green

White Ava & Red-White

White Ava & White-Green

Green Ava & Red-Green

Green Ava & White-Green

These constitute all of the situations in which the contestant might find herself after selecting two doors and after the host has opened a door to reveal a goat in a single iteration, one shot instance of the Reverse Monty Hall Problem.

You (chrysics) said:

(neopolitan's) argument about equally likely options existing, and thus providing a 50% chance of winning, is valid only in specific circumstances (the contestant selected Red and Green; the contestant selected Red and White), and only to somebody with knowledge that is unavailable to the contestant (the location of Mary, even if she is not revealed).

This sort of misses the point, but you seemed to have got the point earlier when you wrote:

If we say that there's a Red Mary game where the contestant has picked Red and Green:

your argument is correct, there are two possibilities (MAC and MCA), each of which is equally likely. The contestant has a 50% chance of winning by switching.

Because all the situations in which the contestant finds herself after the door has been opened are isomorphic, no matter which doors she selected and no matter which door was opened to reveal which goat, she will have a 50% chance of winning by switching.

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With regard to the objections of ChalkboardCowboy and the Law of Large Numbers, this law applies to “the result of performing the same experiment a large number of times”.  Remember that in the situation in which the contestant must assess the likelihood of winning as a result of switching, she has picked two doors, one door has opened and one goat has been revealed.  This opened door and revealed goat tells the contestant which particular subset of mini-games the mini-game that she is playing belongs to (in our example, MAC or MCA).  If we don’t care to distinguish between goats, then simply the opened door tells us (in our example ggC or gCg).

If we try to apply the Law of Large Numbers the way that ChalkboardCowboy is implying, then we don’t get to know which subset of mini-games apply with each iteration, we just know that it is one of the set [ACM, AMC, CAM, CMA, MAC, MCA] (or [Cgg, gCg, ggC] if we don’t care about the goat’s identity).  This means that we are not “performing the same experiment”.

It would be akin to walking into a room with three (fair) gambling machines, one with an average payout of 1/2, one with an average payout of 1/10 and one with an average payout of 1/100, selecting one at random and expecting to get a payout of about 1/5 from a single machine.  However, if you go in and play the 1/2 machine 1,000 times, then you will get close to an average payout of about 1/2.  Do it a million times and you’ll be even closer to 1/2.  (This example is not entirely analogous to what is going on with Reverse Monty Hall Problem, I know that.  I am just highlighting the fact that the Law of Large Numbers won’t work if you are playing different games.)

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So, the question I have is: did I somehow fail to make perfectly clear that I was talking about a decision made by the contestant after the door was opened (and hence after doors were selected) – a situation which, by your calculation, gives the contestant a 1/2 likelihood of winning as a consequence of switching?

And a further question is: how was my scenario substantively different to the original question raised by Craig F. Whittaker?

1. Could you please explain how "the contestant will benefit from a policy of staying, winning 2/3 of the time, if the game is played repeatedly" is compatible with "no matter which doors she selected and no matter which door was opened to reveal which goat, she will have a 50% chance of winning by switching"?

1. Hi Anonymous, this has already been addressed. Please note that I wrote "Please note that this has been written with one reader in mind, so I haven't provided much in the way of context." I suspect that you are not that reader.

That said, I do intend to draw out the explanation that I have presented in a few places already and put it into a separate article, so please check back later (it will be called "Marilyn's Six Games").

2. This comment has to be split in two because apparently there's a character limit....

You've drastically misunderstood my statements that your argument is valid. I believe I made it clear that I'm saying the argument is valid in particular scenarios, for an observer with knowledge unavailable to the contestant. It is invalid in a third, equally likely scenario. There is no possible justification for the latter scenario to not be considered a Red Mary game (Mary is still behind the Red Door). It must be considered as an equally likely possibility.

If Mary is behind the Red Door, and the contestant picks Red & Green, they have a 50% chance of winning by switching.

If Mary is behind the Red Door, and the contestant picks Red & White, they have a 50% chance of winning by switching.

If Mary is behind the Red Door, and the contestant picks White & Green, they have a 0% chance of winning by switching.

We can make these statements because we are observing from outside the game. We require Mary to be behind the Red Door, and as such we have perfect knowledge that we are in a Red Mary game. We do not rely on the host to tell us the location of one goat, we simply know from the start of the game that Mary is behind the Red door. In fact, we don't use the information resulting from the opening of the door at all (if we did, we'd create several extra possible scenarios: some of which would have a 100% chance, others a 50% chance, some 0% - but the average would be 1/3). This is independence of our knowledge from the host's opening of a door is crucial - without it, we cannot arrive at these probabilities. The contestant knows only what is revealed to them, and cannot make the same deductions. Note, though, that the three equally likely possibilities produce an average chance of 1/3 - if we know only that we are in a Red Mary game, and do not know which doors the contestant has chosen, we too would deduce a 1/3 chance of winning by switching. This is much closer (although still not identical) to the information the contestant has.

Now, considering how it's necessary for me to clarify so much what I meant by those statements, and how it does not apply to the contestant, but to an observer with knowledge unavailable to the contestant, I feel that the Revealed Red Mary scenario is a much more valuable tool for considering the question you've asked. After all, you want to know the probability as deduced by the contestant. Considering the set of "Red Mary" games produces a situation in which the calculations we can make are based on the information the contestant has! For all MCA games, even those in which the host opens Green Ava, we (unlike the contestant) know that Mary is behind the Red Door. We are not in the same position as the contestant, we are not making the same deductions. If we consider, however, only Revealed Red Mary games, then we put ourselves in the same situation as the contestant:

[...]

1. So, we shall now consider a Revealed Red Mary game. As the contestant, we have selected the Red and Green doors. Mary has now been revealed behind the Red Door. We know for a fact that we have selected the Red Door, and it held Mary. We know that it holds either Ava (an MCA game) or the car (an MAC game). The a priori likelihoods of each of those, before the door was opened, had been equal (as had the likelihoods of AMC ACM CAM CMA). Now, however, with the opening of the door, we gain new information. And we can deduce the likelihood that we find ourselves in a Revealed Red Mary game due to an MCA arrangement as being 1/3, following the process I outlined earlier. You've stated in the comments under your previous post that you agree with this, and accept that for any given Revealed Red Mary game the probabilities it arose from MAC and MCA are unequal.

Now, this question is equivalent to the one you're claiming has the answer 50%! If the contestant has picked Red and Green, then an MAC game results in them losing if they switch. An MCA game, on the other hand, will see the contestant win if they switch.

So, the billion dollar question:
You do agree that the contestant in a Revealed Red Mary game can deduce that they are, with 2/3 likelihood, in an MAC game.
You do not agree that the contestant in a Revealed Red Mary game can deduce that they will, with 2/3 likelihood, lose if they switch.

Those two statements are identical. If they are in an MAC game, they will lose by switching. If they will lose by switching, they are in an MAC game. How can you agree with one, but not the other?

Now on to your bit about the law of large numbers. Your final paragraph is broadly correct. You can't apply the law of large numbers if you're playing different games, because the probabilities are not the same. You can't sit at the 1/2 payout for a million games and use that to draw conclusions about your expected winnings from the 1/5 payout machine. But you can sit at the 1/2 payout machine for a million games and use it to draw conclusions about your expected winnings from another 1/2 payout machine, because the probabilities are the same.

It's true that we no longer know that Mary is behind the Red Door if we do not restrict ourselves to Revealed Red Mary games. But we will know that Mary is behind the White Door, or Ava is behind the Green Door, or Ava is behind the Red Door, or... etc. There is nothing special about the Revealed Red Mary scenario, any option is equally likely and puts the contestant in exactly the same situation. The labels applied ("Mary" and "Ava" to the goats, "Red", "White", "Green" to the doors) are entirely arbitrary, the game is not sensitive to them in any way.

So all similarly-defined outcomes (Revealed Red Mary/Revealed Green Ava/Revealed White Mary, etc.) are isomorphic. They produce the same set of probabilites. You've said this yourself. In fact, it's only because of this that we're even permitted to focus on a Revealed Red Mary to determine the probabilities deduced by the contestant - if they didn't all produce the same result, we'd have to consider every option individually. Since they all produce the same set of probabilities, you can indeed choose apply the law of large numbers. The result will be the same, regardless of whether you're talking about all possible RMHP games or just the Revealed Red Mary subset of games.

2. chrysics says: "If Mary is behind the Red Door, and the contestant picks Red & Green, they have a 50% chance of winning by switching."

I'm finding it difficult to reconcile this statement of yours and your implied conclusion:

"If Mary is behind the Red Door, and the contestant has picked Red & Green, and the contestant knows that Mary is behind the Red Door, the contestant has a 33.33% chance of winning by switching."

What you are apparently saying is that the contestant's knowledge of the facts affects the outcome (as opposed to their assessment of the likelihood). Remember also the Monty Falls scenario in which Monty stumbles against the door. If the contestant becomes aware that Mary is behind the door because (back in my scenario again) because the producer presses the "Open Red Door" button accidentally, what is the likelihood of winning from a switch? What if the contestant has no way of distinguishing between an accidental opening and a deliberate opening?

Part of the great appeal of the 1/2 answer in my mind is that there is no difference in the outcome from the accidental opening and the deliberate opening. Perhaps this isn't a particularly mathematical preference. :)

3. It alters the chances, because the contestant is only aware that he is in a Red Mary game 3 out of 4 Red Mary games.

In in the one where he isn't aware he is in a Red Mary go he always wins if he switches. That means that only 1 out of the remaing 3 Red Mary games he wins when he switches.

4. I may be a little brusque in this post:

As irishsultan says, that relates to the information available to the contestant. I made it very clear in several of my previous posts that the conclusion of a 50% chance is a conclusion we may make, on the basis of information we possess which the contestant does not. It is a conclusion made from a situation not isomorphic to the situation in which the contestant finds themselves, you should not expect any probability we calculate from our perspective to be the same as the probability the contestant calculates from a different perspective. Note, however, that the average probability we deduce for all potential Red Mary situations is 1/3 - exactly the same average probability the contestant will arrive at.

Meanwhile I shall note that once more, for entirely unclear reasons, you are choosing to ignore the "Red Mary, contestant picks White & Green" scenario in your attempt to determine the probability that the contestant wins an arbitrary Red Mary game by switching. This is choice is entirely without foundation, and I'm not interested in continuing the discussion unless you:
1. acknowledge that you cannot correctly calculate the chance that a contestant in a Red Mary game wins by switching unless you account for this scenario, or:
2. give a clear and valid reason for why this scenario can be discarded from the set of all Red Mary games

"What you are apparently saying is that the contestant's knowledge of the facts affects the outcome (as opposed to their assessment of the likelihood)"

Not at all, in fact I'm saying exactly the opposite. The actual likelihood is entirely independent of the contestant's knowledge. As soon as the contestant chooses their doors, the actual likelihood of winning by switching becomes either 0% (2/3 of the time) or 100% (1/3 of the time). The contestant, however, does not know the entire game state, and can only determine the likelihoods of different scenarios fitting the information available to them.

I actually don't remember the Monty Falls scenario at all, and frankly I'm not interested in sitting down to think about how or indeed if it affects the outcome. What I will say is that if the door-opening is no longer dependent upon the host's knowledge that the door they're opening holds a goat, the rules of the game have been fundamentally changed (it is now possible, even if it did not occur in some given instance, that the host may open the door which holds the car). It should be no surprise to you that a game played by different rules may produce a different set of probabilities.

There's one very big question which I've asked you in each of my last two comments (well, assuming you count the two-parter as being one comment). You've ignored it both times, so I'll try again.

Claim 1: a contestant in a Revealed Red Mary game, who chose the Red and Green doors, can deduce that they are, with 2/3 likelihood, in an MAC game.
Claim 2: a contestant in a Revealed Red Mary game, who chose the Red and Green doors, will lose by switching if they are in an MAC game.
Claim 3: the contestant in a Revealed Red Mary game, who chose the Red and Green doors, can deduce that they will, with 2/3 likelihood, lose if they switch.

It's been clearly established in the comments on your previous blog post that you agree with Claim 1.
I have no doubt that you also agree with Claim 2, which is a priori true from to the definition of the game.
You disagree with Claim 3. How can that be? The final claim is logically equivalent to the combination of Claim 1 and Claim 2, both of which you accept.

This is the third time I've asked you to explain this contradiction. I'm not interested in discussing further if you can't provide a reasonable response.

5. I'm sorry that I didn't respond to your billion dollar question. I did intend to but was called away and by the time I got back my priority was to acknowledge that I messed up with "Monty Two Face" and by then I had forgotten my earlier intention.

"You do agree that the contestant in a Revealed Red Mary game can deduce that they are, with 2/3 likelihood, in an MAC game.
You do not agree that the contestant in a Revealed Red Mary game can deduce that they will, with 2/3 likelihood, lose if they switch.

Those two statements are identical. If they are in an MAC game, they will lose by switching. If they will lose by switching, they are in an MAC game. How can you agree with one, but not the other?"

I don't agree with the first line. I believe that the contestant, knowing that a Red Mary has been revealed, knows that there are only two possible mini-games remaining, MAC and MCA. I believe that given that MAC and MCA were equally likely distributions of the goats and car, and the doors selected were selected at random, and if it's an MCA game then the door opened was selected at random, and the contestant knows that "If Mary is behind the Red Door, and the contestant picks Red & Green, they have a 50% chance of winning by switching", then the contestant knows that the contestant has a 50% chance of winning by switching. This is a very specific situation, with many caveats, I know that. However ...

When the contestant in my scenario has selected Red and Green and the Red Door has been opened to reveal Mary, all of these caveats have been satisfied.

Can you see that I don't actually have the contradiction that you think I have?

6. You are making the same error that you made in the HH/HT example. It's true that both dice are equally likely to be thrown, but once you know a H resulted it's more likely that the HH dice was the one thorwn.

7. Exactly, The HH coin is twice as likely to produce a 'heads' as the HT coin. In the RHMP. Mary is twice as likely to be revealed behind the Red door in a MCA game than in a MAC game.
Neopolitan you are making exactly the same mistake as you did in your Monty Two Face game, and if you do the same conditional probability calculation here as you did there you'd get the correct answer of 2/3.
They are to all intents and purposes the SAME problem.

8. The above argument is based on the assumption you've selected the RED and WHITE doors (or the left and middle doors)

3. There were two questions I demanded answers to, you only answered one of them. It's not a satisfactory answer, for reasons I've already explained multiple times, but I'll accept that for now and give you one final opportunity to answer the remaining question:

You are choosing to ignore the "Red Mary, contestant picks White & Green" scenario (which is not isomorphic to the "Red Mary, contestant picks Red & Green" scenario) in your attempt to determine the probability that the contestant wins an arbitrary Red Mary game by switching. This is choice is entirely without foundation, and I'm not interested in continuing the discussion unless you:
1. acknowledge that you cannot correctly calculate the chance that a contestant in a Red Mary game wins by switching unless you account for this scenario, or:
2. give a clear and valid reason for why this scenario can be discarded from the set of all Red Mary games

I'll also point out that you've apparently got no response to my earlier argument for why the Law of Large Numbers can indeed be applied to this situation. You're yet to provide any explanation for how you believe a strategy can provide a 1/2 chance of winning any single iteration, yet provide a 2/3 chance of winning over many iterations. This is a statement that's fundamentally at odds with basic probability theory, and you keep just brushing it under the carpet.

I've been convinced you're trolling pretty much from the start, but now I'm finding it difficult to even pretend to assume good faith.

"I don't agree with the first line."

You said yourself in the comments on your previous post (I've repaired the typing error which we confirmed in that discussion):
"for any Revealed Red Mary, the likelihood is 2/3 that it is due to there being an MAC configuration of goats and car. I agree. I said as much in Marilyn Gets My Goat (see Q10)". You also made equivalent statements in the 'update' part of the post itself and, as you point out, in your post entitled "Marilyn Gets My Goat".

A real key point here is that when you said it yourself, you were doing so in the voice of Marilyn - the contestant. This leaves no room for ambiguity at all, you can't argue that this is somehow a conclusion we may reach but the contestant may not. You clearly state that the contestant (who has picked Red & Green) in a Revealed Red Mary game can deduce a 2/3 likelihood of being in an MAC game.

Once more, you entirely ignore the fact that the deduction of a 50% chance is made from a privileged position, with information unavailable to the contestant. It is not isomorphic to the contestant's view of the situation, and there is no reason you should expect the probabilities deduced by that observer to match the probabilities deduced by the contestant. You state that all the caveats have been satisfied, but that's simply not true. I've made it clear, repeatedly, that our ability to determine that 50% chance is entirely dependent upon the fact we are not deducing based on the information available to the contestant, but on the basis of our perfect knowledge, unavailable to the contestant that we are in a Red Mary game, regardless of which door is revealed.

So no, I do not see how this contradiction is imagined on my part. What I see is that you are denying a statement you've made quite clearly on more than one occasion.

"Claim 1: a contestant in a Revealed Red Mary game, who chose the Red and Green doors, can deduce that they are, with 2/3 likelihood, in an MAC game.
Claim 2: a contestant in a Revealed Red Mary game, who chose the Red and Green doors, will lose by switching if they are in an MAC game.
Claim 3: the contestant in a Revealed Red Mary game, who chose the Red and Green doors, can deduce that they will, with 2/3 likelihood, lose if they switch.

It's been clearly established in the comments on your previous blog post that you agree with Claim 1.
I have no doubt that you also agree with Claim 2, which is a priori true from to the definition of the game.
You disagree with Claim 3. How can that be? The final claim is logically equivalent to the combination of Claim 1 and Claim 2, both of which you accept."

I don't agree that a contestant can, under those circumstances, "can deduce that they are, with 2/3 likelihood, in an MAC game".

What the contestant knows is that when there is a Red-Green Red Mary game being played, which the contestant now knows to be the case, the Red Mary will be revealed for certain if the Red Mary game is a MAC and only half of the time if the Red Mary game is a MCA. I believe that this knowledge that a revealed Red Mary is less likely in an MCA game counteracts (in a single iteration of the game) the fact that the Red Mary will be revealed less often in an MCA game.

This is supported by the conditional probability equation (which I know is claimed by some to be inapplicable), which gives us:

Pr(Green Car|Red Mary) = Pr(Green Car AND Red Mary) / Pr (Red Mary) = Pr(MAC)|Pr(Red Mary) = 1/6 / 1/3 = 1/2

You've already told me that this is correct. We are just arguing about how to correctly think about a revealed Red Mary in a single iteration, one shot instance of the Reverse Monty Hall Problem. Once the door is opened, revealing the Red Mary, I consider that the fact that the game is a Red-Green Red Mary game is no longer privileged information. The contestant can see the goat and she knows which doors she selected.

2. But you are not asking for Pr(Green Car|Red Mary) in the original post you are asking for Pr(Green Car|User knows he is in Red Mary).

Pr(Green Car AND (User knows he is in Red Mary))/ Pr(User knows he is in Red Marry) = Pr(MAC-H)/(Pr(MAC-H) + Pr(MCA)) = (1/12) / (1/12 + 1/6) = (1/12) / (3/12) = (1/12) / (1/4) = 4/12 = 1/3

3. neopolitan: There are two problems with that.

The first is that that is not my other question. That is the same question you already answered. The second problem is that you've answered it in the same way as last time, and chosen not to address any of the problems I had with that answer. You are yet to explain how my claim:
"a contestant in a Revealed Red Mary game, who chose the Red and Green doors, can deduce that they are, with 2/3 likelihood, in an MAC game" (which you disagree with)

"for any Revealed Red Mary, the likelihood is 2/3 that it is due to there being an MAC configuration of goats and car. I agree. I said as much in Marilyn Gets My Goat (see Q10)" (which you self-evidently agree with)

My other question is the one I quoted immediately after pointing out that you'd ignored it. Here it is again, highlighted in bold for your convenience:

You are choosing to ignore the "Red Mary, contestant picks White & Green" scenario (which is not isomorphic to the "Red Mary, contestant picks Red & Green" scenario) in your attempt to determine the probability that the contestant wins an arbitrary Red Mary game by switching. This is choice is entirely without foundation, and I'm not interested in continuing the discussion unless you:
1. acknowledge that you cannot correctly calculate the chance that a contestant in a Red Mary game wins by switching unless you account for this scenario, or:
2. give a clear and valid reason for why this scenario can be discarded from the set of all Red Mary games

You've also - again - made no attempt to explain why you believe I'm wrong to assert that the law of large numbers applies to this game.

The final section, in which you derive a likelihood of 1/2, is invalid for exactly the same reasons I explained to you multiple times already:

You asked for the probability that switching will win the car, as deduced by the contestant. The most recent comment by irishsultan shows the correct calculation to answer this question.

You did not ask for the probability that switching will win the car, as deduced by an outside observer who knows where Mary is located (independent of her being opened) and who also knows that the contestant has chosen Mary's door as one of their two. This is not the question you asked, nor does it have the same answer as the question you've asked. But this is the question you keep answering, for reasons which remain unclear.

The probabilities deduced by the contestant must be, by definition, based on the information available to the contestant. But the information you are using when you derive the 50% chance does not correspond to the information available to the contestant. It corresponds to the information available to an observer with a perfect knowledge of whether or not they are in a Red Mary game, even if a different door is opened.

This is exactly the issue I am addressing in the question you've still provided no response to. I'll thank you for being unfailingly polite through this whole thing even while many others were not, but at this point I believe you're intentionally refusing to answer this question because you cannot provide any reasonable response. I'm not prepared to continue further even if you do now provide an answer to that question, as I really can't be bothered continuing to rehash the same points while you repeatedly quote the part of my comment from a few days ago which sounds like I agree with you when taken without the context provided by the rest of the comment.

4. I'm not ignoring you, chrysics, I'm just very busy at the moment. I am still thinking about your highlighted text.

irishsultan, yours is a very appealing presentation of what I think that chrysics is also saying. I am also thinking about it.

5. "I'm not ignoring you, chrysics, I'm just very busy at the moment."

That has nothing to do with it. I didn't suggest you were ignoring me on the basis of not replying instantly. I said you ignored that question, because you replied multiple times to comments in which I'd made a big deal of that question without even acknowledging it (nevermind attempting to answer it).

6. It's so much that I was worried that I had ignored you, which is a fair accusation even though it was unintentional, but that I was worried that you would think that I was continuing to ignore you, which simply wasn't the case. I've now addressed the objections of your good self, irishsultan and ChalkboardCowboy.

4. neopolitan said, "However, if you go in and play the 1/2 machine 1,000 times, then you will get close to an average payout of about 1/2. Do it a million times and you’ll be even closer to 1/2."

This is exactly the way I mean to apply the Law of Large Numbers. Run a million iterations of RMH, and divide them into separate groups, such as "Red/Green chosen, Red Mary revealed", corresponding to the complete information that the contestant has available immediately after the host opens a door. Think of these as the "machines".

Then by considering ONLY the "Red/Green chosen, Red Mary revealed" runs, we have in effect conducted the same experiment under identical circumstances a large number of times, and the LoLN tells us that the probability of winning by staying in any single instance should be close to the proportion of instances in which staying was observed to be the winning move.

Note, since the "machines" are defined only by information the contestant has access to, that we always know which "machine" we're playing. The simultaneous belief that the one-time chance of winning (by staying) at the "Red/Green chosen, Red Mary revealed" game is 1/2, while the proportion of wins over many runs is 2/3, cannot be reconciled with the LoLN.

I hope now it's clear exactly how I propose to apply the LoLN, and that there's absolutely no ambiguity arising from "which minigame" the contestant may or may not be playing. It's the "Red/Green chosen, Red Mary revealed" minigame, and the contestant knows it.

1. I see where you are going. I do have another argument for you, but as said above to chrysics, I am very busy at the moment. Consider this a place-holder.

Feel free to comment, but play nicely!

Sadly, the unremitting attention of a spambot means you may have to verify your humanity.