A Response for Charlie

Poor Charlie!  Being another one of numerous people called Anonymous, his comment was filed in the spam directory.  Never mind, it has been extracted, cleaned off and put back where it should be.  However, I am going to call him Charlie rather than Anonymous, just for the sake of the on-going discussion.  (I did toy with calling him Peewee, because he's the "Poisoning / Pimping the Well" guy, but he might have thought I was just insulting him.  We don't want that.)

Anyhoo, here's his comment:

I am the anonymous who reviewed and confirmed that this author resorted to fallacies in the previous blog post.

Since I am partially the subject of the author's blog post here, let me lay out for the reader the context of this discussion.

Everything starts with an innocuous Barnes blog post, where, in the comment section, a seemingly animated commenter going by 'neopolitan sokare', and later 'n30p0litan' investigates Barnes' background, and upon finding him supported by a Templeton grant, becomes agitated by it.

The unfortunate odor of atheist fundamentalism becomes pronounced at that point. This type of atheist, as you know, is deeply angered by and prejudiced against academic work that might possibly be construed as sympathetic to theism, and 'Templeton' is one of their dog whistles.

The more moderate atheists among us have no problem with Templeton because Templeton provides valuable support to a diverse set of initiatives and does a good job of affording research independence.

So how does n30's hostility come to bear at this point? He demands that Barnes explain himself for accepting Templeton support given his neutrality on the question of the theological value of fine tuning.

To n30 and fundamentalists like him, you're not allowed to take Templeton support if you are neutral on the theological question. That's forbidden. To do so, in his own words, is to "act from a covert position as a theist with apologetic leanings". Thus n30 demands Barnes state his public commitment to the fundie line, and promise he's not "one of them".

It's easy to see how atheist fundamentalism poisons academia with such overbearing and paranoid Orwellian demands and pressures to commit to their ideology, which is where I repudiate this kind of extremism. It hurts academia, is totally unnecessary, and ends up making atheists look like lunatics!

But the real issue here is the well-poisoning charge. Why do I think n30 is well-poisoning? Because of n30's absurd Templeton rage.

I don't think taking Templeton money on its own says anything about your allegiance to theism or atheism, nor is it, on its own, a mark against your scholarship. Plenty of atheists I know have taken Templeton money. None of them have seen or spoken of undue influence or produced compromised scholarship.

As I've pointed out, only fundamentalist atheists have this problem. Given the paucity of evidence for it, the only basis for arguing such a point is on faith (of a slightly unusual sort, given the atheism of its proponents).

I stand by the view that charging that someone's work is tainted on the basis of their funding or personal views alone, with no further evidence or engagement with their claims is well-poisoning.

Stop embarrassing the rest of us with this silly ghostbusters witchhunt. Let us, both atheists and theists, get to work in peace.

I'm not entirely sure why WordPress sometimes lists me as "neopolitan sokare" and sometimes as "n30p0litan", the latter was chosen merely because "neopolitan" wasn't available (perhaps for the same reason that it was not available at BlogSpot, hence the use of "wotpolitan", which is just a little play on words).  I have put, at the bottom of various comments, the name I go by (neopolitan) and this blog is titled "neopolitan's philosophical blog", so it should be easy to work this out.  If, on the other hand, you want to refer to me as n30p0litan or n30 or wotpolitan or wot, then I guess you can.  I even accept neo, but I must stress that I've been using this nick (and before that one similar to it) for almost two decades, well before the Matrix came out and almost forced me to abandon it.   No matter which nick you use, I'll probably know who you are talking about.

So, to the crux of your complaint.  Am I unreasonably "agitated" by Barnes' association with Templeton?  I am clearly "agitated" to some extent, although I would characterise it as "being motivated to act".  I don't think this level of "agitation" is unreasonable though, nor do I think that my "agitation" warrants labelling me as a "fundamentalist atheist", whatever that is supposed to mean.

I do have to turn the spotlight on you for a moment, Charlie.  Where are you coming from?  What's your motivation in this?  Why does Barnes deserve or need your protection?  If I have, as you seem to imply, significantly wronged Barnes, why is he hiding behind your skirts and not speaking out?  (He's currently putting a lot of effort into skewering Richard Carrier who is a much bigger fish, so either I am not important enough or he doesn't have the time, either of which is okay.  Especially if he has lapdogs like Charlie to yap at small fry like me.)

You, Charlie, seem to imply once or twice that you are a "moderate atheist", or at least to allow the appearance of such an implication – "the more moderate atheists among us" and "(s)top embarrassing the rest of us".  Which "us" is this precisely?

You make a claim that there is some sort of fundamentalist core within academia, one that demands that everyone toe the line with respect to some sort of atheist ideology.  I'd like to see evidence of that.  I agree that it's built-in as far as engineering and maths goes and, I would argue, proper physics.  You can't design a building and have the load bearing structure be supported by your god, you'll get the sack pretty quickly if you did that.  Your design must have an inherent presumption of the lack of a god.  A mathematical proof has no need of a god hypothesis.  Actual physics doesn’t call on god to explain phenomena, although it is true that some physicists (say cosmologists and astronomers, for example) call on theology to fill the gaps in knowledge that normally would be covered by the phrase "we just don't know".  NASA engineers trying to build the craft for some future manned mission to Mars won't put up with advice from an astrophysicist that includes references to divine intervention.

But none of this is ideology.  It's pragmatism.  The god that Barnes seems to believe in (and the one that I suspect that you believe in, Charlie) doesn't ever intervene.  It is effectively non-existent, and very likely actually non-existent.  Pondering this god while there are real issues to contend with is a massive waste of time.  Treating it as if it were real is worse.

Templeton is a god-bothering organisation.  We both know it.  I don't think you have even tried to deny it.  It preferentially hands out money to people who are furthering its god-bothering agenda – and note that assistance by the recipient doesn't have to be intentional.

You say you know "plenty of atheists (who) have taken Templeton money" none of whom have "seen or spoken of undue influence or produced compromised scholarship".  Hopefully, you'll forgive me if I find this apparently partisan anecdote to be unconvincing.  I don't know who you are, I don't know how many people you consider "plenty" and I don't know in which field(s) these atheists you know work.  There are plenty of people around who do object to the involvement of Templeton (Jerry Coyne, John Horgan, Richard Dawkins, Daniel Dennett, for example).  Note a competing anecdote from John Horgan:

One Templeton official made what I felt were inappropriate remarks about the foundation's expectations of us fellows.  She told us that the meeting cost more than $1-million, and in return the foundation wanted us to publish articles touching on science and religion.  But when I told her one evening at dinner that – given all the problems caused by religion throughout human history – I didn't want science and religion to be reconciled, and that I hoped humanity would eventually outgrow religion, she replied that she didn't think someone with those opinions should have accepted a fellowship.  So much for an open exchange of views.

As to getting to work in peace, feel free to not read what I write.  So, what sort of work were you doing again?

Response to irishsultan (regarding Monty de Sade)

This is an amplification to a point made to irishsultan in the comments to Monty de Sade.

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I'm more certain now:

Variants 2 and 4 are the same with respect to the switches and stays, you just win a goat as well as a car if you get it right and you get nothing if you get it wrong.

Note that the options each of the move variants are based on the distribution illustrated in the "Just Leave the Gap" variant.

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irishsultan seemed concerned about the placements of the goats.  The image below relates to a slightly different variation of Monty de Sade in which we do care about the names of the goats, so a grouping of Mary with the car is distinct from a grouping with Ava with the car.

Therefore we now have the following options:

A = Ava, B = Mary and Car

A  = Mary, B = Ava and Car

A = Car, B = Ava and Mary

A = Mary and Car, B = Ava

A = Ava and Car, B = Mary

A = Ava and Mary, B = Car

(Where Ava just short for "the door behind which Ava is hidden", etc)

The result is still a likelihood of 1/2 of winning by switching or staying..

The Objections of chrysics, irishsultan and ChalkboardCowboy

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 objections of chrysics, irishsultan and ChalkboardCowboy are closely linked (as is Mathematicians), meaning that the likelihood is that they are right and I am wrong.  This said, I still want to make a few points.  First, chrysics wrote a comment in which he highlighted the following:

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.

This relates to an earlier comment in which he wrote:

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.

Note that it was a day later that irishsultan wrote a response that, despite being hugely more compact and using different numbers than I would have used, I actually find more appealing than chrysics’.  (In quoting irishsultan, I have fixed a typo.  While the names of the goats do in fact come from another game, so they are notionally “Marry the Goat” and “Avoid the Goat”, I called them Mary and Ava, not Marry and Avoid.)

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 Mary) = 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

I would have used the following numbers, given that we are talking about Red-Green door selection and a Red Mary being revealed:

MAC  = 1/6                                             MCA = 1/6

SUS = 1/3                                              SUS = 1/3

Mary Revealed= 1/1                                Mary Revealed = 1/2

Red Mary Revealed = 1/18                     Red Mary Revealed = 1/36

Green Ava Revealed = 1/36

Pr(GreenCar AND RedMaryRevealed) = 1/18 = 2/36

Pr(RedMaryRevealed) = (1/18 + 1/36) = 3/36

Pr(Green Car|RedMaryRevealed) = 2/3

Thus, using this approach, it would seem that there is a benefit in staying in a Reverse Monty Hall Problem scenario when the Red and Green doors have been selected and a goat has been revealed behind the Red Door.

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This logic of irishsultan’s talks to me.  He seems to have understood that I am only talking about what the likelihoods are once the door is opened, I’m not interested in scenarios in which the host opened another door altogether or the contestant selected another pair of doors.  This understanding might contribute to my finding the argument so appealing.

Another factor is that irishsultan’s result, in combination with chrysics assertion that if there is a Red Mary and the Red and Green doors have been selected, then the likelihood of 1/2 of winning from a switch, would indicate that there is a change of likelihoods just as the door is opened.  Perhaps I just got them around the wrong way?

I don’t want to leap at that just yet, appealing as it may be since it would bring the hostilities to a close.  I want to think things through a bit more, both in the response to chrysics’ and ChalkboardCowboy’s comments below and also in a couple of articles in which I continue my arguments (arguments which I accept might well turn out to be wrong).

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What I want to do now is highlight why I didn’t find chrysics’ formulation so persuasive, even though it could be said that he was saying the same as irishsultan.  Hopefully, in that process, I will address his question.

In The Reverse Monty Hall Problem I made an effort to make that which was by necessity a hypothetical scenario as real as possible.  I took the reader to a shopping mall, exposed them to three doors, asked them so select two, opened one door and then asked them if they would like to switch doors.

Given some of the comments I received, it appeared that this was not clear enough so I wrote Marilyn Gets My Goat, trying my best to cast the scenario in terms of a very real situation – with a real contestant, a real host, real doors painted real colours, real named goats with real photos (they aren’t my goats, but they are real goats) and a real (albeit concept) car – in which the contestant selects two doors and is then obliged, once a specific door has been opened to reveal a specific goat, to assess the probability of winning if they switch doors.

In the scenario I described, the contestant has selected Red and Green doors and the host has opened the Red Door to reveal Mary the Goat.   My question, in this very specific situation is, given what the contestant knows, what would she assess as the likelihood of benefitting from a switch?

For this reason, I have a problem when chrysics says:

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. 

I have to respond that, in my scenario, the possibility of the contestant having selected Red and White is 0, because the contestant has already selected Red and Green, and the likelihood of the contestant having selected White and Green is 0, because the contestant has already selected Red and Green.  I consider the fact that there are different probabilities in those scenarios for winning as a consequence of switching as irrelevant.

From my point of view, the likelihood of Mary being behind the Red Door and the likelihood of the contestant having selected Red and Green are both 1/1 – once the doors have been selected and the Red Door has been opened revealing Mary.

The fact that chrysics kept going on about possibilities that simply don’t exist anymore after the door is opened (at least in my conception) was just causing problems in communication.  I was very happy to see, however, that single scenario that remains possible is the one to which he attributes a likelihood of 1/2 – which is my answer for what I could see as precisely my scenario.

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Now, that said, what I think that chrysics was saying and what I know that irishsultan was saying is that the contestant, once the Red Door is opened, not only knows that the game is a Red Mary game but she also knows that she knows that the game is a Red Mary game.  Their argument, therefore, is that the knowing that you know makes a difference.

This bothers me.

I tried to put myself into the position of the contestant and tried to think it through from her perspective.  When confronted by three doors and told that there are two goats and a car behind them, placed at random, I would know that there are six possibilities, all of which are equally likely:

ACM, AMC, CAM, CMA, MAC, MCA

When asked to select two doors (and assuming that I would select two doors at random), I would know that there would there would be 3 possibilities, all of which are equally likely.

Red-White, Red-Green, White-Green

Then, before the door is opened, I would know that if Mary were to be behind the Red Door and I had selected the Red and Green Doors, then I would have a 1/2 likelihood of winning from switching.  (I’d know that the same likelihood applies irrespective of which particular goat we are talking about behind which particular door, on the condition that I had selected a pair of doors that would permit that goat to be revealed.)

Now, what chrysics and irishsultan are both effectively saying is that, once I see that Mary is behind the Red Door, the likelihood is no longer 1/2 (the figure that I had just calculated), but is now 2/3.

In the Marilyn Gets My Goat scenario, this would be equivalent to:

Holly Mant: So, Marilyn, you have selected the Red Door and the Green Door.  Pick a goat any goat!

Marilyn: Um, Mary.

Holly Mant: And now of your two doors, pick one only!

Marilyn: Well, ah, the Red Door.

Holly Mant: Excellent.  So, given that you have picked the Red Door and Green Door, let’s examine the possibility that Mary was behind the Red Door.  Ignoring the fact that it would be against the rules, if Mary were to be behind the Red Door, what would be the likelihood that you would win the car if you were to walk up to the White Door right now and open it?

Marilyn (following chrysics’ logic): 1/2

Holly Mant:  Well, you’re in luck, Marilyn … (opens the Red Door to reveal Mary) … There you go, the likelihood of the car being behind the White Door is 1/2.

Marilyn:  No, no, no.  Not anymore.  There’s now a 2/3 likelihood that the car is behind the Green Door and only a 1/3 likelihood that the car is behind the White Door.

As said, I do find irishsultan’s argument particularly appealing, but this situation is still somewhat of an issue for me.  I take “X being the case” as taking precedence over “knowing that X is the case”, facts are true even if I don’t know them to be true.  I understand that this might be no more than a wishy-washy philosophical standpoint rather than a hard and fast mathematical conclusion, but I think that there might just be other relevant arguments in support of the 1/2 result.  One of these addresses Chalkboard Cowboy’s central issue.

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Chalkboard’s approach is to use the Law of Large Numbers, namely the tendency of multiple iterations of the same experiment to produce results that match the likelihoods associated with an individual iteration of that experiment.  In this case, he argues that if the likelihood of winning from switching is 1/2 then if you repeat the Reverse Monty Hall Problem (or Monty Hall Problem) many, many times, the results will tend towards 1/2.  And they don’t, they tend to 2/3.

Chalkboard is right in that this is a major problem in my argument, one which I have to address with more than hand-waving.  I’ve tried to explain in terms of multiple mini-games, which doesn’t seem to convince anyone, so perhaps a Large Number argument might.

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If we strip the game back to basics, what we have are three slots into which a prize can be placed at random (X marks the treasure):

X _ _    ,    _ X _    ,    _ _ X

Then there is some sort of faffing about before one of the empty slots are removed from consideration.

Then the contestant is asked to select one of two slots (~ is the removed slot):

X _ ~    ,    X ~ _    ,    ~ X _    ,    _ X ~    ,    _ ~ X    ,    ~ _ X

If you run this stripped down version over and over again, you’ll get 1/2 as your likelihood of winning with either of the two slots left to choose from.

Now, I agree that the Monty Hall and the Reverse Monty Hall Problems are quite specific instances of this general “Monty Faffs About” game, but would anyone disagree about the 1/2 likelihood result for the overarching “Monty Faffs About” game, a result that would be produced by Large Numbers of instances of “Monty Faffs About”?

The question then is whether, in any specific instance of “Monty Faffs About”, the Law of Large Numbers tells us that the answer is 1/2 or some other answer.

(I know I am being irreverent here, but the underlying point is quite serious.)

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?

The Objections of chrysics - Part 1

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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Over on Reddit, I’ve been engaged in discussions with many “dumb repeaters” and a small cadre of intelligent, highly educated people who have provided useful feedback in the way of objections to what I’ve been writing about recently.

Because I find Reddit particularly unwieldy for more in-depth discussions, I will try to respond to a longer comment here.  It’s not intended that these be comprehensible for all readers, in fact it is pretty much directed at one person, but the challenge was made publically, so the response is presented publically.  If you have not already been engaged in the furore, I suggest looking back at earlier articles here and traipsing over to reddit.com to check things out.  Try r/badmathematics where I am in the running for a Golden Goat Award (I’d like to thank my family …), while you are there, you can soak up the arguments against my position, which I strongly recommend that you do.

The comment is from u/chrysics but this person has also commented on this blog:

neopolitan – 100% of the time when a Red door has been opened to reveal Mary will the Red door have been opened to reveal Mary

chrysics – Nobody is disputing that. But this can occur in two ways:

1.    Red Mary can be the only choice the host was allowed to make

2.    The host had a choice between Red Mary and Green Ava

Now, the probability of either arrangement of goats is equal. But the arrangement of goats required for case two only gives you a 50% chance of a Red Mary game, while case 1. gives you a 100% chance. Two thirds of the time that Red Mary shows up, it's because you were in case 1.

See Q10 of Marilyn Gets My Goat, this just says the same thing.  Also, I address this issue at Monty Two Face.

           

neopolitan – It's a single iteration, one-shot scenario,

chrysics – That has nothing to do with anything.

Naturally I disagree.  I very specifically wrote The Reverse Monty Hall Problem to make it a real situation in which there is a single iteration of the game.  I peopled the scenario with real people (and tried to make that even clearer in Marilyn Gets My Goat).

neopolitan – Personally, I don't really care if the goats are interchangeable or not,

chrysics – Well you should, because - unless you're using the term 'interchangeable' to mean something different from its established meaning in probability - it changes the outcome of the game.

I’m probably not using the term 'interchangeable' in a strict sense.  All I mean is that in a real version of the game, there will be two goats and these goats are not the same goat and thus these real goats are not interchangeable.  (Equally, there is not 2/3 of a goat behind each door and 1/3 of a car.)

neopolitan – I can build my argument on them being the same goat hidden behind two doors, if that is what you prefer.

chrysics – It's not. I don't care about the goats. There is one door with a car and two losing doors. If both losing doors are selected, the host decides which to open randomly; otherwise the host opens the only losing door selected. That is all that is important. The losing doors can hold separate goats, the same goat, nothing, a guitar in one and a performance art troupe behind the other, it's completely and utterly irrelevant.

I agree.  But some people arguing wanted to stick with hypothetical goats, strangely divisible goats, even when I was trying to talk about a notionally real situation.

neopolitan – Which do you actually want me to argue? That the goats are interchangeable or not interchangeable?

chrysics – The interchangeability of the goats is a function of the rules of the game. You have consistently described a set of rules in which the goats are interchangeable. But you keep saying they're interchangeable, and on that one occasion you also described a version of the game in which the rules differ in a subtle-but-important way from your other descriptions, which just so happens to make the goats non-interchangeable.

I don't want you to argue that the goats are interchangeable, nor do I want you to argue that the goats are not interchangeable. I want you to clarify which set of rules you want to play by. Is the host required to open Red Mary whenever that is a possibility, or is the host permitted to open Green Ava even when Mary is behind the red door? If the latter, then I also want you to explain why you insist that the goats are not interchangeable.

Actually, this simply isn’t true.  When I first mentioned the concept of “interchangeability”, I wrote (emphasis added):

(as wotpolitan) It is easy enough to overcome logically. I say my scenario is a single instance, one-shot game. Then someone says, aha, but he could have opened a different door. But my point is that he didn't. He opened a specific door revealing a specific goat and it is this specific scenario that the contestant has to deal with. Some say that the goats are interchangeable, and I say that might be the case with a purely imaginary game, but I put the contestant in a shopping centre with me (the dishevelled amateur philosopher), a relatively real person, conducting the experiment with goats that I keep trying to tell people are real, and patently not interchangeable. They say, if you run it multiple times you'll get 2/3, I agree and say that it's different if you run it just once. I show how it works with conditional probability, they say that that would be fine, but conditional probability doesn't apply - perhaps they are right about that, but it seems to apply fine from what I can see.

I am fine using whichever you prefer, but if you have no preference, let’s go with non-interchangeable goats called Mary and Ava, because it prevents the problem mentioned above.

In the scenario I describe at both The Reverse Monty Hall Problem and Marilyn Gets My Goat, there is no obligation on the part of the host to open the Red Door if Mary is behind it, unless the car is behind the other selected door in a Red-White or Red-Green scenario.  However, if the host has opened the Red Door to reveal Mary, then there is an obligation to consider only scenarios in which the host could open the Red Door to reveal Mary for that particular instance of the game (which is the only instance as I formulated the scenario).

neopolitan – I think your comment there is simply irrelevant because the host makes one decision and one decision only to open one door

chrysics – In potential Red Mary games, the host only make a decision half of the time - the other half of the time, the host has no choice at all. And half of those decisions result in the outcome of the game not being Red Mary. Of the potential Red Mary games, only 3/4 are actual Red Mary games. In 2/3 of those 3/4, you win the car by sticking with your initial choice. The other 1/4 are irrelevant, as they're not Red Mary games at all.

This looks to be based on a misunderstanding of what a Red Mary game is.  I take responsibility if I was insufficiently clear, but a Red Mary game is a game in which Mary is behind the Red door, irrespective of whether she is revealed by the host as being behind the Red Door.

However, if we call these Revealed Red Mary games, you are right in that the host makes the decision whether to open the Red Door or nor half the time and is obliged to open the other half of the time.  I address the problem highlighted here at Monty Two Face.

chrysics – Let me break it into stages, so you can tell me where you think I go wrong. (We'll assume for the sake of consistency and simplicity that the contestant's chosen doors are Red, the leftmost door, and Green, the rightmost door (neopolitan comment – agreed, this is consistent with the scenario as laid out in Marilyn Gets My Goat)

1.    A Red Mary game denotes a game in which the host opens the Red door, revealing the goat Mary to be behind it (neopolitan comment – not agreed, this is a subset of the set of games I call Red Mary games, we could call this a Revealed Red Mary game, I will update your text below to reflect this, updates in brackets)

2.    The goats and the car are arranged behind the doors at random

3.    A (Revealed) Red Mary game may occur when the goats and car are arranged MAC or MCA

4.    Both of those arrangements are equally likely:
P(MAC) = P(MCA)

5.    No other arrangement can produce a (Revealed) Red Mary game:
P((Revealed)RedMary| (!MCA && !MAC) ) = 0

6.    The contestant has no knowledge of the arrangement until the host opens the door

7.    At that point, the contestant has no knowledge beyond what is revealed (Mary is behind the Red door) and anything that can be deduced from that

8.    If the arrangement is MAC, there is a 100% chance that the host will open the Red Door, producing a (Revealed) Red Mary game

9.    If the arrangement is MAC, there is no possible result other than the one given above:
P((Revealed)RedMary|MAC) = 1

10. If the arrangement is MCA, there is a 50% chance that the host will open the Red Door, producing a (Revealed) Red Mary game

11. If the arrangement is MCA, there is a 50% chance that the host will open the Green Door, producing a (Revealed) Green Ava game

12. If the arrangement is MCA, there is no possible result other than the two given above:
P((Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA) = 1, and they are equally likely:
P((Revealed) RedMary|MCA) = P((Revealed)GreenAva|MCA)

13. As MAC and MCA are equally likely arrangements,
P((Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA)

14. Using:
P((Revealed)RedMary|MCA) = P((Revealed)GreenAva|MCA), and
P((Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)GreenAva|MCA), we arrive at:
P((Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA)

15. Using:
P((Revealed)RedMary| (!MCA && !MAC) ) = 0, it can be deduced that:
P((Revealed)RedMary) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MAC)

16. Using:
P((Revealed)RedMary|MAC) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA), this may be rewritten as:
P((Revealed)RedMary) = P((Revealed)RedMary|MCA) + [P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA)] (the square brackets denote nothing other than where the equation has been re-written)

17. The re-written equation P((Revealed)RedMary) = P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA) + P((Revealed)RedMary|MCA) accounts for all (Revealed) Red Mary games, and consists of three equal contributions.

18. The first of these contributions accounts for the (Revealed) Red Mary games arising with the arrangement MCA.

19. The second and third contributions - each of which is equal individually to the first - account for all remaining (Revealed) Red Mary games, in which the arrangement is MAC. Collectively, this MAC. Collectively, this accounts for two thirds of (Revealed) Red Mary games

20. For any given (Revealed) Red Mary game, there is a 2/3 probability that the arrangement is MAC.

I am not well versed enough in this notation to confidently identify the precise step in which the error (as I see it) creeps in.  However, the issue is that we are talking about a situation in which there is a Revealed Red Mary, meaning that the only options are that we are playing an MAC game or an MCA game.  I agree that we will only see the Revealed Red Mary half the time when we are playing a Red Mary game of the form MCA (where Red and Green were selected).  But I disagree that this means that we can eliminate the MCA games that we would not see.  This is the point I am trying to make at Monty Two Face.  I think that we need to compare Red Mary games, not Revealed Red Mary games.

By rejecting Red Mary games that are also Revealed Green Ava games (ie 50% of the MCA games), which is effectively what you are doing, we skew the result.

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Update:

As mentioned in the comment below, I was not seeing the wood for the trees when reviewing chrysics post.  I was convinced that his core argument was that I was wrong, so I completely overlooked that his conclusion was that the probability of a Red Mary being revealed as a result of a MAC distribution of goats and car is 2/3 - a point I made in Marilyn Gets My Goat (see Q10 and also Q16).  Therefore, I actually agree with him.  No wonder that I could not find any error (a fact that was a little disconcerting, since it all seemed pretty straightforward).

That said, my last few comments above still stand, I just wasn't properly addressing what chrysics was saying.