Thursday, 12 March 2015

Saving Monty Fall

As pointed out by KaySen, there is a treatment of Jeff Rosenthal’s Monty Fall scenario by Christopher Pynes which suggests, amongst other things, that in a Monty Fall scenario the correct answer is still 2/3.  Interestingly enough, Pynes is making precisely the kind of error that I was accusing everyone of making when I was first defending my arguments first raised in The Reverse Monty Hall Problem – that is by treating a single instance, one shot game as if it were part of a series.

I am now persuaded that the intent of this argument was misguided in terms of both the Monty Hall Problem and the Reverse Monty Hall Problem, but I consider it to be entirely valid with respect to the Monty Fall scenario.

The whole concept that Rosenthal is trying to convey is that the host accidentally falls against a door, thereby revealing a goat, rather than deliberately selecting a door to open.  This entails that the accident is a one-off.  There is no implication that the host repeatedly slips and falls against a door, and that in each case a goat is revealed – as implied by Pynes:

In cases where the contestant picks the door with the prize behind it in the first place, Monty never reveals it, but reveals another door: in these cases one wins by staying and loses by switching—1/3 and 2/3 respectively. In the case where the contestant doesn’t initially pick the prize door, Monty’s fall reveals the only door it can that isn’t the prize—win 2/3 of the time by switching. This makes Monty Fall and Monty Hall the same in all instances; they are logically equivalent, and thus must have the same probabilities.

There is no “never”, there are no other instances required to make “all instances” a meaningful concept.  It happens once, whoops, and it’s entirely possible that it never ever happens again.  Pynes, I believe, is the one being seduced by his familiarity with the Monty Hall Problem.

There is, however, a way in which Monty Fall can be saved in so much as it can be reliably repeated.

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Monty de Sade was continuing to cast about for a more devious variant of the three doors, two goats and a car game.  He heard about the use of coins to discuss the Reverse Monty Hall problem and a new idea formed in his evil little mind.

His new game goes like this:

There are three doors, with two goats and a car distributed behind them at random just like before.

The contestant is offered one door to choose from, just like the classic Monty Hall Problem.

Monty then offers a more complex continuation.  If the contestant agrees to proceed, Monty will toss a coin and then open one of the unselected doors on the basis of the result.  If the car is revealed, the contestant will automatically lose.  If the car is not revealed, then the contestant will be offered the opportunity to switch doors.

What is the likelihood of benefitting from a switch?

There are three decisions here on the part of the contestant:

The selection of a door, which we take to be random

Whether to risk losing the car on the basis of a coin toss

Whether to switch

As far as I can tell, noting that I have been wrong before, there is a 1/3 likelihood of selecting the car from the outset.  This means that there is a 2/3 likelihood of the car being behind one of the two unselected doors.  However, when Monty tosses a coin, there is 1/2 likelihood of revealing the car, if the car is behind one of those doors, and a 0/1 likelihood of revealing it if the car is behind the selected door.

This means there is a 1/3 likelihood of having the car already and a 1/3 likelihood that the contestant will lose the car in the coin toss and a 1/3 likelihood that the car won’t be lost in the coin toss, but will be behind the one remaining unselected and unopened door.

This means that the likelihood of benefitting as a result of switching, after surviving the coin toss, would be equal to the likelihood of benefitting as a result of not switching.  In which case, there was no point in going through the charade, was there?

This seems counter-intuitive.  It seems to me that if the contestant bothered to go ahead with the coin toss then, once a goat is revealed, if a goat is revealed, she is essentially committed to switching.  By why, if it’s 50-50?  Some might suggest that it’s not 50-50, but the probability tree below indicates that it is – unless I’ve made an error somewhere.


That said, there are two perspectives one could take, and they seem to have equal weight:

Perspective One:

At the beginning the contestant picks one door, the likelihood of having selected the car is 1/3, so there is a 2/3 likelihood of not having selected the car.  If she has the car, then she loses nothing by having Monty toss the coin and open a door.  If she doesn’t have the car, then she gains nothing by sticking with her selection.  So she should proceed.

Then Monty tosses the coin and, phew, doesn’t reveal the car.

Given that there was 1/3 likelihood of the car being behind the selected door and a 2/3 likelihood of it being behind one of the other two doors, it’s more likely that the car is behind the remaining unselected and unopened door and the contestant should therefore switch.

Perspective Two:

The contestant selects a door.  If she has selected the door with the car, then she loses nothing by having Monty toss the coin and open a door.  If she hasn’t selected the car, then she gains nothing by sticking with her selection.  So she should proceed.

Monty tosses the coin and, phew, reveals a goat.  The likelihood of Monty revealing a goat when there are two goats behind the unselected doors is twice that of revealing a goat when there is only one goat that could be revealed.

Given that Monty revealed a goat, it’s more likely that the car is behind the selected door and the contestant should therefore stay.

So, I throw the question out there: what is the best strategy for the contestant?

Note that this time I have no firm answer of my own beyond a gut feeling that the rational contestant should risk a coin toss to make use of the 2/3 likelihood of not having selected the car and switch if a goat is not revealed, having dodged an automatic loss entirely by chance.  I do recognise that the probability calculations don’t seem to support this gut feeling but perhaps someone has some better probability calculations.

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And how, precisely, does this save Monty Fall?  If my 50-50 solution is correct, then it is possible to run the scenario over and over again, as many times as you like, and then extract one iteration of the game in which Monty reveals a goat.  This then represents the single instance involved in the Monty Fall scenario, it’s just that the “luck” is now based on a coin toss rather than a humorously placed banana peel.

Saturday, 7 March 2015

From Two Balls One Urn to the Reverse Monty Hall Problem

So, some might be thinking, how on Earth did I get in this state with The Reverse Monty Hall Problem?  Especially when, only two short weeks or so ago, I was a definite 2/3 answer person.  How did I manage to spend more than two weeks convinced that the answer was 1/2?

It started with balls.  Well, it started with a little thought experiment associated with the Anthropic Principle, but it involved balls.  There are some who think that “Fine Tuning” is an argument for the existence of a god.  Even some closer to normality think that it is something that ought to be explained (or explainable).  I go along with the anthropic principle, because I agree wholeheartedly that, given we are here to observe the universe, the universe must be suitable for beings like us to observe it.  In other words, the posterior likelihood that the universe being suitable for beings like us to observe the universe is 1/1, not 1 in whatever ridiculously large number some apologists come up with.

So, I started thinking about balls.  What, I wondered, would someone say if, after a sequence of 999,999 black balls from a barrel, having been taken at random one by one, I stopped them and asked them to assess the likelihood of the last ball, the one millionth, being white.  It seems ridiculously unlikely, right?  This would mean that every single ball before then had a chance of being a white ball, but they selected a black ball every time.  So, it would have to be 1 in a very large number.  Well, it’s not that large, I thought.  It’s one in a million.  Say that there’s a black ball in the barrel, one black ball, and the balls are removed at random.  It’s a sequence and, if the black ball’s position in the sequence is random, that ball is equally likely to be in the 1,000,000th position as it is to be in the 456,978th position, or the first, or any other specific position.   So, it’s one in a million.

However, I thought, that’s not quite right.  Balls aren’t limited to black and white.  The millionth ball could be a red ball, or a green ball, and so on.  So, it’s more accurate to say that the likelihood of that last ball not being white is one in a million.  So, I came up with another scenario, which is at Two Balls One Urn.  This was a little too complicated, so I adapted it at Two Balls One Urn, Revisited.  In this scenario, there are two million balls in a barrel which is in a pitch black room.  The balls are all various shades, colours and patterns including two that are just white.  Someone goes into the room with an urn, picks a ball at random from the barrel and puts it in the urn.  Then they pick another ball at random from the barrel and puts that in the urn as well.  Then they pick a ball from the urn at random and hold it in their hand.  Finally they leave to the room, they look at the ball in their hand and see that it is white.  What is the likelihood that the ball in the urn is also white?

It works out to be 1/(n-1) = 1/1,999,999 where n=2,000,000 is the number of balls in the urn.

One of the commenters, Anonymous B, suggested doing this with n=3 and pointed out that the answer in this case must be 2/3.  In my response I indicated that, at the time, I didn’t think that an n=3 version of Two Balls One Urn and the Monty Hall Problem can’t be equivalent scenarios, because one comes up with an answer of 1/(N-1)=1/2 and the other comes up with 2/3.

Then I thought about it some more and cognitive dissonance kicked in.

So, because I still find this difficult, I shall go through the Two Balls One Urn with three balls, just to make it perfectly clear what I was doing:

I have an enormous barrel in a pitch black room and I know that in the barrel there are three balls, two of which are entirely white, the other being black.  I take an urn into the pitch black room and, completely at random, I take out two balls from the barrel and place them in the urn.   Because it is so dark in the pitch black room, I cannot see either of the balls when I do this.

Then, while still in the pitch black room, I reach into the urn and, completely at random, I draw out one ball.  Because it is so dark in the pitch black room, I cannot see either of the balls when I do this.

Finally I walk out of the pitch black room and I look at the ball I drew out of the urn.  It is white.

What is the probability that the second ball - the one still in the urn - is white?

Now, following precisely the same logic as when n=2,000,000, the answer should be 1/2.  The white ball in my hand means that either:

the first ball I took out of the barrel was white, and when I took the second one out, I was randomly selecting from two, one white and one black, or

the second ball I took out of the barrel was white, and when I took the first one out, I must have taken one at random from the other two, one of which was white with the other being black.

Therefore, the ball in the urn has a 1/2 likelihood of being white and a 1/2 likelihood of being black.

Now, if the white balls are goats, the black ball is the car, the barrel represents the three doors, the urn represents the selected door, and the ball taken at random from the urn is the opened door, then we have a scenario that is analogous to a variation of the Monty Hall Problem.  And the answer in the Monty Hall Problem is widely accepted to be 2/3 rather than 1/2.

This was a problem.  I thought that it must be wrong, and then I looked up the history of the Monty Hall Problem and saw that prior to 1990, pretty much everyone who thought about it was convinced that the answer was 1/2 (despite the fact that Steve Selvin had written a paper giving the correct answer as far back as 1975, a fact I discovered only quite recently).  I also stumbled across the Monty Falls variation of the Monty Hall Problem in which the answer is 1/2.

There is clearly something strange going on.  So, as is my wont, I looked to see if there was a middle path.  Is it possible that the answer is 2/3 in one sense and 1/2 in another sense?  I thought that it was quite possible, and an answer came to me when I was thinking about the white ball in my hand.

In my scenario, I had "forced" the situation.  In a real situation, it wasn’t guaranteed that the white ball in my hand was going to be white.  It could have been black.  However, in my scenario it quite explicitly states that the ball in my hand is white.

In the Monty Hall Problem, in the treatments of it I had seen, this was not so explicitly stated.  There is plenty of talk about how, before the door is opened, the host might have opened another door, or the contestant might have selected different doors, or the goats and car might have been arranged differently.  But, I reasoned, this is not the situation that the contestant finds herself in once the door has been opened.  Once the door is opened, the goats and car have been placed, the doors have been selected and the host has already opened the door.  Therefore, I thought, this makes the Monty Hall Problem, as posed by Craig F. Whittaker, analogous to me standing there with one white ball in my hand.

Some might argue that in my situation, the decision of the host is not taken into account.  However, I tried to be very explicit in my wording of the Reverse Monty Hall Problem, the host is forced to reveal a specific goat in those circumstances where by not doing so he would reveal the car and otherwise the goat revealed is selected entirely at random.

This, I thought, was directly analogous to my selection of the ball from the urn.  If there is a black ball left in there, then the white ball I removed was the only white ball I could have selected.  If there is a white ball left in the urn, then I could have equally likely have selected that one.  I don’t know which situation I am in, of course, but it is (or rather was) apparently equivalent to the situation that the contestant is in a Reverse Monty Hall game.

And then I wrote about it …

So, now you know how I got to the point I was at when I posted The Reverse Monty Hall Problem, being pretty much convinced of an answer that the vast majority of mathematicians are convinced must be wrong.

Soon I will try to summarise why I was wrong, in what way I was wrong, why so many could not convince me I was wrong and why something quite short and apparently innocuous convinced me that I was wrong.

Friday, 6 March 2015

Marilyn's Six Games

Over the past two weeks I have argued that, at precisely the moment at which a contestant in a Reverse Monty Hall game is required to assess the likelihood of benefitting from switching from the one remaining closed door of her selected two doors to the door she did not originally select, the likelihood is 1/2.  A key part of my argument was what I present in Monty Does Play Dice and another key part was this, my argument that there are as many as six mini-games that a contestant going into the game will face and that when it is time to make an assessment of likelihood, there are only two remaining.  This is still true.  However, it does not change the likelihood of benefitting from a switch, which remains 2/3.

Other than adding this introduction and a note at the end, I have not edited this article from what I intended to write when I was still fully convinced that the answer was 1/2.

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In Marilyn Gets My Goat, which was an attempt to crystallise the scenario introduced in The Reverse Monty Hall Problem, I tried as much as possible to show that when a contestant is playing the game, it’s a real game.  You can do it in real life, with a real contestant, a real host, real doors, real live goats and a real car.  This was because there were a few people who keep saying that, hypothetically, even though a particular door was opened, the host could have opened another door, and revealed a different goat.  Well, yes, this is true but not after the door was opened.

Anyway, another major objection relates to the fact that I agree that, over multiple instances of the Reverse Monty Hall Problem (and the Monty Hall Problem), the contestant will win 2/3 of the time by adopting a policy of staying (with the Reverse Monty Hall Problem, which is equivalent to switching with the classic Monty Hall Problem) while at the same time maintaining that, in a single iteration, one shot instance of the game, there is no benefit in switching (because the likelihood of winning as a consequence of switching or staying is equal at 1/2).

This is a reasonable objection, because it appears to fly in the face of the Law of Large Numbers, a fact which /u/ChalkboardCowboy and /u/OmarDiamond particularly object to.  Note I said “appears”.  Neither Chalkboard nor Omar make that distinction, so when I say something like “my solution doesn’t fly in the face of the Law of Large Numbers” we end up talking past each other.

What I’d like to do here is explain what is going on and how you can go from a likelihood of winning of 1/2 in a single instance to a likelihood of 2/3 over many instances as a consequence.

Note that I’ve already addressed it partially in various comments.

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The thing is that a contestant in the Reverse Monty Hall Problem could be playing one of six different games, they are the game in which she tries to win the car when the distribution of goats (called Ava and Mary, as per Marilyn Gets My Goat) and car are:

Ava-Car-Mary (ACM)

Ava-Mary-Car (AMC)

Car-Ava-Mary (CAM)

Car-Mary-Ava (CMA)

Mary-Ava-Car (MAC)

Mary-Car-Ava (MCA)

If you think that the goats are indistinguishable and/or interchangeable and/or represent only otherwise undefined losing options, then there are still three potential mini-games.

Car-goat-goat (Cgg)

goat-Car-goat (gCg)

goat-goat-Car (ggC)

When you think of multiple instances of The Reverse Monty Hall Problem, you are considering a uniform distribution of these mini-games without the player having any idea which one they will be playing.  You know that if the goats and car are randomly (and uniformly) distributed, then each time you select two doors you have a 1/3 likelihood of getting the car with each door.  As a combination, having two doors gives you a 2/3 likelihood of having the car, so you should (as a policy) stay with your two doors.

However, when you are in the middle of a single iteration, one shot instance of the game, after the door has opened, you have more information.  We’ll use the situation that the contestant Marilyn found herself in in Marilyn Gets My Goat.  Marilyn chose the Red Door and the Green Door and the host (Holly Mant, played by Angelina Jolie) opened the Red Door to reveal Mary the Goat.

When this happens, there are no longer six possible games that Marilyn could be playing.  She can only be playing the two games that have Mary behind the Red Door, namely:

Mary-Ava-Car (MAC)

Mary-Car-Ava (MCA)

Or, if you are not in the habit of naming your goats:

goat-Car-goat (gCg)

goat-goat-Car (ggC)

My argument is that because the goats and car were placed at random and the doors were selected by Marilyn at random, then we have no reason to think that, prior to the game commencing, there was any more likelihood associated with the MAC (ggC) distribution that there should be with the MCA (gCg) distribution.

There is certainly a greater likelihood that the Red Door was opened as a consequence of the car being behind the Green Door – because if there is a car behind that door the host would have been obliged to open the Red Door while he has a choice between the Red Door and the Green Door if Ava was behind the Green Door.  However, prior events do not always affect the present.  If there was a goat behind both the Red Door and the Green Door then the host would open a door at random (strictly speaking on instruction from her producer, who selected a door at random).  However, the likelihood of having opened the Red Door to reveal Mary is 1/1 once that door is opened.

This is a similar situation to tossing a fair coin and getting seven heads in a row.  The a priori likelihood of getting eight heads in a row is 1/256.  However, the likelihood of getting a head after seven heads in a row is 1/2.  It might be less likely to see Mary behind the Red Door in the instances where Ava is behind the Green Door, but once the door is opened and we see Mary behind the Red Door, it’s now no longer a likelihood, it’s an absolute certainty.

This might seem either quite sensible or totally counter-intuitive.  If you feel the former, you should read up on the Monty Hall Problem, because this answer here is not the standard solution.  If you feel the latter, you might want to take a look at Monty Does Play Dice, if you have not already done so.

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In conclusion, the reason why we get 2/3 over multiple iterations is because when we have a Red Mary scenario (being a game in which Mary is behind the Red Door) we only see it half the time.  The rest of the time we will see Ava behind either the White Door or the Green Door.  The same sort of spread happens with White Mary and Green Mary scenarios and the cumulative effect of this over multiple iterations is that there is an increased likelihood to win if you have a policy of staying in multiple iterations of the Reverse Monty Hall Problem (and a policy of switching in multiple iterations of the classic Monty Hall Problem).

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Please remember that while I found this argument quite convincing only a few days ago, it is nevertheless wrong in so much as the likelihoods are 2/3-1/3 and not 1/2-1/2.  I still think that it is true that the contestant is playing one of two specific mini-games, but the ramifications of that appear to be negligible.

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..

Thursday, 5 March 2015

Monty de Sade

The puzzle below was written when I firmly believed that the answer to the question at The Reverse Monty Hall Problem was 1/2 (as argued in my solution as well as in later articles and in many comments, both here and at reddit).

Now I have finally been convinced that the answer is, as many said right from the beginning, 2/3.  That said, I still find the exercise below vaguely interesting.

(For those of you who have suffered through my journey, I am pretty sure that this is an example of a “Monty Faffs About” game.)

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Monty de Sade decides that the Monty Hall Problem is too easy, so he devises a new variant.

There are still three doors (Red to the left, White in the middle and Green to the right).

There are still two goats (Mary and Ava, but we no longer care about their names).

There is still a car.

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

However, the contestant is required to pick between two options, A and B, rather than selecting a door.

A and B may be as follows:

A – G … B – CG

A – GG … B – C

A – CG … B – G

A – C … B – GG

In other words, A may be a door with a goat behind it (the contestant knows not which) while B would be a grouping of two doors, one of which hides the other goat and the other hides the car, and so on.

The contestant makes her choice between A and B (we can assume a random selection is made).

Monty opens a door revealing the goat without telling the contestant whether the door was from A or B, only that it was from the grouping of two doors so that A and B are still viable options for selection (both contain either a car or a goat, not both).

Monty asks the contestant if she wants to stay with her original choice or switch.


Should the contestant switch, or stay?  Why?

Wednesday, 4 March 2015

El Monty's Three Amigos

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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One of my interlocutors recently encouraged me to learn more of conditional probability.  While I was pretty sure that I knew a bit about it, this didn’t stop me from trying to expand my knowledge.  While I was fossicking around the internet I came across an interesting and related little scenario, one that I have modified below:

Say that Dusty, Lucky and Ned are booked to perform for El Monty, who is having a Very Bad Day.  At the end of the performance, El Monty calls over one of his henchmen and says: “I like these guys, these are funny guys!  Just kill two of them!”

The henchman asks which two and El Monty points out that he doesn’t care: “Just pick two of their names at random.  But don’t tell them which ones!  We don’t want them to be sad on their last night, do we?”

The Three Amigos are then led away into the dungeon and thrown into a cell.

Because Lucky is played by the better comedian, they all agree that he should survive.  They come up with a plan to swap nametags once they have wheedled some information.

Lucky then wanders up to the cell door and strikes up a conversation with the guard who promptly refuses to tell Lucky whether he is going to get shot in the morning.  However, after Lucky performs some great gags and signs an autograph for the guard’s niece, the guard agrees to provide the name of one of the other Amigos who is definitely going to be executed in the morning.

Lucky goes back to the other guys and tells Ned the sad news.

Should Lucky and Dusty swap nametags?  Why?

Before they get the chance to decide themselves, however, El Monty is suddenly at the door: “¡Stupido!  ¿Did I not tell you, do not tell them?”

El Monty then takes out a gun and shoots Ned through the forehead, killing him instantly.

Should Lucky and Dusty swap nametags?  Why?

Now, how is this related to the Reverse Monty Hall Problem?

Lucky knows that there is a 2/3 likelihood of being executed (equivalent to a pre-existing likelihood of 2/3 of having a goat behind any door)

The guard refuses to tell Lucky anything about his fate (making Lucky equivalent to the unselected door and making Dusty and Ned equivalent to the selected doors)

The guard identifies Ned as one of the two to be executed (equivalent to being told that a goat is behind a specific door)

El Monty kills Ned (making the probability that Ned will be executed 100%, equivalent to knowing with absolute certainty that there is a goat behind the door, because it's been opened)

Lucky can swap nametags with Dusty (equivalent to switching doors)

So, put it another way, if you are Lucky – do you switch your nametag with Dusty (who is eager to die so that you may live)?  What is the likelihood of you surviving if you do switch?


If you think that this is somehow not equivalent to The Reverse Monty Hall Problem, feel free to explain how.

Tuesday, 3 March 2015

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.)