## Wednesday, 25 February 2015

### Monty Does Play Dice

Please note that since I wrote this article, I have been persuaded that the argument it is intended to model is wrong.  The correct answer for the Reverse Monty Hall Problem is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).  That said, in what it actually does model, the likelihood of winning as consequence of switching or staying is equal at 1/2.  I will try to explain in a future article.

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The saga continues!

I was, perhaps, a little optimistic when I wrote, in the The Reverse Monty Hall Problem and Conditional Probability:

Hopefully this will be my last article on The Reverse Monty Hall Problem – or at least the last on why the solution is “it doesn’t matter whether you switch or stay, the likelihood of benefitting from either choice is 1/2".

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It appears that a number of people disagree with me, quite violently in some cases.  I’ve tried to address the comments of such people as comprehensively as I can, given the limitations of comments on this platform and also over at reddit.  A few times I have encouraged people to run a little role play (and I even wrote Marilyn Gets My Goat for this purpose), but I still don’t seem to get through.  Therefore, I will now try to explain with three relatively simple photos.

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I just happen to have a bunch of different dice, including a black die with pips, a white die with numbers 7 through 12 and a white die with mathematical operators.  These can be used to represent the car (the black die with a single pip, a circle, uppermost), Mary (the die with a multiplication sign uppermost) and Ava (the white die with an 8 uppermost) – but they don’t have to represent the goats and the car, they can just be three slightly unusual dice, two of which are dissimilar white dice.

Here I’ve simulated a Red Mary scenario, in which Mary(X) is behind the Red Door, the Car(•) is behind the White Door, Ava(8) is behind the Green Door and the Red and White Doors are selected.

In terms of my scenario, the contestant must work out the likelihood that the Car(•) is behind the Green Door.  (We can see where the Car(•) is, because we are like gods, but the contestant does not share our powers.)

Here I’ve simulated the Red Door opening, so Mary(X) is no longer behind the Red Door, but is now in front of the door.

As gods, we can see that one of the possibilities has manifested, with the Car(•) behind the White Door and Ava(8), however, it must be remembered that the contestant does not have our abilities.

For the contestant, there is one single alternative, equally likely scenario:

There’s nothing magic involved in picking up the (•) die and swapping it with the 8 die.  There’s no reason, in my scenario (that is a scenario in which the X die, the (•) die and 8 die were placed at random), that the (•) die should be any more or less likely to appear behind the White Door than behind the Green Door, given that the X die was behind the Red Door.

To address the Reverse Monty Hall Problem, given that the contestant makes her assessment of the probability only after the door has been opened, you have to consider the scenario just like we must when considering these real, actual, existent, non-hypothetical dice.

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Again, please note that the scenario being modelled here is not entirely representative of the Reverse Monty Hall Problem.

1. I think it's hopeless.

[Picture of Jean-Luc Picard facepalming]

This does not engage with any of what people have been explaining to you over the last several days and articles. There's no substance here. You've just repeated the problematic part: "there is one single alternative, equally likely scenario", with no justification. Pictures of dice don't count.

1. I think I have to agree. If you can't understand my point from this, then I am at a bit of a loss as to how to convey that point.

Perhaps I can try: In the pictures, I am showing you in a real situation, what the contestant in the game might be exposed to. There are the three doors, the two goats and the car. Two doors are grouped (either selected in the Reverse Monty Hall Problem or not selected in the classic Monty Hall Problem).

The host opens the Red Door revealing Mary. The contestant cannot see behind the White Door or the Green Door, but you and I can. Despite not knowing what is behind the closed doors, however, the contestant now knows that the Car is either behind the White Door or the Green Door. If the original distribution was MCA, then the car will be behind the White Door. If the original distribution was MAC, then the car will be behind the Green Door. MCA and MAC were equally likely. They are still equally likely from the point of view of the contestant, even if the host opened the door because she had too because the Car is behind the White Door, because the contestant doesn't know that. What the contestant does know is that ACM, AMC, CAM and CMA are now no longer possibilities. If some of these were actually still possibilities (which they are NOT), presuming that all we know is that the Car is NOT behind the Red Door, we'd still only have the following possibilities: ACM, AMC, MAC and MCA - meaning that the likelihood that the Car is behind the Green Door, once the door has been opened, is 1/2.

In the time since I started writing this, a few hours back, I stumbled across a new argument that I can run, thanks to Mathematician. I'll put that together formally over the next day or so.

2. Dear Neopolitan,

I am really disappointed that you lost a precious time writing this new article instead of learning probability theory.
It seems impossible to convince you with a precise mathematical argument because you don't understand them.
And it also seems impossible to convince you with heuristic arguments because then you reply with a mathematically wrong statement.

I'll give it one more try and I will focus on the following sentence :

> They are still equally likely from the point of view of the contestant, even if the host opened the door because she had too because the Car is behind the White Door, because the contestant doesn't know that.

This is not mathematically correct. Apparently you think it is. May be I should repeat it again : Your lack of mathematical background on probability theory is the only thing that is keeping you from understanding your mistake.

Indeed the contestant does not know with absolute certainty if the host chose the Red door because he had to. You seem to believe that this "lack of absolute knowledge" is equivalent to an "absolute lack of knowledge". (common misconception)

If you understood a little bit of probability theory you would immediately see the following : even if the contestant does not know with absolute certainty if the host chose the Red door because he had to, he can still compute "in his head" that there is a 2/3 probability that it is the case.

This simple knowledge that the contestant has (and that you always forget, because you are not used to solve probability problems daily), makes the two situations MCA and MAC become not equally likely, from the point of view of the contestant.

Your dice "simulation" is irrelevant. You will never be able to model two situations that are not equally likely with this because the only variable that you can act on, is insufficient.

But I repeat again, this is a very common mistake. You are not saying anything that I did not see multiple times while teaching. I understand perfectly well the point you are trying to make. There is no need to reformulate the thing once again. I only want you to answer to my detailed answer on http://neophilosophical.blogspot.com/2015/02/the-reverse-monty-hall-problem-and.html

3. I am also very eager, neopolitan, to hear your analysis of Mathematician's post. This one, to be clear:

http://neophilosophical.blogspot.com/2015/02/the-reverse-monty-hall-problem-and.html?showComment=1424814096527#c2144284274391674646

4. Chalkboard, I think that Mathematician is more understanding of what I am trying to say now, so I don't want to go over old ground unnecessarily. If Mathematician still wants me to address that comment, I can certainly have a go, but I am going to leave it for now.

2. You are illustrating nothing other than the entirely fallacious idea that because there are two possibilities (let's say A and B), there must be a 50% chance that you are in A and a 50% chance you are in B. But this is not true, because scenario A occurs more regularly than scenario B. The fact the door has now been opened provides absolutely no information of any relevance to that.

This has been very clearly demonstrated to you, multiple times by multiple people. I'll invite you once more to point out which step of my argument in http://www.reddit.com/r/math/comments/2wd66m/the_reverse_monty_hall_problem_and_conditional/cowmvmi you disagree with. I'll also state here that despite your previous response to that comment, the argument presented applies to a single iteration and in no way depends upon the scenario being repeated multiple times- if you believe it does, please explain your reasons.

3. Thank you chrysics and Mathematician,

I've printed out your long treatments and will try to find time to sit down to go through them and give a considered reply. I presume that you would prefer this than an ill-considered shot from the hip?

1. Yes, for sure a considered reply is preferable, and I can understand how it might take some time to go through the argument.

I would think, though, that you could immediately provide some explanation of why you believe the probabilities depend upon whether it is a single iteration or a long series. I can only assume you've given this enough thought to dash off an explanation relatively quickly, as you've made this claim many times despite being told over and over that it entirely contradicts the basic tenets of probability theory. How is it that you believe a strategy can provide a 50% chance of winning each single iteration, yet result in winning 2/3 of the time when used repeatedly? You claimed at some point to have explained this ( http://www.reddit.com/r/badmathematics/comments/2wysyn/the_goat_that_just_would_not_die/covklna?context=2 ) but didn't respond to my request that you point out quite where you'd done so.

2. Because in a single iteration you are playing one of two possible "mini-games", one in which you have (using a revealed Red Mary) MAC or one in which you have MCA. Across a spread of iterations of the larger game, you could be playing any of ACM, AMC, CAM, CMA, MAC or MCA, even though in each individual iteration you come down to one of the two remaining possibles - after the host opens the door.

Because the host selects between two goats at random (so 50% of the time) when the opportunity presents itself, MCA will appear in both the set of games in which Red Mary is revealed and in the set of games in which Green Ava is revealed. Because there is a cross over between those sets, you end up with a 2/3 rather than 1/2 you get with a single iteration. This might not be perfectly explained, but hopefully you can grasp what I am getting at.

3. Honestly, no, I do not grasp what you're getting at at all.

You've said yourself that the choice of considering a Red Mary game is purely to nail down a specific scenario, not because there's anything special about that outcome. No matter whether it's a Red Mary game or a Green Ava game or a White Mary game or anything else, once the door is opened there are two possible arrangements which remain, and the probability of winning by switching will be equal for each outcome. That is, every outcome is isomorphic to Red Mary, and produces the same set of probabilities. And this is something you've said yourself, not something new I'm presenting. So if they're all isomorphic, why does it matter that when you play repeatedly you play more arrangements? How can that change the overall average probability, if each outcome is identical? Why do you believe that the argument you use in the second paragraph does not apply when determining the probabilities for one single iteration?

4. You asked me to respond to your long comment, I have done so - http://neophilosophical.blogspot.com/2015/02/the-objections-of-chrysics_27.html

4. What would it take to convince you that you're wrong?

1. Anon, I asked him the same thing over on reddit, and this is what I got:

"If someone could explain this to me: how is that, when you have three dice in front of you, two being similar but distinct, the third being quite different, each in its own distinct slot and you take away one of the similar dice, a very generalised analysis applies absolutely when you can see right in front of you that this is a very specific situation in which there are two and only two options for how the remaining two dice could be arranged. Explain only in terms of this very specific scenario, don't go into explanations about multiple runs. Don't bother about series of events. No-one, and I mean no-one, has engaged on these terms. If this is "not the right way of thinking about the problem", then we aren't considering the same problem."

5. This model of yours is very interesting. I think I'm starting to see where you want us to go ...

Let me propose the following three sentences to see if I get your point :

A - In all games that resulted in the opening of Red Mary by the host, the probability that the car is behind Green is the same as the probability that the car is behind White.

B - In all games that resulted in the opening of Red Mary by the host, the probability that the contestant selected Red&Green is the same as the probability that the contestant selected Red&White.

C - In all games that resulted in the opening of Red Mary by the host, after the contestant selected Red& Green, the probability that the car is behind Green is the same as the probability that the car is behind White.

It seems to me that you want to prove that sentence C is true, am I correct ?
But sentence C is a little bit complicated to understand in one step. So perhaps, we can decompose it into two sentence A and B, which are easier to analyze, right ?

It seems to me that sentence A is true (that's your convincing little experiment with dice).
And it seems to me that sentence B is also true (because the choice of the contestant was "random", as you said) .
If A and B are true, then it must mean that sentence C is also true.

Do you think that my reasoning is correct, and is this the reasoning that you are using ?

If this is not

1. Hi Mathematician,

I am pathetically appreciative of your more friendly and approachable tone, and the implied compliment! This is not diminished by the high likelihood that you are using the better approach to persuade something that they are wrong. I appreciate your using that better approach.

What you have written is well on the way to being a true reflection of the point that I am trying to get at. Sentence C does have an issue though. I've talked about the contestant being exposed to a single iteration, one shot version of the game repeatedly.

It's not true that over ALL the games that Red Mary is revealed that there will be an equal distribution of White Car and Green Car. As I said, if you run the scenario over and over again, you will get your 2/3 figure, so it will on average benefit you to switch.

I think I've said this before in one of many replies to comment, but perhaps I didn't say it well enough. I will try again.

Red Mary will be revealed when we have either MAC or MCA. Since Red-White (Doors 1 and 2) is equally likely as any other combination of doors selected (they are isomorphic, as someone better educated in this than me wrote recently), we can consider a Red-White version of a Red Mary reveal. We can notate this by separating out the unselected (Green, #3) door - so we could have either MA-C or MC-A.

If we have an MA-C, then Monty could have selected White Ava. The likelihood of our situation arising was therefore 1/2.

If we have an MC-A, then Monty is constrained to selecting Red Mary. The likelihood of our situation arising was therefore 1/1.

I see the standard logic which is that it is twice as likely, given a Red Mary reveal, that Red Mary has been revealed because there is the Car behind the White Door (the other selected door, door 2). I do understand that.

However, what I also understand is that it is less likely that a Red Mary reveal will happen on the basis of an MA-C spread, because 50% of the time that MA-C occurs, the host will reveal a White Ava. This means that even though MA-C is equally as likely as MC-A, we won't see it as often over multiple iterations. On a single iteration, one shot version of the game, (I believe) we should focus on the equal likelihood, not the less frequent appearance that manifests over multiple iterations.

Even if I might be wrong, do you get what I mean here?

2. > Sentence C does have an issue though. I've talked about the contestant being exposed to a single iteration, one shot version of the game repeatedly.

Ok, my mistake. Let me rewrite the three sentence then :

A - In a specific game that resulted in the opening of Red Mary by the host, the probability that the car is behind Green is the same as the probability that the car is behind White.

B - In a specific game that resulted in the opening of Red Mary by the host, the probability that the contestant selected Red&Green is the same as the probability that the contestant selected Red&White.

C - In a specific game that resulted in the opening of Red Mary by the host, after the contestant selected Red& Green, the probability that the car is behind Green is the same as the probability that the car is behind White.

Now, is sentence C the thing that you want to prove ? Is the rest of my reasoning equivalent to your reasoning?

Now, the following sentence is the thing I wanted you to say :

> Red Mary will be revealed when we have either MAC or MCA. Since Red-White (Doors 1 and 2) is equally likely as any other combination of doors selected (they are isomorphic, as someone better educated in this than me wrote recently), we can consider a Red-White version of a Red Mary reveal.

Would you agree that the sentence you just wrote, is equivalent to sentence B that I wrote ?

I should stop there and let you respond to this question, but I cannot resist the temptation to ask you another philosophical question about one of your sentence :

> This means that even though MA-C is equally as likely as MC-A, we won't see it as often over multiple iterations.

On a philosophical level, how would you define the property that two events are equally likely ?
(I don't want a mathematical answer)

3. >> Red Mary will be revealed when we have either MAC or MCA. Since Red-White (Doors 1 and 2) is equally likely as any other combination of doors selected (they are isomorphic, as someone better educated in this than me wrote recently), we can consider a Red-White version of a Red Mary reveal.

> Would you agree that the sentence you just wrote, is equivalent to sentence B that I wrote ? (Sentence B = "In a specific game that resulted in the opening of Red Mary by the host, the probability that the contestant selected Red&Green is the same as the probability that the contestant selected Red&White")

I don't think that they are precisely equivalent, but I think that you can draw from my sentence the conclusion that you state in your sentence.

> On a philosophical level, how would you define the property that two events are equally likely? (I don't want a mathematical answer)

A curious question. Are you telling me that I can't use mathematics? Are you suggesting that I can't look at uniform distributions over multiple iterations? My first impulse given the context is to say that if I run, say, a million iterations of allocating M, C and A to three doors at random, then I would expect to see ~333,333 instances in which C is behind the Red Door, ~333,333 instances in which C is behind the White Door and ~333,333 instance in which C is behind the Green Door. This means, to me, that it is equally likely (in any single instance) that M is behind any individual door. I would say, therefore, that the two events (C is behind the White Door) and (C is behind the Green Door) are equally likely events.

Hopefully that is sufficiently unmathematical for you :)

4. Dear Neopolitan, sorry it took me so long to answer.

I'm really really confused about how your brain works. For me, the definition (which is your own, I don't care if it's the correct one or not) you gave of "equally likely" is perfectly incompatible with the following sentence you wrote :

> "This means that even though MA-C is equally as likely as MC-A, we won't see it as often over multiple iterations."

This is mind blowing for me. Your definition of "equally likely" uses the fact that after multiple iterations the result will be more or less the same. And your sentence states that you have two equally likely events that will give different result after multiple iterations.

Both things cannot coexist in a rational mind. This means that your problem is deeper than a misunderstanding of probability theory. It means that you don't even understand how to use mathematical definitions.

Apparently, you wrote three more posts to explain how such a thing can happen. I am very disappointed !
Instead of spending countless hours to write those articles, you should have open a book on probability theory and started to learn the definitions, the vocabulary, the main theorems and so on ...

I repeat it one last time. I do understand exactly the point that you are trying to make. The various reformulations using dice and "new arguments" and "simulations" are useless for me. You see, I understand your point so well, that I could probably convince some random people that your point is correct. But, there is a subtle mistake in your argument !
You are doing many hand-waves to dismiss it but it's always the same. You consider that two things that are both "completely random" must always be independent (placement of car & selection of doors by the contestant). This is such an obvious fact for you that you are probably wondering why I'm even talking about that. I'm talking about that because, as counter-intuitive as it sounds, when you introduce the third variable (opening of a red door revealing Mary), the two things that were "completely random" and hence equally likely, become not equally likely. A mistake that is well hidden for someone who has no experience of that sort of things but that is as bright as the sun for anyone with experience.

As you cannot see it, you think that your reasoning is correct. I can see the mistake, I can see how to correct it, I did correct it, I gave you a detailed mathematical answer, and you never replied. So apparently you only believe your own arguments. So as you cannot be convinced by a mathematical proof, I am helpless.

This will be my last words (and my last post) : Your arguments are wrong.
You see, you have perhaps a dozen different arguments to say that the probability should be 1/2. I only have one detailed and complete proof. So the battle is unfair, I will win every time ...

Feel free to comment, but play nicely!

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