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.

---

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

**fly in the face of the Law of Large Numbers” we end up talking past each other.***doesn’t*
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.

---

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.

---

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).

---

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.

I know that you are already convinced that your argument is wrong (and that I said that I wouldn't post anymore), but it seems important to emphasize where exactly I think you were fundamentally wrong (not only for you but for other readers as well).

ReplyDeleteThis sentence is the key to your misunderstanding :

> However, prior events do always not affect the present.

First, this is an english sentence, not a mathematical one. The reason formalism and proofs are useful is to avoid using these kind of "obvious" statements that are, in fact, not logically correct.

As I already said, this sentence is perfectly true for independent events. Like in your "tossing a fair coin and getting seven heads in a row" argument. The eighth throw is independent from the first seven ones, so that the probability for the eight throw being heads is 1/2, whatever were the first seven results. The definition of independence is almost exactly saying that the past does not affect the present (it's even more general than that, but anyway ...)

But then, your sentence is not true anymore for events that are not independent ! That's exactly why conditional probabilities are introduced !! Otherwise, if your sentence were always true, the definition of conditional probability would be completely useless. (By the way, you kept repeating that someone said that conditional probability did not apply to this scenario. I don't know who said that, but it is nonsense. You can "apply" conditional probabilities to all the situations you want ! And especially here you have to do it)

So, the opening of the red door is not independent from the other variables of the game (placement of cars, choice by the contestant). Hence, to compute the wanted likelihood of winning by staying, you use the formula for conditional probability : the probability that (the car is behind the Green door knowing that the contestant chose Red&Green and the host opened Red), is equal to the probability that (all of these three events take place (1/9)) divided by the probability that (the contestant chose Red&Green and the host opened Red (1/6)).

A priori, there is no reason to believe that this computation could be simplified ! (and even then, it is often better to do the computation with all possible details if you want to be sure that you are not oversimplifying). So you just compute the answer, and you find P = (1/9)/(1/6) = 2/3.

You've identified an intensely irritating typo (subsequently fixed). It's not really English, in so much as it's grammatically incorrect and it should read "However, prior events do not always affect the present."

DeleteNote that the structure as written was ambiguous, so I disagree that it was not true for events that are not independent - that would only be the case if I had written "prior events always do not affect the present" - which is not a true statement, because there are prior events that do affect the present. Mathematical formalism might be significantly more important to you than for me because you have a higher tolerance for grammatically incorrect and thus ambiguous phrasing in English. And don't get me started on "All oils are not the same". If you think that this is a fine and dandy sentence, then please consider "All people who call themselves mathematicians are not intelligent".

That all said, and correcting what you wrote to account for the typo, you are saying that it is true that not all prior events affect the present (because independent event ones don't affect each other - including simultaneous events, if that was your point) but in this case we don't have independent events, therefore conditional probability. Agreed, but I do think that there is something else going on in there - as exemplified by the dice and the Monty Falls variant.

I went back to see who it was who said that conditional probability didn't apply, it was a user called /u/chrox, and it was only that one person. For some reason, this took on a much larger significance than it should have, but I do note that no-one corrected the user and I interpreted some of the later comments as saying "you, neopolitan, only think that conditional probability applies because you don't understand conditional probability", which of course led to irritation that when you, Mathematician, suggested that I read up on it. I know that conditional probability applies, which you now confirm, but I was being told by a few people (I thought) that it didn't. It was an example of "tunnel vision", I suspect.

It is interesting that intuitively we believe that when a door is eliminated that a new state of affairs is created where we now have 50/50 chance of having the car behind our door. Logically it makes no sense to think this because after the contestant picked a door, no matter what, there was a goat remaining behind one of the other two doors, so eliminating a goat makes no difference at all to the original position. Logically, nothing at all has changed by knowing which of the two remaining door has a goat, yet for some reason we think it has. The information has not changed in the world, but information has changed in the brain because now we know something additional; which other door has the goat. It is a mistake in thinking that what we know has consequences in the world.

ReplyDeleteI think that you misrepresent probability when you speak of information "in the world" versus information in the mind. Quantum mechanics aside, in the world, the goat is completely behind one of the doors, 100%. The only reason that we come to a probability of 2/3 and so on is because we have limited information. If we had a viewpoint "from the world," we would know which door the goat was behind, collapsing the probability to 100%. But, as you suggest, in the reverse Monty, the opening of the door adds no new information to our point of view, and so the probability remains at 2/3.

DeleteBut probability is entirely dependent upon the information we have, not upon the information "in the world."

-B

One philosophical problem us mere humans face is that mathematics is able to describe reality where ordinary language fails. Specifically I am referring to QM. The Monty Hall problem may just give us a glance into what is our cognition doing wrong.

ReplyDeleteSo I put the question up as to why we intuitively fail in the MHP in an analytical philosophy group. The conclusion I reached was that we fail to shift perspective from ourselves to the other side where the two doors after a choice should be treated as 1 item giving a 2/3 probability on the other side.

ReplyDeleteAnonymous I guess that boils down to whether you take ontic structural realism seriously or not. I do.

ReplyDeleteSomething to throw a spanner in the works here, is that in Monty Fall the contestants chances are NOT 50/50. T hey only become 50/50 if Monty can fall on the door with a car and not only the goat door. Here is a paper which outlines the issue;https://www.ffri.hr/phil/casopis/abstracts/9_2_3.pdf

And simulations confirm this.

It does not boil down to the belief in ontic structural realism. I could be an ontic structural realist and still hold that using proper empirical methods and mathematical descriptions, we would be able to describe the relationship between the goat and the door such that we could say which door the goat was behind, 100%. Of course this would be breaking the rules imposed by the Monty Hall situation, which limits our knowledge. But a complete description of the situation, even under a structuralist viewpoint, would not yield probabilities such as 2/3 or 1/2.

Delete-B