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).
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.
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
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.
---
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).
---
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.
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.
---
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.
---
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.
---
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.)
> 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.
ReplyDelete> 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.
No, the chance in this instance is 50/50. This is similar to the Monty Fall setup (and the Monty Faffs about), and different from Monty Hall.
What you are missing is that it is important *how* the contestant learns that he or she is in a Red Mary situation. In this new case he learns it through luck: in 1/3 cases (MAC MCA from a total of 6 equally likely cases) will the host open the red door to reveal a Red Mary. This leaves 2 cases, one which he wins if he switches (MAC), and one in which he loses when he switches (MCA).
Now in the case where it's the host who chooses to reveal a goat there are not 6 cases to consider, but 8 (MCA MAC-OSU MAC-SOU ACM AMC-OSU AMC-SOU CAM CMA), 4 of which are half as likely to occur. After the host has opened a door there are 2 possibilities left, but no matter which door the host opens and no matter which goat he reveals the possibilities will always be of unequal likelihood (i.e. there is always 1 where the host is forced and one where he has a choice)
I hope you can see for yourself that this difference affects the probabilities.
I'm a little sad that you agree so readily that the likelihood associated with Monty Faffs About is 1/2. I'm not sure whether you agree that the Monty Hall Problem and the Reverse Monty Hall Problem are subsets of the Monty Faffs About games (subsets with specific faffing about).
DeleteI think I must agree with Nick's first comment (if I interpret it correctly) to the extent that he indicates that the little chat between the host and the contestant makes no difference. I think you might have been distracted by a turn of phrase "you're in luck". If you look back at Marilyn Gets My Goat you see that the host doesn't choose which door to open, she opens a door at the instruction of her producer. Therefore, the additional chit chat should not have any impact at all on the opening of the door and it's merely coincidence with respect to the talked about door being opened to reveal the talked about goat (this is a 1/3 chance, so it's not a huge coincidence by any measure).
If you maintain that it does have an impact, then we get into the situation in which the contestant could run the conversation in her head. Would that have an impact?
I think you are assuming that because the contestant identified the Red Door and Mary, the host was thus obliged to open the Red Door revealing Mary. That wasn't stated anywhere at all. The question then is, do the assumptions of the contestant affect the likelihoods? I'd suggest not.
Okay, I assumed that in this new scenario the host opened the door always if the contestant guessed correctly that a Red Mary event took place. If that wasn't the case then I still maintain that there is a 2/3 chance that the car is behind the white door (assuming the host is still unbiased and is not trying to be cruel (i.e. he could always choose to open the other door if the contestant guessed the goat behind her chosen door correctly, that would alter the probabilities further if the contestant was aware of it))
DeleteHowever this makes it very clear again why there could be a difference between P(Green Car|Red Mary) and P(Green Car|Host revealed Red Mary). If there is a chance that the host doesn't reveal Red Mary in a Red Mary event then these are fundamentally different questions and there is no a priori reason why the outcomes should be similar at all.
As far as Monty Faff, I got the answer wrong there, I read it incorrectly. It's not the same as Monty Fall (but it's still 50/50). Why is this different from Monty Hall (or the revers Monty Hall problem)? Because here the host always has a choice of which goat to eliminate, this means that all remaining options are equally likely. Why does it differ from Monty Fall? Because there the host is more likely to fall into a door with a goat if the contestant originally selected the door with the car.
Thus monty faffs differs from the situation where the contestant chooses first, with reverse monty hall there sometimes is only 1 goat to eliminate by the host (in which case staying wins) and sometimes there is a choice (in which staying loses). Because the contestant knows nothing about the distribution of the goats he will select 1 goat in 4/6 possible situations, leaving a 2/6 chance that switching wins the game.
Neopolitan, I would have thought that the solution to your "Monty Two Face" problem might have cautioned you against continuing your 50/50 argument for the RMHP - apparently not.
ReplyDeleteMathematically the solution to the RMHP where the contestant has selected the RED and GREEN doors and Monty has opened the RED door to reveal Mary is shown below. Unsurprisingly the answer is 2/3.
Let Car"Door" (where "Door" = R,W,or G) denote the event the Car is behind the RED,WHITE or GREEN door.e.g 'CarG' is the event the Car is behind the GREEN door
Let "Door""Goat" (where Door" = R,W,or G and "Goat" = Mary or Ava) denote the event Monty opens "Door" to reveal "Goat. e.g 'RMary' denotes the event Monty opens the RED door to reveal Mary
We want to calculate P[CarG|RMary] - the probability the Car is behind the GREEN door given Monty opens the RED door to reveal Mary (you having picked the RED and GREEN doors initially)
Bayes Theory tells us: P(CarG|RMary) = P(RMary|CarG).P(CarG]/P[RMary)
The Law of Total Probability tells us: P(RMary) = P(RMary|CarG).P(CarG) + P(RMary|CarR).P(CarR) + P(RMary|CarW).P(CarW)
P(CarG) = P(CarR) = P(CarW) = 1/3 (since the car is equally likely to behind any door)
P(RMary|CarG) = 1/2 (since if the Car is behind the GREEN door Monty MUST open the RED door which hides either Mary or Ava with equal probability)
P(RMary|CarR) = 0 (since if the Car is behind the RED door Monty can't open it)
P(RMary|CarW) = 1/4 (if the Car is behind the WHITE door, then the RED and GREEN door configuration is either 'Mary,Ava' or 'Ava,Mary' with equal probability, and Monty picks a door at random.)
Substituting these values into our orignal equation gives:
P(CarG|RMary) = [1/2*1/3]/[1/2*1/3 + 0*1/3 + 1/4*1/3] = [1/6]/[1/4] = 2/3
So,staying with the GREEN door gives you a probability of 2/3 of winning the car.
Hi Marley, I wrote a version of this above, in the article that you are commenting on and irishsultan wrote a version in the comment that I was responding to. I'm at a bit of a loss as to what you are adding. Are you suggesting that I or irishsultan came up with your answer the wrong way?
DeleteIn the plethora of posts and different blogs I'd assumed that you were still arguing that it was 50/50 in a single instance of the RMHP after Monty revealed a goat. If that assumption was incorrect and you now realise that staying is a 2/3 chance then I apologise..
DeleteWell, I found irishsultan's argument appealing, but I still have an issue as described in the article above.
DeleteThe argument (that staying is a 2/3 chance) is more than appealing, it is correct. I have no idea what you are now disagreeing with, if in fact you are disagreeing with anything.
DeleteHi!
ReplyDeleteWhat we need now is more people in this mess!
So, before I begin, I've thought this through a bit, but I might be hasty in my conclusion, someone will correct me if I'm wrong.
There is an important difference between "you gave information on where one of the goats are" and "you have information the goat behind one of your two doors". In what you call Red Mary games, we get the information that Mary is behind the red door, independent on whether red is a color you chose or not if you first choose two colors/doors and then you learn that Mary is behind one of these, then you have gotten a different type of information.
So if you want to analyze the situation where the host asks you to choose a door and a goat, and gosh darn you guessed correctly, you would have to include the situations where you chose the white (last) door as well, and then the host could not in any situation tell you if you were right or not! In your scenario, the Red Mary guess is not independent on the doors chosen! This turns the situation you describe into a 2/3 scenario as well, once you account for all the probabilities.
What irishsultan says is possibly also correct, it depends on whether the host opens the door you chose if possible or he the host still just opens a random door if there is a choice.
DeleteAlso, that should have been "you have information on where one of the goats are", and "you have the information that the goat is behind one of your two doors". Dang autocorrect.
DeleteWelcome Nick, the more the ... well, more.
DeleteSome people have pointed out that issue here is subtle, but I don't think you've quite got my scenario. In my scenario, the contestant is required to assess the likelihood of winning as a consequence of switching after the doors have been selected and the door has been opened. At this point the "information on where one of the goats are" is precisely the same as "the information that the goat is behind one of your two doors". In my example, it's the Red door after you've selected the Red and Green Doors. You know that Mary is there, you can see her.
The issue I have revolves around whether knowing you know and just knowing are meaningfully different.
Yeah, but how you got to know the information is different, so they end up giving different outcomes.
DeleteYou know where one of the goats are BECAUSE you know it's behind one of your two doors. This gives different probabilities than if you just knew where one of the goats were independent on the doors you chose.
Let's agree on some things, when it comes to the host:
1. The host will always open a door with a goat behind it.
2. The host will always open one of your two doors.
3. If this does not uniquely specify a door, the host will choose a random door.
In the situation you describe, where we're guessing that there's a Mary behind the red door, we start with the following possibilities:
MCA
MAC
AMC
ACM
CAM
CMA
All of these are equally likely, probability 1/6.
Now, if we knew that Mary were behind door 1, independently on which two doors we chose, then we could simplify this situation to the following two scenarios:
MCA
MAC
Both equally likely, so you'd win the car 50% of the time if you ended up on door 2, and 50% of the time if you ended up on door 3.
However, we are not in that situation, because the information we end up getting is dependent on the doors we choose. If we had chosen door 2 and 3 to start with, then we would never get the information that Mary is behind door 1.
So, let's look at what kind of information we have as a contestant. We know that we had 6 scenarios, all equally likely. We also guessed wildly that Mary was behind the red door, and we know we were right. So, how could we end up in this situation?
There's only two possible set-ups for the doors, MCA and MAC. However, in one of these scenarios, the host is forced to open the door with Mary behind it, Let us look at all possible set-ups and remove the ones we know do not apply.
MCA 1/6 chance. The host will reveal Mary 100% of the time. Possible. Probability 1/6
MAC 1/6 chance: The host will reveal Mary 50% of the time. Possible. Probability 1/6*1/2 = 1/12
MAC 1/6 chance: The host will reveal Ava 50% of the time. Impossible.
Rest: Impossible. Mary is not behind door 1.
So, as we can see, it's more probable that we are in our current situation because the original set-up was MCA than because it was MAC. In fact, the probability that we are in our current situation because the original set-up was MCA is 2/3.
I think my point here is that since you are forcing your contestant to guess that Mary is behind one of the two doors chosen, this gives you a 2/3 probability of winning by staying. We have MORE information than just "that door has a goat behind it". We also know that "that door is one of our two chosen doors". We can leverage this to increase our winning-probability up to 2/3.
As I said, I'm not interested in continuing the discussion at this point. But I am still interested enough to keep track of your attempts to evade reality, and I'll just point out that:
ReplyDelete1. That still doesn't answer the question of how a game in which Mary is behind the Red Door but the contestant picked Green and White is not a game in which Mary is behind the Red Door.
2. Your objection to my argument is yet another case of you ignoring facts I've told you multiple times. The argument you quoted applies only when considering the set of Red Mary games (i.e. it does not answer the question you're asking). It does not apply when considering the set of Revealed Red Mary games (the situation applicable to the question you're asking). For some reason you're continually declaring that I "agree" with you (which I don't), on the basis of an argument I've repeatedly told you does not apply to the question you're asking. You still make no attempt to explain why I am wrong in asserting the argument does not apply.
On your first question, does it actually matter much? After all, you can always relabel the doors so that the contestant picks the red door and the white door (which is why it's pretty silly that they are labeled like that anyway), this doesn't change any probabilities.
DeleteYes, it matters. The point is that the "Red Mary" label does not show which door is opened. Neopolitan decided at some point that it would be useful to focus on a specific scenario, rather than the abstract, and chose "Red Mary" as that scenario. At first everybody assumed that this meant "games where the host opens the Red Door, and Mary is behind it", because that's the only rational assumption when we're being asked to judge the contestant's deductions after the host has opened a door. Then at some point a couple of days later neopolitan declared this is actually not what was meant at all, and instead it refers to "the games in which Mary is behind the Red Door" - regardless of whether or not the Red Door is opened. If Mary is in a fixed location and you relabel the doors so that the contestant picked Red and White, you have to also move Mary. And as the scenario doesn't correspond to the contestant's perspective on the game, this actually can alter the probabilities.
DeleteUsing this definition has two important resuls:
The first (and most important, imo) is that Red Mary does not have anything to do with the information available to the contestant. We're talking about the probability deduced by the contestant, after the host opens the door. But the contestant in an arbitrary Red Mary game only has a 50% chance of actually knowing that it's a Red Mary game. It's not actually a useful label in any eway useful for figuring out what the contestant knows, and (unlike the "Revealed Red Mary" situation) specifying that we talk about a Red Mary game doesn't actually give us any real insight into th contestant's thought process, since we can't make any assumption about what the contestant knows.
The second result is that neopolitan's argument about MCA and MAC being equally likely, thus giving the contestant a 50% chance of winning by switching, does in fact hold for a subset of "Red Mary" games (it's complete invalid for "Revealed Red Mary" games). If Mary is behind the Red Door and the contestant picked the Red Door then there are two possible arrangements of goats & car - one of which will see the contestant win by switching, one of which will see them lose by switching. And we have no basis to declare either arrangement more likely than the other, because when we consider "Red Mary" games (rather than "Revealed Red Mary") this doesn't tell us anything about which door was opened. We just know that Mary is behind the Red Door, even if the Red Door is still shut.
Neopolitan's been quoting the part of one of my comments where I said that this produces a 50% chance if the contestant picked the Red & Green Doors (or the Red and White Doors), and repeatedly referred to this as me "agreeing" with the claim that the contestant has a 50% chance of winning in a "Red Mary" game. For some reason, Neopolitan's interpretation of the label "Red Mary" includes the games in which the Red Door is selected by the contestant but not opened by the host, but excludes the games in which the Red Door was not selected by the contestant. Excluding the latter category removes a scenario which is non-isomorphic to the others - and as such, it changes the probabilities. Indeed if you count "Red Mary" games as those in which Mary is behind the Red Door and the contestant chooses the Red Door (but it's not required to be opened) then the contestant does indeed have a 50% chance of winning. Choosing to look at this subset of games makes absolutely no sense, the "Revealed Red Mary" scenario is far more useful. But Neopolitan seems to prefer the (entirely flawed) "Red Mary, assuming the contestant picked the Red Door" category because that produces Neopolitan's desired answer (though only because it no longer resembles the question being asked).
(second comment because I hit character limit. which doesn't work properly, fwiw: claims there's a 4096 character limit but actually seems to kick in somewhere in the region of 3900)
DeleteUltimately you're right if what you mean is that the question doesn't matter at all to the issue of "what's the probability that the contestant will win by switching". But it does matter if you're (as neopolitan has been) trying to argue that it must be 50%, on the basis that Red Mary games give a 50% chance of winning - because they only do so if you refuse to count the Red Mary, contestant picked White & Green category, but there's no logical justification for that. And I'm sure neopolitan knows it because I don't believe anybody accidentally ignores the same question this many times.
I accept that I was not completely clear as to what a "Red Mary" game is at the start. I'm going to try to explain again, but I am not certain that I will be successful this time either:
DeleteWhen the host opens the Red Door to reveal Mary, the contestant knows, at that time, that she is in a Red Mary game in which she (the contestant) selected the Red Door and the Green Door. I fully understand that at that time, the contestant does not know whether she is also playing a White Ava-Green Car game or a White Car-Green Ava game.
Similarly, I understand that if the situation were different (which it isn't in my scenario as described in Marilyn Gets My Goat), the contestant could have been in a Red Mary game without knowing it, because either the host could have opened potentially have opened the Green Door to reveal Ava (in a similar Red-Green selection scenario) or the contestant could have selected White and Green or Red and White doors potentially allowing the host to open the White Door to reveal Ava. The point that I don't seem to be able to make clear to you is that in my scenario, the host opening the Green Door didn't happen and the host didn't open the White Door after the contestant had selected the White Door and one other.
In my scenario, the contestant selected the Red and Green doors and the host opened the Red Door revealing Mary. We can work out the probabilities (which I am swinging around to believe amount to 2/3 likelihood of winning from staying) while considering only the scenario that actually happened. Therefore, I am now almost totally convinced that your result is right, but I think you arrived at it incorrectly - given my scenario. The approach the irishsultan used was more appropriate for my scenario.
Re the 1/2 result, as I wrote in response to someone else, I still find this very interesting because it's what the dice tell me. I think that what you have said elsewhere might be very useful for working out why the dice tell me it's 1/2 even though the result in the scenario I describe, as I am beginning to accept, is 2/3. I need to ponder it some more though.
"When the host opens the Red Door to reveal Mary [...] The point that I don't seem to be able to make clear to you is that in my scenario, the host opening the Green Door didn't happen and the host didn't open the White Door after the contestant had selected the White Door and one other."
DeleteNow you're back to defining a Red Mary game as one where the Red Door has been opened with mary behind it. Which is exactly what everybody believed you meant in the first place, until you made a big kerfuffle about how despite all evidence to the contrary that totally wasn't what you meant and you were actually talking about any game in which Mary is behind the Red Door even if the host doesn't open it.
"but I think you arrived at it incorrectly - given my scenario"
I've made it very clear several times that I don't believe the Red Mary (no door-opening requirement) scenario is a good choice, because it does not correspond to the knowledge the contestant has and does not pin down a subset of games with uniform probabilities equivalent to the situation the contestant finds themself in. But if you're using it, you can either consider all games where Mary is behind the Red Door (including when the contestant didn't pick the Red Door at all) or calculate the wrong answer. Or more sensibly just pick a better scenario (such as "Revealed Red Mary") to consider in the first place.
" The approach the irishsultan used was more appropriate for my scenario."
Indeed it is. Irishsultan's approach (which calculates possibilities with the prior condition that the contestant knows they are in a Red Mary game) turns it into the Revealed Red Mary scenario. Which I've repeatedly told you is the correct way to approach it.
I stand by what I wrote in the link. I don't see how what I wrote above is inconsistent with what I wrote in what you linked. Perhaps we are focussing on two completely different things. I really don't know why I am finding this so difficult to explain to you, I think it might be because you don't read the whole comment since in my comment I wrote "I understand that if the situation were different (which it isn't in my scenario as described in Marilyn Gets My Goat), the contestant could have been in a Red Mary game without knowing it" - this indicates quite clearly that the door doesn't have to be opened on a Red Mary for it to be a "Red Mary game".
DeleteNote that in my first response specifically to you in article form I wrote "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." In that article I also wrote "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 now notice the typo ...)
I'm willing to accept that it's possible that you were repeatedly (albeit implicitly) telling me to use the Revealed Red Mary scenario, but I couldn't simply see it because in my own muddled head I was reading you as mentioning scenarios in which the host didn't open the Red Door and scenarios in which Mary could not have been behind the opened door, scenarios which I (in my ignorance) could never justify referring to as a "Revealed Red Mary". It is mere fortune that, in discussion with irishsultan, I was able to arrive at what you had been telling me from the start, much later, via another route.
I don't see it in the actual text, but it's probably there between the lines.
And anyway, at least I made it eventually, huh?
(I know that I have been stubborn and arrogant and boundlessly stupid, characteristics that many dislike. I personally dislike it intensely when someone attempts to takes credit for something that they don't have the right to take credit. We're all different like that.)
OK, on reading again I can see that I misinterpreted that part of your post. Apologies for that.
Delete(I disagree that there was anything implicit or 'between the lines' in me saying that Revealed Red Mary was the appropriate scenario - it seems to me the final paragraph of this comment made that point quite explicitly, but that's not worth arguing about - it's easy to miss something when the comment is so long.)
Your analysis of "Monty Faffs About" reveals an assumption that you didn't make explicit when describing the game, namely the non-treasure slot which is removed from consideration is chosen at random. The (implied) fact that the six (_,X,~) arrangements are equally likely, and the fact (again, implied, not stated) that the contestant's strategy has no bearing on the result, requires this assumption.
ReplyDeleteSo while you've attempted to make a "general", "overarching" game, what you've actually done is to specify exactly one game, which manifestly does not correspond to MH or RMH. You might as well call it, "Monty Opens a Random Goat Door". To answer your question, I do agree that in THIS game, the contestant has a 1/2 chance to win in any single instance, regardless of strategy, and that the long-run proportion of wins must approach 1/2.
I disagree that MH or RMH are "instances" of this game. The rules of all three are known, and it's easy to see that MH and RMH are equivalent to one another and inequivalent to MFA.
On the other hand, if you really want MH and RMH to be instances of MFA, then you can no longer assign any probability to winning at MFA, since it's no longer a well-defined game. It contains both MH/RMH (where the long-run win percentage with best strategy is agreed by everyone to be 2/3), and MORGD, which is clearly 50/50.
Deleteneopolitan said, "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."
ReplyDeleteOh, come on. Nobody is suggesting a distinction between knowing and knowing that you know. I hope we can all agree that, if we know Mary is behind Red, there is an infinite family of facts we also know, of the form, "I know that I know that I know that...I know that Mary is behind Red". This has no bearing on anything and we should move past it as quickly as possible.
When the host opens the Red door to reveal Mary, we learn two things: first, that Mary is behind the Red door (which we knew already) and second, that the host CHOSE to open the Red door. It's that second piece of information that alters the probabilities, as has been discussed endlessly already. In one situation he'll be forced to open Red (staying wins), while in the other (switching wins) he'll also have the option to open Green.
I agree that there is a problem with knowing that you know that you know that you know that you are falling into endless repetitions of "that you know". I think, therefore, that it is not purely that you know that you know (which is the core idea behind the "Revealed Red Mary" concept) but rather how you know. This seems to be already accepted because if you know because you are semi-god like and only know that Mary is behind he Red Door, or you know that it's a Red Mary game because Monty fell over, or Mary revealed herself in a fit of rage, then the likelihood that you will benefit from a switch is 1/2. So, yes, it is that we know that the host chose to open the Red Door. But that's not enough in itself either. For the result to be 2/3, the host must make the decision based on the selection of doors by the contestant and the host must choose between two goats at random (and fortunately that is built into the scenario).
DeleteAs you can see, I am pretty much settled on the conclusion that I was wrong. I will update my articles appropriately once I've resolved the "what does Red Mary game mean?" dispute with chrysics.
When do you think you'll be finally settled to the conclusion that you wrong, and will you update reddit admitting such?
DeleteI am settled, I have started the process of trying to convince those I convinced that 1/2 is the answer that 2/3 is the answer (and explaining why), and I just posted another article in which I do announce that I am now convinced of the 2/3 answer. As to updating reddit. I can update those three instances where I posted a link, but it's not really my responsibility to address those where someone else posted a link. What I intend to do is top and tail my articles with my admission that I was wrong, once I've resolved that one final issue with chrysics or when I've given up on him responding, which will likely be after the weekend.
DeleteReddit is now updated in those instances where I posted a link.
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