Please
note that since I wrote this article, I have been persuaded that the argument
it contains is wrong. 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
following is tediously pedantic, because I have to walk people through the
process step by step. Please try to read
all of it and go with it. Don’t jump
ahead, ask yourself each question and see if you get the same answer as
Marilyn. If you don’t, then we can
discuss that.
---
It’s pretty universal from all the
comments (both on the blog and over at reddit) that I must have
it wrong. I’ve tried repeatedly to
explain that, also pretty universally, the criticisms are based on a
misunderstanding of my scenario. That’s
what this article is all about.
It’s for those people who seem to almost
deliberately misunderstand what I have written.
I don’t think people deliberately
misunderstand. I think that it’s because
what I am talking about is so close to something else that is widely
understood, most people stop really listening and presume that I am talking
about the thing that they are familiar with.
There have been some people who have
raised other, excellent points. I will
try to address those in a later article, so that this can focus on one thing –
the actual scenario of the Reverse Monty Hall Problem. I’ll do it in the form of a little role-play.
---
Post Burnage Comment:
I
would like to stress that this scenario could be performed in real life. To do so, all we would have to do is:
- engage the services of Angelina Jolie and Marilyn vos Savant,
- engage the services of two goats, one called Mary and the other called Ava,
- establish a broadcast studio called “the Craig F. Whittaker Memorial studio”,
- purchase three doors, one Red, one White and one Green, and
- organise for the concept designer to donate the car as a prize (but we could conceivably use a different car).
---
Holly Mant (played by Angelina Jolie) is
the host of the single episode game show Marilyn Gets My Goat, broadcast
from the Craig F. Whittaker Memorial studio.
There is one single contestant, Marilyn vos
Savant, who plays one single round.
Holly explains to Marilyn the key
features of the game:
There
are three doors:
- A Red door the left,
- A White door in the middle, and
- A Green door to the right
Holly shows Marilyn pictures of the doors:
 |
| Note the the doors are different styles (images taken from various stock catalogues) |
Behind
each of the doors is hidden one of two goats (Mary and Ava) or a car:
The placement of goats and car is
completely random
Holly shows Marilyn pictures of the goats
and the car (note carefully the non-interchangeability of the two goats):
Marilyn
will be asked to assess various relevant probabilities during the process (so
all the questions below are posed to Marilyn and the answers are Marilyn’s).
If
at any time, Marilyn is uncertain about anything, she can ask any question she
likes of Holly.
The
game will proceed with the following steps:
Step
1. Marilyn will be asked questions about
initial likelihoods
Step
2. Marilyn will select two doors at random
Step3. Marilyn will be asked questions about “post-selection”
likelihoods
Step
4. Holly will open a door on instructions from her producer and will offer
Marilyn
the opportunity select from the two remaining closed doors (either switching or
staying)
the opportunity select from the two remaining closed doors (either switching or
staying)
Step
5. Marilyn will be asked questions about “post-door-opening” likelihoods
Step
6. Once Marilyn provides the correct answer, she is permitted to open the door
of her choice.
of her choice.
The game commences
Initial likelihoods
Q1
What is the likelihood of the car being behind the Green door?
There is an equal likelihood that the car is behind any individual door, irrespective of the colour, so the likelihood of the car being behind the Green door is 1/3
Q2
What are the possible distributions of goats and car behind the Red, White and
Green doors, using the terms M for Mary, A for Ava and C for the Car?
- Red-A, White-C and Green-M
- Red-A, White-M and Green-C
- Red-C, White-A and Green-M
- Red-C, White-M and Green-A
- Red-M, White-A and Green-C
- Red-M, White-C and Green-A
Q3
How likely are each of these distributions?
They are all equally likely at 1/6
Marilyn selects her two doors at random
– she selects Red and Green
“Post-selection” probabilities
Q4
What possible choices of doors did you have, Marilyn?
There were only three choices:
And these are functionally equivalent to:
- Red and White
- Red and Green
- White and Green
- Not Green
- Not White
- Not Red
Q5
Given that you selected your doors at random, what was the likelihood of each
of these choices?
Each choice was equally likely, therefore 1/3
Q6
Now that you have selected Red and Green, what is the probability that you
selected Red and Green?
100%
Q7
What are the possible distributions of goats and car behind the Red and Green doors
(using A for Ava and M for Mary)?
- Red-A and Green-C
- Red-A and Green-M
- Red-C and Green-A
- Red-C and Green-M
- Red-M and Green-A
- Red-M and Green-C
Q8
How likely are each of these distributions?
All equally likely at 1/6Q9 What is the likelihood that the producer will instruct Holly to open the Red door?
The likelihood of Red C is 1/3 and the likelihood of Green C is 1/3. In either of these cases the producer will instruct Holly to open the other door. Therefore we have:
There is also a 1/3 likelihood that there are goats behind both the Red door and the Green door. In this case, I don’t know the basis on which the door to be opened will be selected, so the best I can say is that it will be random.
- a 1/3 likelihood that the Red door must be opened in accordance with the rules, and
- a 1/3 likelihood that the Green door must be opened in accordance with the rules.
(Holly takes this as a
clarifying question and confirms that, in these circumstances, the door will be
selected at random.)
Therefore, if there are two goats behind the Red and Green doors, there is a 1/2 likelihood that the Red door will be opened and 1/2 likelihood that the Green door will be opened. Therefore we have:Q10 What is the likelihood, that the Red door would have to be opened, in accordance with the rules, on the basis that that the car was behind the Green door?
Totalling this up, we have:
- a 1/6 likelihood that the Red door is opened at random because there are goats behind both the Red door and the Green door, and
- a 1/6 likelihood that the Green door is opened at random because there are goats behind both the Red door and the Green door
- a 1/2 likelihood that the Red door will be opened, and
- a 1/2 likelihood that the Green door will be opened
There is a likelihood of 1/2 that the Red door will be opened. There is a likelihood of 1/3 that a goat is behind the Green door and a likelihood of 1/6 that there is a goat behind both doors. Using conditional probability, we can say that if the Red door is opened, then the likelihood that it was opened because there is a car behind the Green door is 2/3.
This
is, effectively, equivalent to the answer to the classic Monty Hall
Problem. However, I note that Holly
has not yet opened the door.
---
Please stay with me. This is where it gets interesting!
When Craig F. Whittaker posed his
question in 1990, he asked this:
“Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?”
Note that Whittaker states that the host
has opened the door and that only then is the contestant offered the
opportunity to switch. The contestant
must calculate the likelihood of benefitting from a switch after
the door has been opened.
Does it make a difference?
---
Holly opens the Red door revealing Mary
the Goat and offers Marilyn the opportunity to switch from the Green door to
the White door
Q11
What is the current likelihood that there was a goat called Mary hidden behind
the Red door?
1/1Q12 Assuming that Holly has told the truth, what is the current likelihood that there is a car behind either the White door or the Green door?
1/1Q13 What are the possible distributions of goat and car behind the two remaining closed doors, knowing that Mary was behind the Red door?
- White-A and Green-C
- White-C and Green-A
Q14
How likely were each of these distributions before the goats and cars were
placed?
They were equally likely distributions, out of six possible, so 1/6
Q15
How likely are each of these distributions now?
They remain equally likely distributions, but now out of two possible, so 1/2Q16 What about the likelihood that the producer would instruct Holly to open the Red door being 2/3?
That was before Holly was instructed to open the Red door. The likelihood of the Red door having been opened now is 1/1 irrespective of the motivation. Similarly:
In these two instances, we know that prior probabilities have changed as a consequence of the door being opened. The same applies to the likelihood that the producer might choose the Red door.
- the likelihood of the car being behind the Red door was 1/3. But now it is 0, and
- the likelihood of Mary being behind the door was 1/3, but now it is 1/1
Q17
So, Marilyn, in conclusion, what is likelihood of the car being behind the
White door, the single door that you did not originally select?
Exactly the same as the likelihood of it being behind the Green door, 1/2
Holly offers Marilyn the opportunity to
switch
Given
15 years or more of being convinced that a switch in a Monty Hall game is the
best strategy, despite the logic above - or because she realises that it doesn't make a difference, Marilyn does not switch.
Holly opens the Green door
If
you want to know if Marilyn got Holly’s goat, please toss a coin.
Heads
she did, tails she didn’t.
---
---
As mentioned above, this argument is wrong. Precisely which bit is wrong is a little vexed. I originally thought the biggest issue nestles in Q10 and Q16. I still think it does, but many below argue that the issue is in Q15.
I have finally got around to addressing this in (My) Ignorance Behind Marilyn Gets My Goat, hopefully without sparking huge controversy this time. See also The Whole Monty Hall Debacle.
>Q15 How likely are each of these distributions now?
ReplyDelete>They remain equally likely distributions, but now out of two possible, so 1/2
This isn't correct. To demonstrate why I'll quickly introduce my notation R1-W2-GC to indicate the event that the red door hides Goat #1, the white door hides Goat #2, and the green door hides the car.
Now the contestant has selected the white door, and the host has then revealed Goat #1 behind the red door.
So we have either R1-W2-GC or R1-WC-G2.
Currently we have probabilities:
P(R1-W2-GC)=P(R1-WC-G2)=1/6
giving
P(host reveals red AND R1-W2-GC)=1*1/6=1/6 &
P(host reveals red AND R1-WC-G2)=1/2*1/6=1/12.
So P(GC IF host reveals red)=1/6/(1/6+1/12)=2/3.
Your mistake is a fairly simple one; you're failing to take into account the fact that the host treats the two goats differently to the car.
I agree that the host treats the goats differently to the car but you're just recalculating the result of Q10. I agree with the result of Q10, at the time that the result of Q10 is applicable. Q15 is applicable after the door is opened.
DeleteYou will almost certainly not agree with me on this. For my part, I will provide you with a visual argument in support of my position in the next day or so.
You seem to consider that everyone has Alzheimer's disease ...
DeleteAfter the Red door is opened and Mary is revealed, the candidate seems to completely forget that she chose Red and Green (and not Red and White) and that the host could not reveal a Car. So of course, if Marylin doesn't remember which door she selected, or the rules of the game then she will think that she has a 1/2 chance ...
But Marylin remembers that she selected Green and that the host could not reveal a Car. In all your arguments, you seem to think that it's not important. I made a very detailed answer on the other post to explain you that it IS important, and exactly how much it is important.
To finish, may be there is something subtle that you don't understand :
You have two independent variables for the Green door, which are the answer of these questions :
* Is it one of the two doors selected by Marylin ?
* Does it contain a car ?
Then you say that the host open a Red door revealing Mary. And in your computations, you act like the two first variables are still independent. They are not !!
You have here a third variable (which door is open), that is not mutually independent from the other two. So there is absolutely no reason to believe that the situations that were a priori equally likely (when you don't consider the third variable), are still equally likely when you take this third variable into account.
It seems that no argument will be able to convince you that you are wrong, if you continue to refuse to learn some basic (not always intuitive) results about probabilities. For example you can read (http://en.wikipedia.org/wiki/Pairwise_independence) to see how the introduction of a third variable can make two independent variables give non independent results.
Um, no. I'm not suggesting that the contestant forgets anything. I'm merely suggesting that an event that has past doesn't have a residual probability attached to it. A series of 8 heads in a row with a fair coin is highly unlikely. An eighth head after a series of 7 heads has a probability of 50%, despite how unlikely it might have (a priori) been to have 8 heads in a row. (This was my friend's contribution, by the way.)
DeleteWhat Marilyn (not Marylin) knows is that there are, with a Red Mary, only two possible distributions of Car and Ava behind the White Door and the Green Door. The fact that the host may have had to make a decision (a decision which has already been made) will not magically move the Car across from the White Door to the Green Door, not even 2/3 of the time.
Perhaps I can ask you a counter question in response to your two questions: What is the likelihood of the Car being behind any specific door (totally ignoring for the moment the fact that there is a selection of door(s) and the opening of a goat door)?
How would pairwise independence apply to the location of the car behind a specific door?
> I'm merely suggesting that an event that has past doesn't have a residual probability attached to it.
DeleteIf somewhere in your previous arguments you used that sentence to prove a point, then you HAVE to prove that the variables that you are talking about are independent. (Note that in your examples of coin flips, the successive results are independent)
Do you understand something when I talk about "independence of random variables", or is it all gibberish to you ?
Because, while your sentence is true for independent variables, it is ABSOLUTELY FALSE for variables that are not independent. So if independence of variables means nothing to you, we might as well end the discussion here because obviously you don't have enough basic knowledge of probabilities to understand what I'm saying.
> What is the likelihood of the Car being behind any specific door (totally ignoring for the moment the fact that there is a selection of door(s) and the opening of a goat door)?
1/3
> How would pairwise independence apply to the location of the car behind a specific door?
Probability 101 again ...
Translate everything in mathematical terms. Forget about doors, cars, contestant, goats, Marylin. I will only talk with letters so that no possible confusion will arise.
You have three random variables.
* Variable C : It can take three values : 1, 2 or 3
* Variable U : It can take three values : 1, 2 or 3
* Variable O : It can take three values : 1, 2 or 3
Individually the three variables have a uniform probability distribution. The first two variables are pairwise independent and the third one satisfies P(C=O) = P(U=O) = 0.
That's the only "rules of the game". If you think that this mathematical model does not accurately describe the Monty Hall situation, then explain why.
What you want to compute is P(C=2 | (O=1)&(U=3))
(Translating to : the contestant chose door 1 and 2 and the host opened door 1, the probability that the car is behind door 3)
So you apply conditional probability formula and you compute the two following terms :
* P(O=1) = 1/3 (uniform probability distribution)
* P((C=2)&(U=3)&(O=1)) = P((C=2)&(U=3)) - P((C=2)&(U=3)&(O=2)) - P((C=2)&(U=3)&(O=3)) (Law of total probability)
= P((C=2)&(U=3)) -0 - 0 (by hypotheses on my variable O)
= P(C=2)*P(U=3) (pairwise independence of the variables C and U)
= (1/3) * (1/3) (each variable has uniform probability distribution)
= 1/9
You apply the formula and you get :
P(C=2 | (O=1)&(U=3)) = (1/9) / (1/3) = 1/3
According to you, I'm doing something wrong here. Because according to you, I should get 1/2 at the end.
Apparently you don't trust me when I say that the only thing that is missing in your reasoning is the notion of "independence of random variables", but really it's the only thing that is keeping you from seeing your mistake. Because you made a mistake. The fact that the mistake leads to a (perhaps) true result is irrelevant. I couldn't care less about the result being 1/2 or 1/3. I care about the correctness of your argument. You use incorrect arguments and you don't realize it. Use correct argument and perhaps I will accept your result.
I'm going to make an admission.
DeleteI don't have a PhD in mathematics. For this reason, my method of argument isn't going to be as rigorous and well-developed as yours. I apologise for that, but I am not going to spend many years of my life getting the qualifications that you have in order to argue the point with you. I hope you understand that.
Thanks for laying out the process of dealing pairwise independent variables. It's a little arcane, but I think I follow it. However ... I think this applies more appropriately to analysis of the decision making process of the host across the whole spectrum of possible mini-games. This, I agree, gives you a result of 2/3 when the question is "what is the likelihood that the host will open the Red Door revealing Mary because the Car is behind the White Door and Ava is behind the Green Door?" (This is the covered at Q10.)
Can you please explain why you think that this isn't the case, remembering of course that I am not a Maths PhD.
> I don't have a PhD in mathematics. For this reason, my method of argument isn't going to be as rigorous and well-developed as yours. I apologise for that, but I am not going to spend many years of my life getting the qualifications that you have in order to argue the point with you. I hope you understand that.
DeleteWait, what ?
I never implied that you should get a phd to discuss the subject with me. I'm merely suggesting that if you want to argue about a probability problem, you should at least know what the words "random variables", "independence" and "conditional probabilities" mean.
Those are the most basic objects of probability theory. You don't need a phd, you just need a few hours to read any book called "probability theory".
You still did not answer to the very detailed answer that I made on the other page yesterday. http://neophilosophical.blogspot.com/2015/02/the-reverse-monty-hall-problem-and.html
I think the detailed answer is simple enough for you to understand. But I am under the impression that you don't want to understand ... Prove me wrong !
It's reassuring that I don't need to get a Maths PhD! I do understand that I run the risk of using terms that have very specific meanings within your field, in ways that don't precisely align with those meanings. I've not used the term "random variables" as far as I can recall. Nor have I talked about "independence" myself, others have certainly raised it. However, I accept that I did use the term "conditional probability" and a deep understanding of conditional probability would likely necessitate considering the terms "random variable" and "independent".
DeleteHere's what I meant by each term - if you can tell me if there is a better term or description, I will try to use that term (when I can remember it in the heat of the moment):
random - I've used it in terms of the cars and goat are distributed at random (sometimes "completely at random") by which I mean that it is equally likely that a specific goat or the car was placed behind a specific door. I would say that the distribution of heads and tails you get from a fair toss of a fair coin is random with heads and tails being equally likely with each toss.
independent - noting that I've not really gone into this ... if events are not causally linked, then they are independent. Two subsequent (or indeed two totally separate) fair tosses of a fair coin are independent. Getting a heads with one toss won't affect the likelihood of getting a heads with the other toss. In my scenario, the distribution of goats and car is independent of the selection of the door(s) by the contestant. The opening of the door is not independent, because the host is constrained by the requirement to not reveal a car. I do understand this.
conditional probability - this I might be using a lot less stringently than you are ... by a conditional probability, I merely mean that there is a probability of an event happening, say the opening a door out of a range of doors in various rooms, and there is a probability of that event happening when something else (something related!) has happened, say the locating of the door opener (Fred) in one of the various rooms. So, I mean what is the likelihood that Fred will open door 12 out of 20 doors, noting that there are four rooms with five doors and it's equally likely that Fred will go into each of the rooms and equally likely that each of the doors will be opened once Fred is in a particular room. That'd be 1/20. But once Fred is in the third room, it becomes 1/5.
There's probably a much better, more stringent explanation of conditional probability and I could have stolen one from from the internet, but I am trying to explain where I am coming from.
Note that I am fully aware that conditional probability (in my not stringent enough definition) does not apply when events are independent. That said, you can still use the equation Pr(A|B) when A and B are independent, but if you find that you get a value of Pr(A|B) that does not equal Pr(A), then you've got a problem. (The same applies to the reverse, Pr(B|A)=Pr(B) if A and B are independent.)
I did see this comment - http://neophilosophical.blogspot.com/2015/02/the-reverse-monty-hall-problem-and.html?showComment=1424814096527#c2144284274391674646 - which I think is the one you mean. I've been a bit busy, but my intent was to respond to you by pointing to the next article. I'll put that up as soon as I can (should be in the next half hour, unless I am interrupted). I do appreciate your effort to explain in detail, but again, I think that this applies before the door is opened. Hopefully that new article will help (although the a priori likelihood of that is low! <- weak attempt at humour, not a serious assessment of likelihood).
I agree with Anon that Q15 is the first one you, er, "Marilyn" got wrong. Also, can you clarify this seeming contradiction?
ReplyDeleteFirst you say, "if the Red door is opened, then the likelihood that it was opened because there is a car behind the Green door is 2/3".
Then, once the red door is in fact opened, you say the likelihood has become 1/2.
Since the entire scenario is hypothetical (i.e. I'm reading an account of it on the page), I don't see the difference between the "if the red door is opened" in the first sentence, and the "once the red door is opened" in the second. Yet you've assigned those different probabilities. Maybe one of those sentences doesn't mean what I (or you) think it means?
neopolitan, where you say in your original post "this is where it gets interesting", should read "this is where I show my ignorance of conditional probability"
ReplyDeleteUntil you can explain how these 2 statements:
1) In a single game the chance of winning (if you stay) is 1/2
2) In multiple games the chance of winning (if you stay) is 2/3
can both be true as you assert, then it's a waste of time providing either a logical or mathematical proof refuting your position.
At what point (number of games) does the probability change from 1/2 to 2/3, or is it a gradual process such as 1/2 then 7/12, then 15/24, then 31/48...........2/3?
Yeah, sorry neopolitan, Marley52 (together with the many of us who have told you the same thing on r/philosophy, r/math, and r/badmath) is right: your position that a single iteration can have a different probability than the same thing iterated multiple times is pure nonsense. You can put the words together into a grammatical sentence, but the mathematical meaning is nil.
DeleteThe fact is that you have gotten two different answers (2/3 and 1/2) to the same problem, and instead of saying, "huh, I must have missed something", you decided that you'd discovered something new. Then when people showed you exactly what you missed, you doubled down. When people pointed out that your interpretation implied serious contradictions, you called them "fascinating insights". When a PhD mathematician gently suggested that your lack of mathematical training is probably why you're having trouble seeing the error, you started talking about how you were going to write a paper! The reddit threads dedicated to these blog posts have over 200 combined comments, and not a single one of them agrees with your analysis, despite your heroic expository efforts. Your conclusion is that we're all (still) "missing your point".
Do you have a trusted friend who is (almost!) as smart as you are? Someone who could read this stuff and who you would trust to tell you, "hey bro, it's time to climb down from there"?
CCowboy,
Delete"Do you have a trusted friend who is (almost!) as smart as you are? Someone who could read this stuff and who you would trust to tell you, 'hey bro, it's time to climb down from there'?"
I don't know anyone who would use the term "bro", I simply don't mix in those circles. However, I do have a few trustworthy friends who are as bright as me or brighter, so I tried it out on someone. This is what happened:
When asked, he said that he had not heard of the Monty Hall Scenario. So, I hit him with the Reverse Monty Hall Problem. His immediate impression was that the result had to be 1/2, and he zeroed in on the fact, even without me mentioning it, that once the door is opened, the previous probabilities are gone. I talked him through the 2/3 answer and let him know that this is by far the most commonly accepted answer. I even tried arguing with him that the answer must be 2/3. I did this by selecting three small things behind my back, one being the target item, holding two in my left hand and one in the right, then asking him which hand was more likely to contain the target item. He correctly identified the left hand. Then I dropped out the non-target item in my left hand and asked him which hand was more likely to contain the target item. He said they were equally likely.
Later, he looked up the Monty Hall Problem at wikipedia and came back stating that he had read it and now understood it and that the answer must therefore be 2/3. At that point, I showed him variations of my work and also talked through a three ball version of Two Balls, One Urn, as well as running him through a role play variation with three dice. At one point during this discussion, he looked confused, said something to the effect of "This is doing my head in, it's right. But it can't be right." After going through it again, he was again convinced that it must be 1/2 and the last I saw of him was when he was explaining the conundrum to someone else.
Now, perhaps I am just very persuasive in the flesh, or my friend (who has been IQ tested with a result above that of the higher percentiles for medical occupations (per Hauser)) is dumber than the testing suggests. I suspect however that I am just not explaining myself clearly enough, although I do wonder what, precisely, I need to do to be more persuasive in writing.
This phenomenon is another reason why I haven't folded, there is no-one that I have talked to face to face, to whom I have shown the little dice based role play, that has not been persuaded. What I am doing, therefore, is trying to find the trick to turn this face to face persuasion into written persuasion. When I get back to my other computer, I will try again ... this time with photos.
So again:
Delete"can you explain how these 2 statements:
1) In a single game the chance of winning (if you stay) is 1/2
2) In multiple games the chance of winning (if you stay) is 2/3
can both be true as you assert?"
Anything else is a waste of time.
neopolitan, I wasn't as clear as I should have been. I didn't mean find a friend who has never heard of the MHP and try to convince him of your position. It's my understanding that you've done that already, and that your friends were convinced. That's not surprising.
DeleteI meant, find a friend you trust, let them read the comments here and on reddit, and then give you advice about whether you're wrong or the rest of the world is.
Hi CCowboy,
DeleteI sort of suspected that you meant that, but it was an interesting exercise anyway. And note that I didn't know that my friend didn't know about the MHP, I was a bit surprised that he hadn't. I've already discussed the comments on here and reddit with some people, once I've presented my argument and little dice based proof, they are bemused about all the objections and how virulent they are. I guess some of my friends might be a bit cynical about argumentum ad populum.
In any event, I started this little thing pretty convinced that I might well be wrong. Since then, I've noticed consistent problems in the responses to what I have tried to say - constantly reverting back to multiple iterations, reverting to potential decisions made before the door is opened, keeping possible outcomes which are no longer possible, discarding outcomes that are still possible, and the provision of mixed messages about whether the goats are interchangeable or not. It's almost like there's a general consensus that I absolutely must be wrong, but not about precisely where, which leads people to make various, in some cases, conflicting claims. The closest that we have come to a consensus on where I am wrong is with regard to Q15 and the answer thereto. My answer to that will be posted here soon (it doesn't mention Q15 directly though, but I am treating people with respect and not pointing out the obvious - if I've misjudged, yet again, then people can always pose questions in the comments).
Hi Marley52,
DeleteWhen you consider multiple iterations, you could be about to play any out of six mini-games before you start. You are selecting doors at random, the goats and car are placed at random and the host does what the host does (randomly selecting only when she can). Therefore, the possible mini-games you could be playing are ACM, AMC, CAM, CMA, MAC and MCA. The spread of the six mini-games gives you an overall 2/3 chance of benefitting from a policy of switching.
However, in a single game, once the door has been opened, you are only playing one out of six possible combinations of two mini-games (MAC or MCA), (AMC or CMA), (ACM or CAM), (ACM or AMC), (CAM or MAC) and (CMA or MCA) depending on whether the host revealed a Red Mary, a White Mary, a Green Mary, a Red Ava, a White Ava or a Green Ava respectively. This gives you a 1/2 chance of winning from a stay decision.
Now, while I accept that you might think that I am wrong on this, can you confirm that you understand what I am saying. If you are saying that I am wrong on the basis of not understanding what I am saying then, with all due respect, time is indeed being wasted.
I don't think you are wrong on this, I KNOW you are wrong on this, and I do understand what you're saying. You're saying that the average probability of winning by stating is 2/3 and the individual (game) probability of winning by staying is 1/2 - which is mathematically impossible (unless at least some of the individual game scenarios give a probability of winning by staying of more than 2/3).
DeleteYour answer to Q15 is incorrect because you don't understand how conditional probability works.
You appear to think that winning is somehow magical, because it's the outcome that we want.
DeleteI know how conditional probability works. The argument, as it applies to that, isn't really over my manipulation of the numbers, which I think has already been accepted as correct by someone - but about the applicability of conditional probability. If it were applicable (which, for the sake of the argument, it might be in some similar but not totally analogous situation), then my use would be correct.
If you knew how conditional probability worked you wouldn't keep getting the incorrect answer. And you still haven't explained how the average probability can be 2/3 if the individual probability is 1/2 - which as I said is mathematically impossible and thus makes your conclusion WRONG.
Delete> However, in a single game, once the door has been opened, you are only playing one out of six possible combinations of two mini-games (MAC or MCA), (AMC or CMA), (ACM or CAM), (ACM or AMC), (CAM or MAC) and (CMA or MCA) depending on whether the host revealed a Red Mary, a White Mary, a Green Mary, a Red Ava, a White Ava or a Green Ava respectively.
DeleteApparently, you make the assumption that the initial choice of the contestant is no longer important once the door is open.
As the choice of the contestant no longer appear in your possible scenarios, it must mean that you think it's no longer relevant. I'm not saying that you are wrong here, I'm just trying to understand your argument.
> This gives you a 1/2 chance of winning from a stay decision.
And then given this assumption, your conclusion is that the probability is 1/2. In other words, you found that given the assumption, then the initial choice of the contestant is not important.
You seem to think that this constitutes a proof that your assumption was correct. That's not a proof, that's circular reasoning.
Circular is the only sort of reasoning neopolitan appears capable of.
Deleteneopolitan said, "I've noticed consistent problems in the responses to what I have tried to say - constantly reverting back to multiple iterations, reverting to potential decisions made before the door is opened, keeping possible outcomes which are no longer possible, discarding outcomes that are still possible, and the provision of mixed messages about whether the goats are interchangeable or not. It's almost like there's a general consensus that I absolutely must be wrong, but not about precisely where, which leads people to make various, in some cases, conflicting claims."
ReplyDeleteI'll try to address this piece at a time:
"consistent problems": Yes, you've been hearing the same things over and over. But you're the only one who thinks the objections are problematic.
"reverting back to multiple iterations": You're the only one who thinks this could conceivably be problematic. Everyone else realizes that the probability of an outcome in a one-shot scenario is, necessarily, identical to the limiting proportion of that outcome over many iterations.
"reverting to potential decisions made before the door is opened": Yes, because we must examine each different path we could have taken to the observed situation, in order to compute the relative likelihood of each one. It's the "proportionality principle" in the paper you linked to me earlier.
"keeping possible outcomes which are no longer possible, discarding outcomes that are still possible": You seem to have no reservations about throwing out all the games where Mary is behind green, because 0% of those games become Red Mary games. And you agree that 100% games where the player chooses Red (with Mary) and Green (with the car) become Red Mary games. But you have never taken account of the fact that 50% of the games where the player chooses Red (Mary) and Green (Ava) become Red Mary games. The other 50% are Green Ava. Your analysis has consistently counted this scenario as contributing 100% to the Red Mary outcomes.
"mixed messages about whether the goats are interchangeable or not": Your analysis is wrong, whether the goats are distinct or "the same goat" (which can be imagined as the two losing doors opening into the same empty room). The fact that people have tried to engage you under each assumption is not evidence of any inconsistency or confusion on their part.
"general consensus that I absolutely must be wrong, but not about precisely where": This one is the furthest from reality of all, I think. There is a consensus that you ARE wrong, not that you MUST be (which sounds like we don't want to accept it, but we just can't figure out why it's wrong). You have been wrong in multiple ways, and different people have tried different approaches to help you see it. The fact that these approaches differ does not mean we're confused about why you're wrong.
"conflicting claims": I haven't seen any conflicting claims. Nobody is arguing with each other about whether you're wrong. We've just been trying different angles to get you to see it.