# The Monty Hall Problem – Part 3

I’m going to make two objections to standard justifications of the correct answer to the Monty Hall problem. The conclusion to an unsound argument can still be true, so if I’m successful at showing that there is a problem with the reasoning supporting the accepted answer to the problem, this will not show that the accepted answer is false, just that the justification of the answer is faulty.

At a high level, the two objections are that the standard justification (1) begs the question, and (2) commits the fallacy of equivocation. I’m more confident of the first objection, but the second objection might well turn out to be the more interesting point. It is unclear to me, at this point, whether either objection will be successful.

It might be the case that the objections I raise point to unstated assumptions, and that the argument can be fixed simply by making explicit an unstated assumption. It has been pointed out by others that there are often various unstated assumptions in the presentation of the Monty Hall problem that are required to make the reasoning for the accepted conclusion deductively valid.

For example, one must assume that the placement of the car prior to the selection of a door by the contestant was done randomly, and that each door had an equal chance of having the car placed behind it. If, contrary to this assumption, the car was always placed behind door #1, then it would obviously be best for the contestant to always select door #1 and always stick to that initial selection.

Similarly, if the placement of the car was done in such a way that there was an 80% chance that it would be placed behind door #1, then always selecting door #1 and always sticking to door #1 would be the best policy for a contestant. Thus, in order to prove that switching is the best strategy, the assumption must be made that the placement of the car is done at random and that each of the three doors has an equal chance of having the car placed behind it.

One must also assume that there is no switching of the location of the car and goats after the initial placement of them prior to the game. One must assume that Monty Hall knows which door the car is behind. One must assume that the contestant does not have X-ray vision (like Superman) or infrared vision (to detect the body heat of the goats) or super-sensitive hearing (so that, as Jim Lippard pointed out, one could hear a goat behind one of the doors).

One must assume that Monty Hall will not lie to, or blatantly deceive, the contestant, for example by opening a door with a life-size picture of a goat that blocks the contestant’s view of the car behind the picture (although some misleading of the contestant is allowed). One must assume that God does not intervene and transform the car into a goat or vice versa. These very specific assumptions need not all be made explicit, because more general assumptions can cover a multitude of sins or, rather, preclude many odd ways of messing up the problem, so that the accepted answer will follow from the stated assumptions.

Let me start my first objection with a critique of the probability tree diagrams. The diagrams abbreviate a sequence of events. A more detailed sequence would look like this:

1. A car is placed behind one of the three doors, and a goat is placed behind each of the two other doors.
2. The contestant makes an initial selection of a door (in this case, door #1).
3. Monty Hall opens one of the other two doors, revealing a goat behind the door (in this case, door #3)
4. Monty Hall offers the contestant the option to switch to the other remaining door (in this case, to door #2).
5. The contestant makes a final selection of a door (in this case, choosing between door #1 and door #2).
6. The door chosen by the contestant in the final selection is opened, revealing whether the car is behind that door.
7. If the car is revealed to be behind the door chosen by the contestant in the final selection, the car is then given to the contestant.

There are thus, at least seven different events that occur in temporal sequence, and thus the entire event occupies at least seven different moments or points in time. Since each of the seven events requires a measurable period of time to occur, there are at least seven periods of time here.

There is no indication of the passage of time in the probability tree diagrams. However, with the passage of time, come the possibility of new information. Assuming that the contestant is a normal human being and is conscious during each of the seven events, the contestant is constantly having experiences during the seven events, and thus is constantly receiving new information throughout the duration of the seven events. As the information available to the contestant is constantly growing, the probabilities of various events are also changing, from the point of view of the contestant.

Most people recognize that at the time the contestant makes the initial selection of a door (in this case, selecting door #1) the probability that the car is behind that door is 1/3, and many (most?) people believe that the information received by the contestant after the initial selection changes the probability that the car is behind the door that was initially selected. This is, on the face of it, in keeping with the general principle that new information can affect the probability of an event (as with my example of the prediction that it will rain tomorrow).

This suggests to me that the probability tree diagrams are ambiguous, in that it is unclear at what point the probability of 1/3 is being assigned to the statement that “The car is behind door #1″. Was this probability assigned prior to the initial selection of door #1? immediately after the initial selection of door #1? or after Monty Hall has opened up door #3 to reveal a goat behind that door?

If the information the contestant gets from Monty Hall is irrelevant to the probability of the statement “The car is behind door #1″, then I suppose it does not matter which of the above three points in time is intended, since the probability would be the same whichever point in time is intended. But it seems to me to beg the question to simply assume that the information received by the contestant when Monty Hall opens door #3 is irrelevant to the probability of the statement “The car is behind door #1″. This is the point of disagreement between the many who are inclined to say that the probability of winning by sticking with door #1 changes from 1/3 to 1/2, and the few who insist that the probability of winning by sticking with door #1 starts out as 1/3 and remains 1/3 even after Monty Hall has revealed a goat behind door #3.

In other words, since the disagreement appears to be over whether the information received by the contestant when Monty Hall opens door #3 is relevant to, or has an impact on, the probability of the statement “The car is behind door #1″, it is incumbent upon a defender of the accepted answer to the Monty Hall problem to show that this information is irrelevant or has no impact on the probability of the statement “The car is behind door #1″.

Since the probability tree diagrams make no reference to the passing of time, and fail to distinguish between probability assessments made at different points in time during these events, I don’t see how the diagram can possibly address the main question at issue. I suppose a verbal explanation of the diagram could provide the missing temporal aspect of this problem, but such an explanation, I believe, would show the diagram to be ambiguous.

Now to address the verbal reasoning in support of the accepted answer. I thought I was going to object to the validity of the logic of the argument, but on a closer look, my objection seems to focus on a specific premise:

6. If the contestant sticks with the door that was initially selected, then the probability of the car being behind the finally selected door is equal to the probability of having initially selected the door with the car.

This premise appears to bridge the gap in time between the initial selection of a door by the contestant, and the final selection of the door by the contestant. What assumption warrants the bridging of this period of time? the idea that the probability remains stable through the period of time in question?

Is the assumption that the contestant receives no information during that period of time? That would be a bizarre assumption, and it would contradict any straightforward reading of the problem (since the contestant at the very least needs to hear Monty Hall offer the opportunity to switch to the remaining other door). Is the assumption that the new information received by the contestant–during the period of time between the initial selection and the final selection–is irrelevant to the probability of the statement “The car is behind door #1″? In that case, premise (6) begs the main question at issue.

It might well be the case that (6) is true, and I think I know how to defend the truth of (6), but as the argument stands, without further elaboration and justification, it appears to me to commit the fallacy of begging the question.

I forgot to mention a twist on the Monty Hall problem that led me to this objection:

Suppose exactly the same scenario, with one additional event: After Monty Hall reveals a goat behind door #3, Monty tells the contestant that "The car is behind door #1", and then Monty Hall gives the contestant the option to switch to door #2.

Suppose the contestant knows that Monty has never lied to a contestant before, and that Monty is a generally hontest person, and the contestant has no reason to suspect deception by Monty in this particular instance.

Clearly, the contestant should NOT switch to door #2, but should stick with door #1.

This is a case where the information received by the contestant is clearly relevant to the probability of the statement "The car is behind door #1".

• http://www.blogger.com/profile/11009570389150569819 Frédéric

Hi,

The information given by Monty Hall is about doors #2 and #3. If we take the situation you describe (the contestant selects door #1), the probability of the car being behind door #1 is 1/3, and the probability of being behind either door #2 or door #3 is 2/3.

When Monty Hall opens #2, the probability of the car being behind door #2 is now shown to be 0. But the probability of the car being behind either door #2 or door #3 does not change, so the new information has in effect changed the probability of door #3 to 2/3.

Door #1's probability is not impacted by Monty Hall new information because that information was not about that door in the first place.

So while that line of argument is not my favorite, it does not seem incorrect to me.

• http://www.blogger.com/profile/12197516145154863792 Enkidu

I think that we intuitively think that there are two independent events. 1) three doors = probability 1/3, and 2) two doors = probability 1/2.

However, the events are not truely independent because Monty Hall knows which door hides the car and is unkikely to spoil the game by revealing the car. (of course if the car is behind door one, it doesn't matter which he opens, but if it is not behind door one then it is certainly behind door two)

Going back to the librarian version, the liklihood of initially pickg the corrrect book is very small, but, assuming the librarian is not cheating, when she removes all but one other book the liklihood of that book being correct is overwhelming.

I'm putting this in general terms because I haven't gone back to check the actual numbers.

• http://www.blogger.com/profile/12244370945682162312 NAL

6. If the contestant sticks with the door that was initially selected, then the probability of the car being behind the finally selected door is equal to the probability of having initially selected the door with the car.

I don't like that explanation. When a contestant is asked to stick with the initial selection or change, the contestant is, in effect, making a new selection. This new selection is between two doors, not three. Therefore, 50-50 chance.

Frédéric said…

Door #1's probability is not impacted by Monty Hall new information because that information was not about that door in the first place.
============
Response:

What do you mean by saying the new information is "not about that door" other than that the new information is irrelevant to (has no impact on) the probability of the statement that "The car is behind door #1"?

If you are asserting that the new information is irrelevant to the probability of the statement that "The car is behind door #1", then I think you are right, but I also think that your claim begs the question against those who are inclined to think that the new information IS relevant to the probability of the statement that "The car is behind door #1".

You appear to simply be asserting that they are wrong and you are right, rather than giving a good reason why someone should side with you rather than those who hold the opposing point of view.

Also, if the new information is that door #3 does not have the car behind it, then would you say that this new information is about door #3 and NOT about door #2?

But obviously, this information about door #3 is relevant to the probability of the statement that "The car is behind door #2". So, since information that is directly about door #3 is relevant to claims about what is behind door #2, it is not at all obvious that information about what is behind door #3 must be irrelevant to the question about what is behind door #1.

• http://www.blogger.com/profile/11009570389150569819 Frédéric

NAL: sorry, you're wrong. Run a simulation if you don't believe the logic.

Bradley: yes, I suppose I assert that those who think the probability of door #1 changes are wrong.

For a more detailled reason: note that once the participant has selected a door (here #1), Monty Hall is constrained to the other two (#2 and #3).

Let's name the various events #1, #2, #3 (that stands for the car being behind door number…), and G2 and G3 (for the event that MH reveals a goat behind door 2 or 3. As said above, G1 is now impossible).

We have the a priori probabilities P(#1) = P(#2) = P(#3) = 1/3, and P(G2) = P(G3) = 1/2 (as MH could open either 2 or 3, and without more info it's a 50/50 chance a priori).

Using the formula for posterior probabilities (http://en.wikipedia.org/wiki/Posterior_probability), assuming we witness G2, we can compute P(#1|G2) and P(#3|G2) (P(#2|G2) is trivially 0).

P(#1|G2) = P(G2|#1)P(#1)/P(G2) = (1/2) * (1/3) / (1/2) = 1/3.

P(G2|#1) is 1/2, as if the car was behind #1, either G2 or G3 can be observed with 50/50 chance.

Now P(#3|G2) = P(G2|#3)P(#3)/P(G2) = 1*(1/3)/(1/2) = 2/3.

P(G2|#3) = 1 because if the car is behind door 3, MH cannot open any door but 2.

The fact that P(#1|G2) = P(#1) shows that the events are independent. In other words, observing G2 does not change the probability of #1 (that's the definition of independent).

Ok, that was tedious. Obviously, I much prefer the basic approach of considering that the switch strategy always convert failure into success (and success into failure), then adding that initial chances of failure are 2/3, meaning that using the switch strategy ensures a final chance of success of 2/3.

Frederic

Frédéric said…

Let's name the various events #1, #2, #3 (that stands for the car being behind door number…), and G2 and G3 (for the event that MH reveals a goat behind door 2 or 3. As said above, G1 is now impossible).

We have the a priori probabilities P(#1) = P(#2) = P(#3) = 1/3, and P(G2) = P(G3) = 1/2 (as MH could open either 2 or 3, and without more info it's a 50/50 chance a priori).
===========
Response:

Thank you for your clear analysis in terms of Bayes's theorem.

It might well be the case that the analysis that you present avoids begging the question and avoids the fallacy of equivocation. However, you are basically throwing out a Red Herring here.

Even if your justification of the accepted answer to the Monty Hall problem avoids the fallacies of begging the question and equivocation, that does not show that the justifications I'm talking about avoid those fallacies.

One might argue that a non-question-begging justification in terms of Bayes's theorem would prove that there must also be a non-question-begging justification in terms of some other ways of calculating probabilities, such as the use of probability tree diagrams and associated calculations. I suspect that there is some mathematical equivalence or translation possible from Bayes's theorem to other ways of calculating probabilities.

Nevertheless, proving that it is POSSIBLE to come up with a non-question-begging probability tree diagram justification of the accepted answer is not the same as proving that some specific attempt at providing such a justification is in fact successful. And that is the question at issue here.

I do appreciate your carefully laying out a Bayes's theorem justification, and I will look it over to see if it avoids the problems that I see in other sorts of justifications of the accepted answer.

• http://www.blogger.com/profile/11009570389150569819 Frédéric

Ok, I think I see your point. I'm a math major, so I go for the math as soon as possible and usually does not bother with the logic (in the philosophic sense). A professional bias of some sort.

Let me try another, less mathy approach.

First I need to clarify something: in your article, you used #1 as the one selected by the participant, #3 as the one opened by MH, and #2 as the one the participant could switch to. I'll just change this a bit (as I have implicitly so far), but go a bit further and state that these tags are assigned after the various actions are executed:

Door #1 is the one selected initially by the participant
Door #2 is the one opened by MH to display a goat behind it
Door #3 is the final door

Note that in doing so, the doors do not have any label until assigned by the actions (in a sense, the time aspect is reflected in these labels).

Those who have a problem with that should think of it as just focusing on the doors independently of any preassigned label. We don't care which door is on the left, center or right, we ignore the colours, numbers, or anything else.

Let me further add a few assumptions: the car is randomly placed before the game starts; and MH will not communicate anything to the participant except by opening door #2. And the participant knows the rules.

Now back to the argument: the participant will select door #1. Then obviously, door #2 will be opened. That's the rule (and door #2 is the one opened by MH). So the fact that door #2 is opened is not adding any new information about door #1. This should address the "beg the question" part.

What about door #3, then? Well, we now know something about this door, something we didn't know before. It is not door #2. Maybe it could have been, but now it is not.

At this point, to compute the new probability of door #3, we can go with the basic forward reasoning I used in my first answer, or the posteriori one I used in the second, but the end result is that P(#1) is still 1/3, and P(#3) is now 2/3.

What do you think?

Frederic

• http://www.blogger.com/profile/03513589609857845105 J

I want to address the what I believe to be a logical flaw in your argument and for the most part leave the actual statistical proof alone (I think Fredric did a great job on the conditional probability test for independence).

Where I think you err is in the statement, "The car is behind door #1", it is incumbent upon a defender of the accepted answer to the Monty Hall problem to show that this information is irrelevant or has no impact on the probability of the statement "The car is behind door #1". In stating this you have shifted the burden of proof from the person making the positive claim, “the probability of event #1 is changed by X.” to the person making the negative claim, “the probability of X has not changed.” Carrying that methodology forward, it would be incumbent on the negative claimant to prove why an infinite number of unrelated possibilities don't affect the probability of event #1 (in effect to prove a negative). Now it does become incumbent on the negative claimant to show why the probability of the alternate choice changed to 2/3, but that is rather axiomatic: Since the it can't be shown that the probability of #1 changes, and the probability of the sample space must be equal to 1, the probability of the alternate must be 2/3 (although this may look tautological it's actually not, the change in the alternate probability is driven by the change in sample space, i.e. combining the probabilities of #2 and #3 into one box.)

J said…

Since the it can't be shown that the probability of #1 changes, and the probability of the sample space must be equal to 1, the probability of the alternate must be 2/3…
===========
Response:
This looks like an appeal to ignorance. The failure of someone to prove that the probability of "The car is behind door #1" has changed is NOT a good reason for believing that this probability remains the same.

I agree that the sum of the probabilities of the alternative possibilities needs to add up to 1, but that does not establish the truth of the controversial claim that the probability of the car being behind door #1 remains the same even after Monty reveals a goat behind door #3.

J said…
Where I think you err is in the statement, "The car is behind door #1", it is incumbent upon a defender of the accepted answer to the Monty Hall problem to show that this information is irrelevant or has no impact on the probability of the statement "The car is behind door #1". In stating this you have shifted the burden of proof from the person making the positive claim, “the probability of event #1 is changed by X.” to the person making the negative claim, “the probability of X has not changed.”
========
Response:
The burden of proof should rest on whoever is making a claim, whether the claim is positive or negative in nature.

Defenders of the accepted answer generally ASSUME that the probability of the car being behind door #1 remains unchanged, but this is a controversial assumption, so until some good reason is offered to support the assumption, it seems to me that the use of this assumption to justify the accepted view commits the fallacy of begging the question.

51 PM, June 10, 2011

Frédéric said…

Ok, I think I see your point. I'm a math major, so I go for the math as soon as possible and usually does not bother with the logic (in the philosophic sense). A professional bias of some sort.

Let me try another, less mathy approach.
============
Response:
Thank you for your interest and contributions on this topic.

I'm still working on understanding your first more "mathy" approach, so it may be a while before I get around to responding to your most recent comments.

• http://www.blogger.com/profile/03513589609857845105 J

"The burden of proof should rest on whoever is making a claim, whether the claim is positive or negative in nature."

I agree, that is why is I said, "Now it does become incumbent on the negative claimant to show why the probability of the alternate choice changed to 2/3"

And that has been done. When you change the problem from one of pure random probabilities, to one where one the possibilities becomes fixed, you have changed the sample space. It does become a case of combining the probabilities box #2 and box box#3. The proof is much too long for a blog comment, I would suggest Jason Rosenhouse's, "The Monty Hall Problem: The Remarkable Story Behind Math's Most Contentious Brainteaser"

The argument doesn't rely on the claim that p(1) remains 1/3; it relies on that p(2) and p(3) have been combined and that p(2)+p(3)=2/3. And once that claim is made and demonstrated, the claimant has absolutely no burden of proof to show why p(1) remains 1/3.

Now where all of probability theory (along with all of mathematics) has a proof problem? Why must p(s)=1? From everything I have seen, it's circular logic, but hey it provides testable theories.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

I am unfamiliar with probability trees. I've looked at your earlier post and I think I understand what you mean about the tree's implying that initial probs can't change. YOu start out with 1/3 prob that prize is behind door #1 (left branch) and 2/3 prob that it is behind any other door (right branch). Now stuff happens, Mony messes around entirely within the right branch by opening all the doors but one, but nothing changes the starting 1/3-2/3 split. I don't love it, but I guess it is OK.
However, in your first response to this post you suggested "What if after opening the empty door Monty then flat out says that the prize is behind door #2 or door #3?" This.. this is a killer argument. Monty's actions have definitely changed the prob that prize is behind door #1. So the tree is wrong to imply that Monty's actions can't do this. You might say, well, Monty used special illegal information. But Monty only used the exact same information–knowledge of where the prize actually is– that he used to flip the empty door, and supposedly that was a perfectly legal move. It is key that Monty knows where the prize is since if he opens a random door instead of an empty door then the problem changes significantly.
I completely accept your objection. This proof is awful. Especially so because, as you pointed out, the entire point of this problem is Monty's offering new information and that is exactly what the tree is ccompletely ignoring. Where did you get this proof from? Is this really what students are being taught these days? Horrible.
I can't do Bayesian stuff and I don't know probability trees. I can only do naive probability where you start out with a universe of equally likely evetns, label the events with outcomes, and count the proportion of good outcomes. However, that's eactly what your trees seem to be doing: they seem to be partiitioning up the universe, putting labels on it.
In naive probability new information destroys the universe. You can't creatively label the universe to account for new information. The labelling assumes a static universe; after new information is introduce the universe must be recreated. I will show you what I mean. I will try to make a good probability tree instead of your awful one:
For concreteness, assume you always choose door #1.
The top of the tree has 3 branches, all equally likely (so all with prob 1/3), corresponding to the prize's being behind door #1, door #2, or door #3.
Branch prize1: 1/2 of the time Monty flips open door #2 and 1/2 of the time he flips door #3. SO the endpoints are prize1-flip2 with prob 1/6 and prize1-flip3 with prob 1/6
Branch prize2: 100% of the time Monty flips door #3. So the single endpoint is prize2-flip3 with prob 1/3
Branch prize3: 100% of the time Monty flips door #2. So the single endpoint is prize3-flip2 with prob 1/3
The tree is done, the universe has been partitioned into 4 endpoints with probs 1/6,1/6,1/3, 1/3.
Now comes the event! New information comes and it will mutilate the tree! MONTY DEFINITELY FLIPS OPEN DOOR #3. Aghh! Instantly all endpoints of my tree that do not contain "flip3" are obliterated. Big red X's cross them out. What is left?
Two endpoints are left: prize1-flip3 with size 1/6 and prize2-flip3 with size 1/3.
YOu see that the second endpoint has twice the size of the first, in the universe of equally lilely events it takes up twice the area of the first endpoint. So when you rescale the universe so that relative sizes are preserved but the probs add up to one and the rubble of impossible outcomes has been cleared out, you get the prob of the prize being behind door #1 is 1/3 and the prob of the prize's being behind door #3 is twice that, ie 2/3.
I know this is in fact Bayesian–but it is very simple posterier probability so that is why I could handle it.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

I want to take a different tack. The awful probability tree proof implies that whatever Monty is doing in the right hand branch of the tree (prob that prize is behind any door but door #1) has no effect on the left branch (prob that prize is behind door #1), so the left branch continues to have prob 1/3. That's the 6) boldface sentence in your post. It is in fact true. The reason the prob of the prize's being behind #1 stays at 1/3 is that Monty's actions can't affect that prob. The problem is not that the fact is false but that the tree offers no proof.
This is interesting, that Monty's actions did not affect the door #1 prob. The proof of calculating posterior probabilities obscures this. When you use that proof then the introduction of new information causes all probs to be recalculated and you find that the new prob for prize-behind-door1 happens to be the same as the old prob. Saying that you ran through the numbers and ended up at the same place is not exactly the same as saying that you needn't have bothered to do the calculations in the first place since it was impossible that you would move.
The reason Monty's action did not affect the prob of prize-door1 is that his actions were irrelevant. What does that mean?
For concreteness let's suppose that you choose door #1 and Mony flips open door #3.
Information is something you didn't know ahead of time, something that as far as you were concerned could have come out differently. If you already know what an action is going to be then it has no information. A way to think of it is that before you got the information you had a list of possible things it could be–and didn't know which one would be chosen– and that getting information is choosing from the list. If you tell me "it's raining" then that is information for me. I didn't know you would say that; you might have said "it is sunny" instead. The information changes my universe–the probability that I will go on a picnic today goes way down. If you go on to say "some people are using umbrellas", tha tis not infmormative. Once I knew it was raining I assumed umbrellas were present. Your information was completely predictable, it did not surpise me, it did not change my universe, it had no informative content.
For the Monty Hall problem, the information is that Monty flips door #3. Before he did it I did not know whether he would flip door #2 or #3. Once I know that it is door #3, my universe has more information and all probabilities of all things must now be revised accordingly. Actually, I guess Monty could flip open both doors, strip off all his clothes, and run from the room while laughing maniacally. That would be really informative because I totally didn't predict it! It wasn't even on my list of possible outcomes! It would be so informative it would makes the newspapers:"Hall Flips Out In Middle Of Game Show". It's fine that it wasn't on the list; the list is just a concept to convey the idea that information has to be unpredictable.
I'm not sure what "relevance" is but I think it has to be something like: information is relevant to an object if the state of the object changes what the information is going to be. If you think of information as a choice from a list, I mean that that the state of the object changes the probabilities of the various choices. If the state of the object has no affect on what information you get, then the information is irrelevant.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

I agree that the probability tree is no good since it can't incorporate change over time. Now let's talk about the "relevance" proof. By that I mean the proof that goes like this:
For concreteness, assume you choose door #1 and Monty flips open door #3.
At the beginning, all doors are equally likely to contain the prize so the probs are 1/3, 1/3,1/3.
Now we get the information that Monty has definitely flipped #3. This information is IRRELEVANT to the prob that the prize is behind door #1 but RELEVANT to the probs for door #2 and door #3. So the prob for #1 in unchanged, but the probs for #2 and #3 might have changed.
So after the flip the probabilities change and the new probs are:
Prob for #1 is still 1/3 since the info was irrelevant.
Prob for #3 is oviously 0, since the door has been opened and, look, we can see the prize is not there.
Prob for #2 is… I don't know. Seems difficult to calculate. The info was relevant so it might have changed, but did it? And if so, what did it change to? However, I can deduce it. The probs must add up to 1, and I know the other probs are 1/3 and 0, so this prob must be 2/3.
So the new probs are 1/3,2/3,0

I don't like this proof because it doesn't constructively calculate the new prob for door #2, but it does interestingly make clear that the reason the prob for #1 did not change is that it couldn't. If you go for the proof where you reestimate all probs after the new info comes in, then it looks like it is a coinicidence that you do all these calculations and the new prob for #1 comes out the same as the old prob. It is not clear that you could have just skipped the calculations since the prob couldn't change. I think this is a perfectly good proof.

The problem is that it is too terse. Some people think it is obvious that Monty's info is irrelevant to door #1, so they don't need this point expanded. However, apparently lots of other people don't find it obvious. It's reasonable for them to ask for more details. I have no idea why people are refusing to give you these details. The irrelevancy is not an assumption:it's something to prove. This isn't a secret. Some of your responders talked about "independence" so they clearly know this.
I could go on at length, but the short answer is that I think a definition that well formalizes the intuitive idea of relevancy is: "information is relevant to an object if, before we get the info, the state of the object affects the probability of how the info turns out".
In our case the info was the Monty definitely flipped door #3. before we got that info, we did not know whether he would flip #2 or #3.
The probability of flipping door #3 is 50% both when door #1 has the prize and when door #1 does not have the prize. So knowing that door #3 has been flipped can't tell us anything about whether door #1 has the prize. It's irrelevant.
On the other hand, the probability of flipping door #3 is 100% when door #2 has the prize and 75% when door #2 does not have the prize. So the info is relevant. Knowing that door #3 was flipped increases the prob that door #2 has the prize.
In more mathy terms, my statement about relevancy is:
IF P(B|A)=P(B|~A) THEN P(A|B)=P(A) (if the probability of B is unaffected by the state of A, then the probability of A's state is unaffected by knowledge of how B turned out). This statement is easy to prove if you accept the formula for posterior probabilities P(A|B)=P(A&B;)/P(B)

• http://www.blogger.com/profile/03513589609857845105 J

Happy Spider,

I think I can do this now without a blackboard (we'll see, it's tough to do with a blackboard).

Initial state: Picture a circle (the sample space) with three boxes inside of it.

There is an X placed inside of one of the boxes at random and favorable outcome is selecting the box with the X in it. Since the selection of the box with the X was random, the probability of favorable outcome for any one of the boxes is now 1/3.

The contestant is allowed to choose one of the boxes and Monty gets to keep two of them. The sample space just changed. It is now subdivided. Picture drawing a circle around one of the boxes (the contestant's) and then another circle around the other two (Monty's). The probabilities are now: The contestant 1/3 and Monty 2/3.

Now information is introduced. “One of the boxes in Monty's subdivided space does not have the X.” But this is NOT NEW information. If the X is only in one box and Monty's space has two, then obviously the X is not in one of the Monty boxes. The probabilities remain; contestant space 1/3, Monty's space 2/3.

Now one of Monty's boxes is opened. There are two possible outcomes: 1. The opened opened box contained the X, this IS new information, the probability of the opened box becomes 1 and the other two becomes 0; or the box is empty, this IS NOT new information, because we already knew that one of the two boxes in Monty's space didn't contain the X. In the empty open box case, someone must make a conclusive argument that Monty's sample space probability changed based on NO NEW information. Otherwise, the probabilities for the sample spaces remain 1/3 and 2/3.

And now Monty asks you do you want to trade boxes, but what he is really asking is, “Do want to trade sample spaces,” since he has eliminated one of the boxes in his sample space without changing the probability of the space.

The key to the logic is recognizing when there is NO NEW information.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

Here is another thought against your probability tree proof. Some people seem to think that at the beginning there is a 1/3 chance of being correct and a 2/3 chance that the prize is under any other door. When Monty then flips open all doors but one on the right hand side of the tree, that can't change the fact that the set {all other doors} had prob 2/3, so the single remaining door on the right side of the tree has prob 2/3.
This is wrong. Let's play around a bit to demonstrate.
Let's make one slight change in the problem: Let's change the problem so the initial probs (of a door's having the prize) are not 1/3,1/3,1/3. Let's instead make them 48%,48%,4%.
Assume you choose door #1.
Now if you construct the tree as before, you get that the left branch has prob 48% and the right branch has prob 52%. Now have Monty flip open an empty door on the right. Then by the logic above, you would say that the remaining door has 52% chance of being correct so you should switch to it.
This is not true. First of all, the final probs will now depend on which door exactly was flipped, so break it down into two cases:
Suppose Monty flips door #2. Supposedly we should think the door #3 has a 52% chance of having the prize. But that is obviously false. We just have two doors left–#1 and #3– and the initial probs were that #1's chance of having the prize dwarfed #3's chance. It was 48% vs 4%. So it is much more likely that #1 has the prize than #3. You should not switch. If you don't buy that, what if I made it even more extreme? What if the initial odds were 49.99999%,49.99999%,.00002%. Do you still think that if I flip door #2 then door #3 suddenly has a 50.00001% chance of having the prize? How can that be when door #2 is flipped all the time and door #3 never ever has the prize in the whole history of the game?
Suppose Monty flips door #3. Well, since door #1 and #2 started out with about the same initial prob then I guess they stayed the same, so maybe the 48%-52% thing works for Monty flipping door #3.
Let's crunch the numbers. That's using posterior probability, as one of the other repliers did and as I did with a previous reply about probability trees.
Not enough space to go through it here, but the result is:
If Monty flips door #2, then 6/7 prob that #1 has prize and 1/7 prob that door #3 has prize. So, just as intuition promised, the initial prob for door #3's being so low means that sticking with #1 is clear winner.
If Monty flips door #3, then 1/3 prob that #1 has prize and 2/3 prob that #2 has prize. Huh. That so surprised me that I wrote a quick computer program to run a million trials and count how many ended with door #3 flipped and how many with door #3 flipped AND prize behind #1. 33% it is. Not enough space to explore this.
What went wrong? I've said that I agree that in the original problem Monty's flip is irrelevant to the prob that door #1 has the prize so it is true that the final prob for door #1 stays at 1/3.
The problem is that in the modified problem Monty's flip is no longer irrelevant. Calculate it yourself. I like to use "information is irrelevant to an object if the state of the object does not affect the chance that the information will turn out the way it does". In the original problem, Monty has a 50% chance of flipping door #2 both when the prize is behind door #1 and when it is not behind door #1. So knowing which door he chose is irrelevant. In the modified problem, Monty has a 50% chance of flipping door #2 when the prize is behind door #1 but a, umm, .48/.52*0+.04/.52*1=.04/.52=1/13 prob when the prize is not behind #1. Different probs, so the results of the flip are NOT irrelevant to the prob of #1.
If you work with the numbers for a bit you see that what you need for the flip to be irrelevant is that the initial probs for #2 and #3 must be equal. They don't have to be 1/3–initial probs of, say, 1/5,2/5,2/5 are just fine–but they must be equal.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

J,

Could you go through your proof again but this time let the probability of having an X in a box be 49.9999%,49.9999% and .0002% instead of 1/3,1/3,1/3.
At what point does the proof for the new case diverge from the proof for the original case? It must do so somewhere. If I choose box #1, then my prob starts at 49.9999%, but after Monty opens an empty door in the other sample space then the prob changes. This is intuitively. Imagine that you really did play this game hundreds or thousands of times. Then Monty would be opening door #2 all the time, but the prize will never be behind door #3. So even though the sample space (door#2,door#3} had an initial prob of 50.0002% and I knew that at least one door was empty, if Monty opens door #2 then the prob for the space changes and goes way way down.
Or do you think the proof goes through exactly the same for my modified problem and that I am wrong about the probs' changing?

• http://www.blogger.com/profile/03513589609857845105 J

Spider,

I looked at your question 2 ways (I used 0.49, 0.49, and 0.02 because I like numbers I can work in my head) and either way it doesn't change the the proof.

The first way is pretty quick: The .49's and the .2 were the initial probabilities that it would land in a given box. If so nothing changes, because once it is in a given box the probability that it is in any given box when the contestant chooses is still 1/3.

The other way of looking at it starts out looking more interesting . Let's say there is 49 silver coins in two of the boxes and 2 silver coins in one of the boxes, and the contestant is asked to pick one at random.

Now the contestant's expected payoff is 33 “1/3” coins and Monty's is 66 ”2/3” (1/3*49+1/3*49+1/3*2) Now Monty opens one of the boxes (here is where the anticipation of interesting results leads to disappointment). If Monty opens the box with 2 coins in it we now have new information (we couldn't have known before that Monty held a box with 2 coins in it) and the new outcome becomes fixed at the contestant has 49 coins in his box. However, if Monty opens a box with 49 coins in it, we have NO NEW information (If two of the three boxes had 49 coins in them, and Monty held two of the boxes, one of the two of the boxes Monty controlled had to have 49 coins in it). In the latter case, the odds for the sample spaces remained fixed and the contestant should still switch by a margin of 2/3 to 1/3.

The proof (sadly from the point of finding something new and interesting) remains valid.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

This comment has been removed by the author.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

J,
I'm sorry, but the problem you analyzed isn't the one I am proposing. I was unclear. When I said to change the initial probs from 1/3,1/3,1/3 to .48,.48,.04 I meant that the contestant knows which box has which probability. If I initially choose box #1, I do so knowing it's initial prob is .48. If Monty later opens box #3, I know he opened a box with initial prob .04.
The problem you analyzed is that the initial probs have changed to .48,.48,.04 but you don't know which box has which prob. I think this is so because you said that the contestant has a 1/3 prob of initially choosing the correct box.
I agree with you that if the boxes are unlabelled then, since
you are randomly choosing one box in a situation where there are 3
boxes and exactly 1 has the prize, you have a 1/3 chance of choosing correctly. Sort of odd, isn't it? The probs are .48,.48,.04 but somehow I end up with prob 1/3 which isn't even one of the three available options. I could go further: suppose Monty ALWAYS puts the prize in box #1 so the probs are 1,0,0. If the boxes aren't labelled then we are still at the original problem of 1/3,1/3,1/3. Same with the box flip: Suppose Monty is biased to always flipping flip #2 if it is unchosen and empty. Then if box #2 and box #3 are unchosen and Monty flips #3, you know for sure that box #1 has the prize. However if it is unlabelled which box is which, then your knowledge of Monty's bias does you no good and you are in the original case of 50% chance of flipping either box.
This shows how relative probabilities are. They sound so
absolute when stated but they always implicitly rely on how much
information the actor has. If often happens that if the actor doesn't have info then things end up unbiased: in this case, if the contestant doesn't know which box has which of the .46,.46,.04 probs, then the probs end up 1/3,1/3,1/3. But not always! Suppose boxes are labelled, the probs are .48,.48,.04 and the contestant doesn't know how the probs are distributed (they are in the order given). just by coincidence, the contestant loves the number 1 and so has a 100% of picking box #1. Then by chance he has lucked into having a .48 chance of the correct box instead of 1/3.

The second problem you analyzed was, sort of, one where the boxes are unlabelled with contents 48,48,4 but after the flip you find out what the contents of the flipped box are. The prize seems to have vanished and the new goal is to get the most coins? If so, then I'm afraid I disagree with your analysis. The initial prob is that there is a 2/3 chance that the box you chose has 48 coins.
If Monty flips a box with 4 coins then this a new info and now
the prob that your choice has 48 coins is 100%. I agree.
If Monty flips a box with 48 coins then you say that is is not
new info. I disagree. If both unchosen boxes have 48 coins then
there is a 100% chance that Monty will flip a box with 48 coins,
but if one unchosen has 48 and the other 4 then there is only a
50% chance that the flipped box has 48 coins. So Monty's flipping
a box with 48 coins increases my belief that both unchosen boxes have 48 coins. I tried calculating the conditional probs and I get
that my prob of having chosen a 48 box goes down from 2/3 to 1/2 after Monty flips a 48 box.
I used this formula (let "C"="choose", "F"="flip, "|"="given")
P(C48|F48)=numerator/denominator
numerator= P(C48 and F48)=P(F48|C48)*P(C48)=1/2*2/3=1/3
denominator= P(F48)= P(F48|C48)*P(C48) + P(F48|C04)*P(C04)=
1/2*2/3 + 1*1/3= 2/3
so P(C49|F49)=(1/3)/(2/3)=1/2
Does this show your analysis of the original problem is false? You seem to think just because one of the unchosen boxes must have 48 coins means that you get no new info when Monty flips a box with 48 coins. True in original problem; false in new coin problem.

• http://www.blogger.com/profile/09204088792462782905 Happy Spider

J,
Oops, I see that in your second problem you were trying to maximize expected value, not pick the box with the most coins.
You start off with three boxes with 48,48,4 coins but you don't
know which box has which amount.
So originally your expected value is 33. I agree.
If Monty opens the box with 4 coins then you know your box has 48 coins so your expected value is now 48. I agree again.
The problem is, what if Monty opens a box with 48 coins? You argue that the info is irrelevant, so the expected value remains at 33 coins (and so you should switch to the other box for an expected value of 66). I argue that the info IS relevant because if both of the unchosen boxes had 48 coins than Monty has a 100% chance of opening a box with 48 coins but if one box has 48 and the other 4 then Monty only has a 50% chance of opening a box with 48 coins. In fact, I said that the probability of your box having 48 coins is now 1/2, so your expected value is 1/2*48+1/2*4=26. Granted, that's close to an expected value of 33, but it's not exactly the same, is it? So I argue the expected value has changed.

Here is another way to look at it. I have some experiment,sample space, universe with a probability. When I get new information I sort of stay with the same sample space but now I carve up the sample space into pieces depending on how the information turned out. So the ignorant sample space is related to the knowledgeable sample space. To go from ignorant to knowledgeable, I carve up the space. To go from knowledgeable to ignorant I "forget" the information I knew and merge the pieces together. The ignorant prob is the weighted average of the little knowledgeable probs all added up.
The point is, if I take my original prob and divide it into two pieces, but one piece proportionately weighs more than expected, than the other piece has to weigh less. Otherwise when I add the two pieces together the average has to be more than the original prob instead of equal to it.
A little more formally, suppose I am interested in P(X). Now I get some info, and the set of all ways the info can turn out is A,B,C,D,… So my P(X) subdivides into P(X|A),P(X|B),P(X|C)…
And the formula that relates P(X) with all the little probs is to take the weighted average:
P(X)= P(X|A)*P(A) + P(X|B)*P(B) + P(X|C)*P(C)
Oh no, but I am working with expected values and not probs. Maybe it's the same formula?
E(X)= E(X|A)*P(A) + E(X|B)*P(B) + E(X|C)*P(C) ?

In our case X is the number of coins. Info can turn out in two
ways, A=open box with 4 coins and B=open box with 48 coins. I think P(A)=1/3 and P(B)=2/3? I think you stay with the original probs; I don't think it matters that Monty is opening an unchosen box because, unlike in the original problem, Monty just randomly opens a box and doesn't use secret information about knowing which box has the prize. So there is a 1/3 chance that the box Monty opens will be the one with 4 coins. So the formula is:
33.33= 48*1/3 + E(X|flip48)*2/3
100=48 + 2*E(X|flip48)
See? The expected number of coins given that Monty opens a 48 box can't be the original 33.33 because the number has to go down
to compensate for the number going up to 48 when the 4 box was opened. If you solve for E(X|flip48) you see it is 26.

Happy Spider said…

The probability of flipping door #3 is 50% both when door #1 has the prize and when door #1 does not have the prize. So knowing that door #3 has been flipped can't tell us anything about whether door #1 has the prize. It's irrelevant.
On the other hand, the probability of flipping door #3 is 100% when door #2 has the prize and 75% when door #2 does not have the prize. So the info is relevant. Knowing that door #3 was flipped increases the prob that door #2 has the prize.

In more mathy terms, my statement about relevancy is:
IF P(B|A)=P(B|~A) THEN P(A|B)=P(A) (if the probability of B is unaffected by the state of A, then the probability of A's state is unaffected by knowledge of how B turned out). This statement is easy to prove if you accept the formula for posterior probabilities P(A|B)=P(A&B;)/P(B)
===============
J said…

Now information is introduced. “One of the boxes in Monty's subdivided space does not have the X.” But this is NOT NEW information. If the X is only in one box and Monty's space has two, then obviously the X is not in one of the Monty boxes. The probabilities remain; contestant space 1/3, Monty's space 2/3.

======================
Response:

Nice explanations.
Thank you both for the thoughtful comments and discussion.

I'm going to make some comments on the conditional probability justification of the claim that the revealing of a goat behind one of the unselected doors has no impact on the probability (of 1/3) that the contestant's initially-selected door was the one with a car behind it.

Frédéric said…

The fact that P(#1|G2) = P(#1) shows that the events are independent. In other words, observing G2 does not change the probability of #1 (that's the definition of independent).
============
Response:

The above is just one key piece of the conditional probability justification of the accepted solution to the Monty Hall problem.

Note that if Frederic's justification is correct, and does not beg the question, this does NOT mean that either the tree diagram or the verbal reasoning that I outlined in a previous post are good and correct.

What it would mean is that an attempted justification that commits the fallacy of begging the question could be repaired by adding this bit of reasoning to show that the revealing of a goat behind one of the non-selected doors does not change the probability of the contestant's initially-selected door being the winning door.

Given the apparent similarity of Frederic's justification to the plausible justifications provided by J and by Happy Spider, I supsect that I will not be able to find a question-begging premise in Frederic's reasoning, but I will make an effort to look for such a premise.

Here is a bit of clarification on the fallacy of begging the question.

1. I don't have in mind the accusation of circular reasoning, such as:

1. P
Therefore,
2. P

Such blatantly circular reasoning is deductively valid, but the premise is not acceptable, at least if the point of the argument is to persuade someone who has doubts about the truth of P, that P is in fact true. Such a person would be irrational to accept P as being a true premise or as a premise known to be true.

Of course such circular arguments are sometimes given, but the premise is stated in other words, hiding the fact that the premise is the same claim as the conclusion of the argument.

2. I also don't have in mind less blatantly circular reasoning, such as:

1. P
2. If P, then Q.
Therefore,
3. Q
4. If Q, then R.
Therefore,
5. R
6. If R, then P.
Therefore,
7. P

Again, such reasoning may be perfectly valid, but anyone who has doubts about the conclusion of this argument, ought to have the same doubts about premise 1.

3. I have in mind the idea of a premise which is acceptable from one point of view, but which is doubted or controversial from another point of view, and the point of view in which the premise is doubted is a point of view which also rejects or doubts the conclusion being advanced.

For example:

1. If Jesus performed miracles, then God exists.
2. Jesus performed miracles.
Therefore,
3. God exists.

If "miracles" is defined in such a way that a miracle is something that only God can bring about, then premise 2 makes this argument a bit of circular reasoning.

But if "miracles" is defined more broadly as some sort of supernatural or natural-law-defying event, then premise 2 would not make this argument circular.

However, atheists, skeptics, and naturalists not only have doubts about the existence of God. We also typically have doubts about miracle claims too. So, although the argument here might avoid the charge of circular reasoning (by defining "miracles" in a way that does not entail action by God) it cannot avoid the charge of begging the question, because it assumes the truth of a premise that, while acceptable or plausible for religious believers, is doubted or rejected by people of a more skeptical point of view, namely the very sort of people that this argument is supposed to persuade to become believers.

The sort of question begging that I have in mind occurs when a controversial premise is used, and when no plausible justification is provided in support of that controversial premise.

Such question begging is typically the result of a failure on the part of the arguer to be sensitive to how the world looks from other points of view, esp. from other points of view with which the arguer disagrees.

Religious believers often make use of premises that are controversial from the point of view of atheism, skepticism, and/or naturalism.

Of course, atheists and skeptics can also be insensitive to how the world looks from a Christian or Muslim point of view, and offer skeptical arguments that are based on premises that are controversial from a religious point of view.

Question begging of this milder, non-circular type, is common in arguments about God and basic religious beliefs.

• http://www.blogger.com/profile/06056410184615941086 M. Tully

"failure on the part of the arguer to be sensitive to how the world looks from other points of view"

Oh, well then the problem becomes solved. The Monty Hall problem is not decided on a point of view, but on the empirical evidence. If you want to just argue why the premises are wrong, then show a case where they argue to an empirically false conclusion. Then we have an interesting question to be answered.

Frédéric said…

Let's name the various events #1, #2, #3 (that stands for the car being behind door number…), and G2 and G3 (for the event that MH reveals a goat behind door 2 or 3. As said above, G1 is now impossible).

We have the a priori probabilities P(#1) = P(#2) = P(#3) = 1/3, and P(G2) = P(G3) = 1/2 (as MH could open either 2 or 3, and without more info it's a 50/50 chance a priori).

Using the formula for posterior probabilities (http://en.wikipedia.org/wiki/Posterior_probability), assuming we witness G2, we can compute P(#1|G2) and P(#3|G2) (P(#2|G2) is trivially 0).

P(#1|G2) = P(G2|#1)P(#1)/P(G2) = (1/2) * (1/3) / (1/2) = 1/3.

P(G2|#1) is 1/2, as if the car was behind #1, either G2 or G3 can be observed with 50/50 chance.

Now P(#3|G2) = P(G2|#3)P(#3)/P(G2) = 1*(1/3)/(1/2) = 2/3.

P(G2|#3) = 1 because if the car is behind door 3, MH cannot open any door but 2.

The fact that P(#1|G2) = P(#1) shows that the events are independent. In other words, observing G2 does not change the probability of #1 (that's the definition of independent).
==========
Response:
I don't detect any question begging in this justification of the accepted answer to the Monty Hall problem. The conditional probability calculations provide a good reason for concluding that the initial probability of door #1 having the car is not changed by the information that Monty Hall has opened one of the other doors revealing a goat (door #3 in my example). Since the probability of door #1 being the one with the car does not change, and since the combination of the probabilities of door #1 having the car and door #2 having the car must total up to 1 (since door #3 has been eliminated by Monty Hall), the probability of door #2 having the car must be 2/3 (after we find out the information that there is a goat behind door #3).

Although conditional probability calculations confirm the claim that the probability of door #1 having the car is unaffected by Monty Hall revealing a goat behind door #3, this is not the most perspicuous justification of the accepted answer.

Another way of making the same key point is to argue for two sub-points:

A. We know from the start that Monty Hall will open one of the remaining doors to reveal a goat (whether or not there is a car behind door #1; there are two doors with goats, so even if there is a goat behind door #1, there must be one other door with a goat).

B. In terms of the probability of the car being behind door #1, it does not matter whether Monty Hall reveals a goat behind door #2 or door #3 (because either way, there is the same chance/probability that the other remaining door has the car. Both remaining doors have the same logical relationship to door #1 in terms of the issue of which door has the car behind it.)

If we can persuade someone of the truth of points A and B above, then it follows that the information that "Monty Hall has revealed a goat behind door #3" has no impact or affect on the probability that the car is behind door #1.

Thus the probability that the car is behind door #1 remains 1/3, and the probability that the car is behind door #2 becomes 2/3, since the combined probabilities of the car being behind one of the three doors must add up to 1 (certainty), and because the probability of the car being behind door #3 has become 0 (certainly false).

M. Tully said…
"failure on the part of the arguer to be sensitive to how the world looks from other points of view"

Oh, well then the problem becomes solved. The Monty Hall problem is not decided on a point of view, but on the empirical evidence.
=================
Response:
Empirical evidence does not interpret itself; it must be interpreted by a person, specifically by a person who draws inferences from the empirical data.

Whenever a person draws inferences from empirical data, one's assumptions, principles, and theories are relevant and influence what conclusion is drawn. If we understand 'point of view' to encompass such assumptions, principles, and theories, then inferences to conclusions involve a 'point of view'.

In some cases there is no legitimate alternative 'point of view'. If I have three buckets, and there are 25 strawberries in each bucket, then I have a total of 75 strawberries in those three buckets. There is no legitimate alternative to elementary math, to answer this question.

However, elementary math does involves assumptions, principles, and concepts, so it can be thought of as a point of view. Elementary math provides a logical background for the inference made in this example.

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