I fear that people may have gotten lost in the details of the data and code for the football won/lost example, so I wanted to clarify why I wrote the post.
In general there are three ways of attacking a statistical problem:
1. Probability calculation. Set up a probability model and crank it through. This will require a bunch of assumptions, and you’ll also need to set parameters in your model to reasonable values.
2. Direct empirical calculation. This will work if you have enough data, and if these data are not subject to selection.
3. Statistical modeling. Kind of like method 1 above, except that you fit (“learn”) the parameters from the data; as a result you can fit a more complicated model. I include machine learning in this category too.
In statistics classes, we focus on method 3. No surprise, right? Statistical modeling encompasses probability calculation and direct empirical calculation; indeed methods 1 and 3 can be viewed as special cases of method 3. Method 1 is method 3 but with a simple model and a crude method of setting the parameters. Method 2 is method 3 but with a simple model assuming stationarity in all directions.
So, yeah, statistical modeling. There’s a reason my colleagues and I have written multiple books on the topic, spent innumerable person-hours developing and using Stan, etc.
But . . . it’s good to know about methods 1 and 2 as well.
Why? Four reasons.
First, methods 1 and 2 are simpler, and sometimes they work just fine.
Second, even beyond simplicity, methods 1 and 2 have fewer requirements. Method 1 does not require the data (which is how we were able to get a good answer to that football question in the first place), and method 2 does not require a data-generation model. In contrast, method 3 requires data and a model.
Third, even when estimates based on methods 1 and 2 are seriously flawed, they can be useful starting points and comparison points to better approaches. Indeed, sometimes when a probability calculation gives a ridiculous result, this can be useful in developing intuition. For example, the notorious calculation of a probability of a tied election as 10^-90 came from an inappropriate application of a binomial-distribution model, which motivates the development of models for statistical dependence among voters, while the failure of straight-up empirical estimates motivates models that combine probability modeling and empirics.
Fourth, when people are informally estimating things, they’re often using some version of method 1 or 2. Which is fine! But then I think it’s important to be aware of what you’re doing and to ask, What is the probability model you are assuming, or What is the frequency calculation you are making?
Those four reasons–that’s the point of yesterday’s post.
I appreciate explicitly laying out these approaches, particularly because, as you say, our teaching focuses almost entirely on method 3. I think we would do better to provide more explicit training on methods 1 and 2.
I find that in my classes, both undergraduate and graduate students tend to default immediately to some version of method 3. That is, when given an inference problem, they think of some relevant data and an off-the-shelf statistical model like regression and try to work from there. That is also the case for many of my professional colleagues. But while method 3 can be reviewed as encompassing all 3 approaches, it also offers many more choice points. What is the appropriate model? What kinds of parameters would it need to include and for what reason? What data is appropriate to the inference situation and why? By immediately jumping into method 3, one is liable to miss out on other options and end up with an inference that is wrong or only tangentially addresses the situation.
By forcing one to consider the problem in the abstract (ie, without data), method 1 forces you to think about the most important theoretical constructs to include in your model, how they interact with one another, and how they are coupled to observables. By dealing only with raw data, method 2 forces you to consider not only what data would ideally speak to the inference, but also how that data in practice may be biased by selection. This latter point can then lead you to design a better study or a better model to address selection bias. In the end, by following methods 1 and 2 first, as you did in your earlier post, one is in a better position to adopt method 3 in a more thoughtful manner that is more likely to result in meaningful inferences.
Your simulation assumes that the games are independent, which they obviously are not. In the NFL, each win for a team is accompanied by a loss for the opponent. And in the first 4 games of a season, there will be common opponents. Does this dependence change the intuition for how large the effect should be?
Also, the size of the effect with sigma_a =1.0 was larger than with sigma_a=0.7. And the effect as seen in the 10-year reporting of the simulation was larger than what is seen in the 10-year actual data. Is this because of dependence in the data (as above) or because sigma_a isn’t low enough?
Seems there are a lot more moving parts in this problem than it first appeared…
Bjs:
We’re talking about expectations so I don’t think dependence will be an issue.
It seems that there is an issue when you are “replacing the normal(0,1) distribution by a normal(0,0.7) on the positive side and normal(0,1.0) on the negative” because the expected numbers of wins and loses are not equilibrated.
Carlos:
No, you’re wrong on this. Each game has a probability that team i wins, which is 1 minus the probability that team j wins. An asymmetric distribution of the ability parameters doesn’t change that identity, which is there by construction.
Andrew, your simulation only records the outcome for team i.
Consider a 2-team league. No matter if A and B have equal ability or one is drastically better than the other, the probability of a 0000 start to the season must be exactly the same as the probability of a 1111 sequence.
In a million team league, where the teams play an independent set of 4-million other teams, the probability of 0000 can be different than 1111, depending on the shape of the ability distribution. This is what your simulation shows.
The question remains, what happens in a 4-game sequence in a 32-team league?
Bjs:
Sure, and I can tell you that for this purpose a 32-team league is much more like an infinity-team league than a 2-team league. I’m very sure of this from my probability intuition, formed by working on many probability problems over the years. If someone wants to do a simulation with 32 teams and an NFL-style schedule for the first 4 games of the season, then they should go for it. Just in general, much can be learned from the effort of trying to set up a realistic simulation.
> Carlos: No, you’re wrong on this
Thanks, I was indeed wrong. I lost sight of the difference applied later (I think I got confused in part because of your mention of the issue you encountered with the home advantage, but that was applied after the difference).
I checked if there was any bias in the “simple” simulation compared to “proper” simulations with a low number of teams with fixed abilities being paired week after week but apparently there is no bias (or it’s low enough that I got bored before detecting it). For simplicity I didn’t include the home advantage correction in my checks (by the way, that correction reduce the frequency of all-wins and all-losses sequences compared to the no-home-advantage case).
Fascinating. I have created the taxonomy specifically for the Monty Hall Problem here:
https://trissimondsen.wordpress.com/2026/03/27/prodromus/
Your thoughts?
I thought this was a nice way to portray in detail the various way the Monte Hall problem might be defined and analyzed. However, one puzzle remains: given that Monte is a human, and accepting his denial of using a random choice, what information is contained in his choice of door to open that would have any impact on the probabilities the contestant would attach to the remaining doors? It is hard for me to imagine that Monte’s choice could give me any relevant information about the remaining doors – except in the case where Monte has a secret signal that the contestant is aware of. Except for such cases, doesn’t it make sense to treat Monte as if he were a mechanized process?
You are absolutely correct – it doesn’t. The table illustrates that. The only way the problem would NOT be 1/2 is if Monty is mechanized (q=1/2). It is also noteworthy to point out that 1/2 is not a Bayesian derivative. Bayes ends at underdeterminate – the two doors are neither better nor worse for the player. 1/2 is the unavoidable posterior for the player.
There is a good discussion of the Monte Hall problem at
https://terrytao.wordpress.com/2022/10/07/a-bayesian-probability-worksheet/
Tao shows that one must be careful in defining the problem. Slight changes in the description of Monte Hall’s behavior can switch the result from (1) switching moves probability of winning from 1/3 to 2/3 to (2) not switching gives probability of winning of 1.0.
Here’s a different variant of the problem that may be easier to understand. There are one million doors. A car is behind one of them and goats behind the others. You pick one using a random number generator. Monte Hall opens all but one of the other doors—revealing only goats. Would you switch?
The one million door analogy is the mechanized q=1/2 amplified. You now expect Monty to open 999,998 empty doors with even more than the already humanly impossible 50/50 randomized position. The actual problem when correctly identified through the taxonomy table, is clear that the protocol is unknown and therefore inadmissible to the agent. I have reached out to Tao on multiple occasions but he has remained silent on the matter.
> If the host’s selection protocol is not explicitly defined as a symmetric random variable, the problem is underdetermined.
> The ONLY internal rule from Monty which tilts the posterior to 2/3 is his tie-breaking rule being random, which 2/3 uses, but is not included in its problem statement.
That’s not correct. If the host’s selection protocol is not explicitly defined the lack of information about the tie-breaking rule can be represented by a 50/50 split because all the doors are identical as far as we know. Our probability assignment should be invariant if we relabel the doors. Whatever tie-braking rule is actually used is irrelevant because we don’t know anything about it and it cannot be part of the solution. The 50/50 probability doesn’t represent the unknown tie-breaking rule.
> Bayesian conditioning cannot proceed uniquely without an explicit mechanism.
It surely can, when one accepts that those probabilities represent uncertainties and not mechanisms. If that’s an issue for frequentists the problem may be frequentism rather than Bayesian conditioning.
You’ve hit the nail on the head regarding invariance, but you’ve missed the landing. If the doors are truly identical and we “relabel” them, then by definition, the probability must be 1/2. To reach 2/3, you are forced to break that symmetry by assuming a specific, hidden mechanism (q = 1/2). You claim the host’s protocol is “irrelevant” to the solution, yet your 2/3 solution is mathematically dependent on that exact protocol. You are essentially using “Bayesian uncertainty” to mask a frequentist assumption of a symmetric host. As the taxonomy table shows, without an explicit mechanism, the problem is underdetermined – you’ve simply chosen to “fill in the blank” with a 2/3 bias.
> To reach 2/3, you are forced to break that symmetry by assuming a specific, hidden mechanism (q = 1/2).
No, I’m not.
> You claim the host’s protocol is “irrelevant” to the solution, yet your 2/3 solution is mathematically dependent on that exact protocol.
No, it isn’t.
When I pick a door I have 1/3 probability of picking the car. If I know that one of other doors will be open I doesn’t matter what protocol is used to choose when possible, the probability that the car is behind the door I choose is still 1/3.
The probability that the car is behind one of the other doors was 2/3 and it remains 2/3 all along. It is 1/3 for each of those doors before one is opened, afterwards it’s 0 for the one that is opened and 2/3 for the one that remains closed. If there was a choice, it’s completely irrelevant how it was made.
There is no explicit mechanism required to arrive to the 2/3 solution. I’m not filling any blank.
> If there was a choice, it’s completely irrelevant how it was made.
To be clear, if the problem was a completely different one the fact that one particular door was opened could be informative and impact the probability of having selected the car changing it from 1/3 to something else.
But we have agreed that I cannot distinguish between the doors and thus, just like I cannot prefer one to the others when picking, I cannot learn anything by seeing which one is opened.
> If the doors are truly identical and we “relabel” them, then by definition, the probability must be 1/2.
No, what we want to calculate is the value p(car=A | picked=A, opened=B) and its complement p(car=C | picked=A, opened=B). They sum to one but they are not 1/2 “by definition”. Relabeling the doors we get p(car=B | picked=B, opened=C) and p(car=A | picked=B, opened=C), etc. with the same values as before.
> To reach 2/3, you are forced to break that symmetry by assuming a specific, hidden mechanism
To _not_ reach 2/3 you’re forced to break the symmetry between the two doors that you didn’t pick and assume that one is more likely to be opened than the other. If you could assign different opening probabilities to them they wouldn’t be identical. The opening probability for both must be the same if it’s invariant under relabeling of the doors.
I appreciate your responses. Unfortunately you have not correctly identified which problem you are solving. Your reasoning contains both the vanilla FSSP and the SSP rationale in defense. This illustrates the 50-year-old paradox perfectly.
> Unfortunately you have not correctly identified which problem you are solving.
In case it was not clear, I’m solving the problem where
– I’m presented with three doors and I know that there is a car behind one of the doors but I know nothing about how the placement of the car was decided (so I have no reason to think that the car is more likely to be behind any of them, the doors are exchangeable as far as I’m concerned)
– I pick one (it doesn’t matter which one, see above) and I know that one of the two remaining doors will be opened and the car won’t be there but I know nothing about how the choice is made if a choice can be made (so I have no reason to think that one of the doors is more likely to be opened than the other, the doors are exchangeable as far as I’m concerned)
– One door is opened, as expected, and the car is not there, as expected, and the probability that the car is behind the door I picked is still 1/3 and the probability that the car is behind the other door is 2/3
I didn’t assume anything about how the placement of the car was chosen or about how the door that would be opened was chosen if a choice had to be made. The reasoning is based only on the fact that I don’t know anything that distinguishes one door from the others. There is nothing in the problem statement that would justify treating the doors differently.
I understand that you may want to solve or not solve other problems but I don’t think that this creates a paradox.
It absolutely does. If you did not assume “about how the door that would be opened was chosen if a choice had to be made”, HOW precisely IS it determined?
The case is precisely not clear on your end. You seem to be purposefully avoiding the table, which has every detail blatantly crystallized. Let me try this another way, is the problem you’re solving NOT on that table? If it IS, which is it?
> If you did not assume “about how the door that would be opened was chosen if a choice had to be made”, HOW precisely IS it determined?
I don’t know! Making that choice is the host’s job, not mine.
I do not make any assumption about how the car is hidden in the first place either. Doesn’t that bother you?
“I don’t know” – q is unknown; inadmissible to the agent. The problem is underspecified.
> q is unknown; inadmissible to the agent.
There is no q in my problem. I don’t know who’s the agent either.
> The problem is underspecified.
In whatever solution you find acceptable to whatever problem you want to solve, what’s your assumption regarding which hand does the host use to open a door? If you don’t have an explicit assumption HOW precisely IS it determined?
There’s no q in your problem and you don’t know who the agent is, then that problem has no domain in our discussion. I am specifically referring to The Monty Hall Problem where BOTH are CRUCIAL.
Is whichever hand Monty uses a factor mentioned in the taxonomy table? If it is not, it is irrelevant. We can also discuss which door has a closer proximity to where Monty stands that day, was he in a good mood, perhaps his contract just got extended that morning. Irrelevant.
> Irrelevant.
How he chooses which door to open is also irrelevant. Unless you assume that it depends on whether we picked the car and that observing which door he opens provides valuable information, of course, but then the hand he uses to open the door is also relevant if you assume that it depends on whether we picked the car and that observing which hand he uses provides valuable information. If the choice of door and hand doesn’t provide any information about whether we picked the car or not the initial probability (1/3) will remain unchanged.
> I am specifically referring to The Monty Hall Problem where BOTH are CRUCIAL.
I see. I was referring to The Monty Hall Problem which is what I described (I’m presented with three doors and I know that there is a car behind one of the doors; I pick one and I know that one of the two remaining doors will be opened and the car won’t be there; One door is opened, as expected, and the car is not there, as expected.) and can be solved as I explained.
Given that Your Monty Hall Problem is apparently not the same as My Monty Hall Problem it’s not surprising that we cannot reach an agreement.
Hence the 50-year-old “paradox” persists.
To avoid future circular logic without an escape, PLEASE refer to the table. Identify which problem you are defending FIRST. If it isn’t there, you are NOT discussing the Monty Hall Problem.
Consider another version of the MH problem.
One million doors. Car behind one door; empty behind the others.
You pick one door using a high-quality random number generator.
Monte Hall says that you can have (1) what is behind the door you chose or you can have (2) everything behind all the other doors.
Q1. Would you switch?
Q2. How is this different from the million door case when MH gives you the choice of everything behind all the other doors but through a different mechanism?
This would be most accurately referred to as another Fully Specified Stochastic Process (FSSP), but as a “Let’s Make a Deal” variant – not the Monty Hall Problem.The distinction I am making through Priority Rule and historic chronology is that the MHP is quite specific in its problem statement.
To answer your scenario:
A. Switch.
Because no doors have been opened, no host selection has occurred. The probability is strictly a matter of initial set distribution (999,999/1,000,000).
B. This problem requires zero host agency.
The q (Monty’s selection protocol) is completely removed. Monty is now a static cashier offering a bulk package – there is no filtering. You are using a no-selection-mechanism to justify the 2/3 result in a scenario that is defined by its selection mechanism. You’ve bypassed the “HOW” by changing the game entirely.
I don’t think you answered my question. If MH opens 999,998 doors and you see than they all reveal nothing, would you switch? You would switch if he let you switch from 1 door to 999,999 doors. But, after he opens 999,998 doors, he is still offering you the same deal.
Consider another variant. You pick one door. MH offers the opportunity to switch to all the other doors. You accept. Then he opens 999,998 doors, all revealing nothing, and he again asks you if you still want to switch. As I understand your view, it is reasonable to say “No. There is a 50:50 chance that the car is behind the door I picked.”
That cannot be right under any reasonable approach to probability based on betting preferences.
That is already a different question from the one you asked yesterday. You asked yesterday “you can have (1) what is behind the door you chose or you can have (2) everything behind all the other doors.” This is BEFORE Monty does anything or any doors being opened. Of course you would take the 999,999 offer. That’s like saying would you like one door or 999,999 doors to win the car.
You are now asking a different question, where there are TWO offers AFTER your initial pick.
So let me walk through this with you.
The first offer is 1 door or 999,999 doors, AFTER you make your initial pick to separate into two groups. You switch to take the 999,999 doors.
That is a different question from the one you asked yesterday. Yesterday, you asked: “you can have (1) what is behind the door you chose or (2) everything behind all the other doors.” This is before any doors are opened, so of course you would take the 999,999 doors – it’s literally choosing one door versus 999,999 doors to win the car.
Now you are asking a new question: there are two offers after your initial pick.
First, you pick 1 door versus 999,999 doors – you switch to the 999,999 doors.
Then Monty opens 999,998 doors from your 999,999 doors and offers you a second chance to switch.
At this point, the problem requires knowledge of how Monty selects doors to open. Without specifying the removal mechanism:
– if Monty avoids the car, the remaining door keeps nearly all the probability.
– if Monty removes doors randomly (and luckily doesn’t hit the car), the initial weights collapse, via renormalization and the final two doors are 50/50.
Without this information, the problem is underspecified; there is no uniquely determined probability for the last choice. As a “betting” man and the player, your choices are neither better or worse from each other. 1/2 is the only mathematically defensible baseline for an agent who lacks a specified protocol.
“Reasonable betting preferences” don’t resolve this – they only force you to acknowledge that the probabilities are undefined.
In the canonical Monty Hall problem, the parameter q BAKES IN how Monty chooses which doors to open. Without specifying q, any probability you assign to switching or staying is simply assuming a particular strategy, not deriving it mathematically. The problem is underdetermined until q is defined. I recommend you actually do the math to see this. I have learned that most people who cling to 2/3 have never actually done the math.
> First, you pick 1 door versus 999,999 doors – you switch to the 999,999 doors.
> Then Monty opens 999,998 doors from your 999,999 doors and offers you a second chance to switch.
> At this point, the problem requires knowledge of how Monty selects doors to open. Without specifying the removal mechanism: if Monty avoids the car, the remaining door keeps nearly all the probability
Interestingly, I think we all agree on that. (Maybe I completely misunderstood what you wrote, let’s see?)
> Without this information,
Without what information?
> the problem is underspecified; there is no uniquely determined probability for the last choice.
There is no uniquely determined probability for the last choice in what you wrote but the problem is no underspecified because after you noticed “if Monty avoids the car, the remaining door keeps nearly all the probability”.
> As a “betting” man and the player, your choices are neither better or worse from each other. 1/2 is the only mathematically defensible baseline for an agent who lacks a specified protocol.
So what is it? “1/2 is the only mathematically defensible baseline for an agent who lacks a specified protocol” or “if Monty avoids the car, the remaining door keeps nearly all the probability”?
> Without specifying q, any probability you assign to switching or staying is simply assuming a particular strategy, not deriving it mathematically. The problem is underdetermined until q is defined.
It seems that you didn’t specify q when you wrote “Without specifying the removal mechanism, Monty avoids the car”. So what is it? “Undefined” or “the remaining door keeps nearly all the probability”?
If Monty doesn’t avoid the car, you still find out what’s behind the opened doors…
So, there’s 3 doors. You pick one. Monty picks one other one at random not avoiding the car, and opens it… now:
1) 50% of the time you see a car behind the door, and you select the car door
2) 50% you see nothing behind the door, and you have a 50% chance of having picked the right door first, and a 50% chance of wrong door… If you either switch or stay, you still win 50% of the time.
Overall you win 75% of the time
Or, Monty avoids the door:
1) You pick a door at random, Monty picks one of the remaining doors avoiding the car and you see what’s behind it.
2) 1/3 of the time you picked the right door at first, and 2/3 of the time you picked the wrong door first, and Monty showed you which of the remaining doors has the car (the one he DIDN’T pick).
3) If you switch, you win 2/3 of the time.
In both scenarios you win more than 1/3 of the time which is what would happen if you just picked a door at random and they opened it.
Monty’s selection algorithm DOES matter. However, in every version of the problem statement I’ve seen it’s clear that he will never pick a door with a car, and so “the Monty Hall Problem” as stated in textbooks etc wins 2/3 of the time if you switch.
Ack… it’s early morning, and somehow I assumed you couldn’t pick the car in the first place in the first scenario… that was wrong.
here’s the rewrite:
If monty picks at random without regard to the car:
1/3 of the time you picked the car first. Monty will show you a door, and if you switch you’ll lose.
2/3 of the time you picked not a car. in that scenario, 50% of the time monty will show you a car, and you switch to that door and win… 50% of the time you see an empty door, and if you switch to the closed door you win
So p(win | switch) = 1/3 * 0 + 2/3 * (1/2 + 1/2) = 2/3
The main difference is whether you switch to the car door that he reveals for you, or you switch to the closed door that he doesn’t reveal for you. If he chooses at random you have to choose to switch to the car door if he reveals it. Whereas if he avoids the car, then you have to always switch to the closed door.
Sorry for confusion, I should have breakfast before answering probability questions on the internet.
I take it that you are referring to the canonical Monty Hall Problem with your explanation, and not Bob’s variations. Bob’s variations have a hypothetical two switch offer. So it is quite different but the development for verification is similar.
As for the canonical, you ARE precisely at one of the fault lines of the problem. I have compiled a historical timeline of how the problem became the canonical via textbooks. Through Priority Rule, the problem was left WITHOUT Monty’s internal decision mechanism, specifically his tie-breaking rule. That is why when we disagree, you are referring to a problem where Monty’s internal rules are fully specified, but more importantly, artificially manufactured. Because Monty’s mechanism is contextually-based, not algorithmic. So you can tell already that the problem has been altered.
There are three different “variations” of the process which people conflate as THE Monty Hall Problem, hence the “paradox”. Once you’ve clearly identified which problem SHOULD BE the Monty Hall Problem, there is only one unique correct posterior which corresponds to the problem as stated, while retaining the correct agent asking the question. Because the player definitely does not have two or a family of possibilities to act upon to complete his path and quantify the uncertainty for the expected payout.
Note, “he will never pick a door with a car” alone does not tilt the posterior to 2/3. Monty picking between two empty doors at “random” or precisely 50/50 is required. 2/3 bakes this into the calculation. It is often purposefully omitted in the problem statement, but when the math calculation is requested to be shown, the truth is revealed. Why? Because they want the answer to be “2/3 switch”.
So to answer your question, the textbooks say so – no, they do not. Only the later editions where the problem is altered say so.
Tris,
Your statement “Note, “he will never pick a door with a car” alone does not tilt the posterior to 2/3. Monty picking between two empty doors at “random” or precisely 50/50 is required” seems incorrect.
Suppose Monty always avoids the car, and always picks the left non-picked door when possible. Then if you always switch:
1/3: You picked the car, Monty opened the left door, you picked the right, you lost
2/3: You did not pick the car: p the car was behind the right door, Monty opened the left, you picked the right, and you won; 1-p the car was behind the left door, Monty opened the right, you picked the left, and you won.
Therefore, your chance of winning when switching is 2/3*(p+1-p)=2/3. So it did not matter that Monty picked between two empty doors with 50/50 chance. I also added that the distribution of the car between the two doors does not need to be 50/50. And it is still 2/3.
You missed the operational word in both of those bullet points: IF.
Those two scenarios were conditional hypotheticals demonstrating why the removal mechanism dictates the math. I explicitly wrote:
– IF Monty’s protocol is to avoid the car, the results is X.
– IF his protocol is random, the results is Y.
The entire point is that the problem text does not specify which protocol is in effect. Because the mechanism is unknown, the problem is underspecified, and Bayesian updating terminates at underdetermination.
As an analyst, your work stops there. The mathematically rigorous answer is “Undefined.” You cannot completely the posterior without MAKING UP A RULE that wasn’t in the text.
However, the player, or in Bob’s scenario, the bettor, does not have the luxury of remaining in the abstract underdetermination. They must quantify their uncertainty for the expected payout in a one-shot decision. This is where the Observational Sufficiency Principle (OSP) bridges the gap.
Because the bettor’s admissible information (F) contains no defined mechanism to weight one door over the other, applying an asymmetric probability (like 99.9%) requires smuggling in an assumption (a specific q-parameter) that isn’t observable, and, more precisely, STATED. Without that hidden mechanism, the remaining doors are observationally indistinguishable. Neither is better nor worse.
1/2 is not a “guess” – it is the unavoidable baseline for an agent forced to act without a specified protocol.
> You missed the operational word in both of those bullet points: IF.
I don’t know what would you think so.
> I explicitly wrote: – IF Monty’s protocol is to avoid the car, the results is X.
I just focused on this one because that’s the standard Monty Hall Problem: “Certainly Monte Hali knows which box is the winner and, herefore, would not open the box containing the keys to the car” [Selgin, 1975].
I accept that, Bob76 accept that, the standard 2/3 solution accepts that and even your negative to accept the standard 2/3 solution is not about that but about something else.
I do agree with you that “IF Monty’s protocol is to avoid the car, the results is X (>1/2)” – however, I’m afraid you won’t agree with yourself because you’ve repeatedly say before and after that “Monty avoids the car” is not a completely specification of the protocol and the problem is undetermined.
I really wonder if you stand by what you wrote there
“IF Monty’s protocol is to avoid the car, the results is X”
or whether you want to say instead that
“IF Monty’s protocol is to avoid the car that still gives no defined mechanism to weight one door over the other when the car is elsewhere so it does not fully specify which protocol is in effect and because the mechanism is unknown, the problem is underspecified, and Bayesian updating terminates at underdetermination.”
Is Tris a bot? The Monty (not Monte) Hall problem is based on a quiz show. People watch quiz shows for the suspense and TV companies make the shows that people will watch. Monty will always open a door without the prize to sustain the suspense. This is meta-knowledge hence I doubt that Tris is a human.
This reply is very much the level of someone who believes the correct answer is “1/3 of the time you picked the car and 2/3 of the time you don’t so you should switch”. ICYDK, the very first mention of the problem from Steve Selvin, the host was written as “Monte”, meaning “Monty”. Dale was simply reminding us of how the problem originated. And if you believe “Let’s Make a Deal” is a QUIZ show, I recommend you look further into where precisely the fault line of the dispute is before replying.
To all:
Enough on the Monty Hall problem! It’s an interesting problem but the thread is getting out of control. Thanks for understanding.