The Monty Hall paradox is well known, even in this site there are several discussion about it (for example: How can I apply Monty Hall problem correctly? ; Monty hall problem, getting different probabilities using different formulas? ), however my point here is a bit different.

One of the most important point of causal inference, at least as presented in Pearl (and colleagues) literature’s, is that standard statistical tools are adequate for passive observation but not for intervention. The last is the core of causality, therefore standard statistical tools cannot face properly causal issues. For example we cannot deal with intervention by standard conditioning; tools as do-operator was proposed ad hoc.

Now in the Monty Hall problem, intervention appear; indeed the problem is precisely to evaluate one probability/choice after the host intervention. However one solution, probably the more shared, invoke precisely the standard conditioning tool and it seems work. Read here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Solutions_using_conditional_probability_and_other_solutions or even the solutions proposed in the discussions suggested above.

However this fact seems me in contrast with the main argument of causal inference. It is so? If yes, because it seems work? What is the solution proposed in causal inference?

  • 1
    $\begingroup$ What are examples of inadequate standard statistical tools? Is computing a conditional probability as in the Monty Hall problem (which btw I would not call a paradox) a standard statistical tool? "However this fact seems me in contrast with the main argument of causal inference" That argument is very vague because it speaks about 'standard statistical tools' which is a bit of a catch-all term. $\endgroup$ Dec 5, 2020 at 16:15
  • $\begingroup$ It can appear vague but I used Pearl’s terminology. He suggest a “demarcation line” between statistical and causal concepts. Joint probability distribution is the main statistical concept/tool. The most important example of causal concept is the structural equation. The most conflation involve conditioning; an statistical concept. It is wrongly used also here? $\endgroup$
    – markowitz
    Dec 5, 2020 at 16:53
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    $\begingroup$ Maybe the problem with this situation is that this Month Hall problem is not an example of inference. When you roll a dice or have some other sort of gamble (with known parameters that describe the sample distribution) then it is just a problem relating to probability theory and it has not to do with statistics. $\endgroup$ Dec 5, 2020 at 17:45
  • $\begingroup$ "It can appear vague but I used Pearl’s terminology..." a problem with some terminology and concepts is that the practice might get hidden behind the concepts (and this could be an example). The question revolves around 'what does all that terminology actually mean'. It is hard to complain for me since I haven't read any of Pearl's books, but the texts where the terminology is being used often read fuzzy to me and are most often without a pragmatic example. Terminology can be a way to commit argumentum ad verbosum. $\endgroup$ Dec 5, 2020 at 17:48
  • $\begingroup$ I see your points. Maybe you are right. I don't know if the tools of causality deal with the problem above, my question come from that too. About “inference” maybe you are right again, maybe the above problem is off topic; however this last can be a false problem. I’m not knife deep in Pearl literature but I know that him introduce “interventional probability” concept too. I have thinked that it could match the Monty Hall problem, but is possible that Julian Neuer right. $\endgroup$
    – markowitz
    Dec 5, 2020 at 18:43

3 Answers 3


The problem with this situation is that this Monty Hall problem is not an example of inference.

When you roll a dice or have some other sort of gamble (with known parameters that describe the sample distribution) then it is just a problem relating to probability theory and it has not to do with statistical inference or causal inference.

In terms of a structural model you might have something like:


The model contains four independent variables

  • The door behind which is the prize (categorical with equal probabilities for each door)
  • The door chosen the first time (categorical with equal probabilities for each door)
  • An additional variable for the case when the 1st door chosen equals the door with the prize. In that case the quizmaster must choose a random door that he leaves open.
  • The strategy

There's some variables that are functions of others

  • The doors opened by the quizmaster is a function of the three random variables. If the prize and first choice are different then the quizmaster opens all other doors. If the prize and first choice are the same then the quizmaster opens all other doors except the one determined by the additional variable.

  • The second choice by the contestant. Depending on the strategy

    1. The contestant will switch to a door depending on the quizmasters actions.
    2. The contestant will not switch.
  • The outcome is whether we win a prize or not.

The Monty Hall problem is to compute whether the strategy of switching is a good strategy.

So the intervention is not the quizmaster, but it is the strategy. The computation, the solution, is eventually the old fashioned statistics giving $$P(\text{success $| do($strategy$=$switch$)$}) = 2/3$$ and $$P(\text{success $| do($strategy$=$no switch$)$}) = 1/3$$ The causal inference is not really doing anything here to solve the Monty Hall problem. What it does is giving us a method to describe the structure of relationships and determine whether statistical correlations between outcome probabilities $P(\text{success})$ and interventions $\text{$do($strategy$=$no switch$)$}$ can be interpreted causally.

  • $\begingroup$ As I said in the comments, this point can be a false problem or just a terminological one. We can speak about causal calculus, more in general. Even if the probabilistic model is fully specified the distinction between causal and statistical concepts remain; them hold at population level also. The solution here move around the concept of “intervention”. In some sense it appear in the problem but we have to verify if it match the sense intended in causal literature and, if yes, how to deal with it here. $\endgroup$
    – markowitz
    Dec 10, 2020 at 16:56
  • $\begingroup$ @markowitz it is not inference. The intervention happening here is not to be regarded in the frameworks that are used to approach causal inference. $\endgroup$ Dec 10, 2020 at 17:38
  • $\begingroup$ I feared something like you suggest. The point is that here the intervention is made by the host and not the player? Or something else? I know that gambling is the realm of Probability Theory but causal inference/calculus and causality in general, at least as presented by Pearl, seems a quite general tool. I ask myself if that permit to address problems like above, with different point of view in comparison to the standard. Anyway, can you give us simple examples where the permitted and unpermitted kinds of interventions, in causal sense, appear clearly? $\endgroup$
    – markowitz
    Dec 11, 2020 at 9:40
  • $\begingroup$ @markowitz I am currently puzzling how we could force this in the framework by Pearl, but I have to dig further in the literature. The introductory literature leans heavily on structural equation modelling, related paths or graphs, but that is all linear stuff. $\endgroup$ Dec 11, 2020 at 9:47
  • $\begingroup$ I also wonder whether it is really different from regular statistical tools; it seems to be the same thing cast in a different (more formal) language. The core remains probability theory and the theories about causal inference are not different from statistical methods, they just provide an extra layer in relation to the causal interpretation. You do not need Pearl's framework to do causal inference. It is more like a tool that helps you to do it. Without a directed acyclic graph or without a do-"operator" you can also reason about experiments and relationships between theory and observation $\endgroup$ Dec 11, 2020 at 9:51

That is a very interesting question. My two cents:

Intervention as defined by Pearl and others means altering the causal DAG by removing an arrow --- i.e., removing the direct effect of a variable A on a variable B --- and setting the value of B "manually" so to speak.

In the Monty Hall problem, you say the host intervenes; I disagree. When the host opens the door to show the goat, I don't see any variables being manually set in the causal DAG.

Anyway, what would be the causal DAG for the Monty Hall problem?

I haven't built it, so this is just speculating, but this is what came to mind when I read your question:

When the host opens the door, the only variable I see him intervening on is the contestant's belief of which door has the car, but I think this variable would not appear in the DAG I would draw.

In short: the host's intervention is on the epistemic state of the contestant, and only if we include this state as a node in the causal DAG can we talk about causal issues here.

It would be interesting to hear other opinions.

  • $\begingroup$ From your interesting points one question come in my mind. The intervention intended in the do-operator (then DAG), deal only with observer intervention? No matters what others make (here the host)? $\endgroup$
    – markowitz
    Dec 6, 2020 at 13:34

I discovered that this story is used also in: CAUSAL INFERENCE IN STATISTICS A Primer - Pearl Glymour and Jewell (2016). Monty Hall problem can be addressed in both ways: purely statistical and causal one.

Authors deal with Monty Hall problem with standard conditioning and related Bayes rule. Indeed at pag 14 it is said that:

Another informative example of Bayes’ rule in action is the Monty Hall problem, a classic brain teaser in statistics. In the problem, you are a contestant on a game show, hosted by Monty Hall. Monty shows you three doors—A, B, and C—behind one and only one of which is a new car. (The other two doors have goats.) If you guess correctly, the car is yours; otherwise, you get a goat. You guess A at random. Monty, who is forbidden from revealing where the car is, then opens Door C, which, of course, has a goat behind it. He tells you that you can now switch to Door B, or stick with Door A. Whichever you pick, you’ll get what’s behind it. Are you better off opening Door A, or switching to Door B?

Switching is better

We can prove this surprising fact using Bayes’ rule. Here we have three variables: X, the door chosen by the player; Y, the door behind which the car is hidden; and Z, the door which the host opens. X, Y, and Z can all take the values A, B, or C.

In the text the solution is not outlined but I find it in this webpage http://bayes.cs.ucla.edu/PRIMER/

enter image description here

However them note also that (pag 15):

The reader may find it instructive to note that the explanation above is laden with counterfactual terminology; for example, “He could have opened,” “because he was forced,” “He was about to open.” Indeed, what makes the Monty Hall example unique among probability puzzles is its critical dependence on the process that generated the data. It shows that our beliefs should depend not merely on the facts observed but also on the process that led to those facts. In particular, the information that the car is not behind Door C, in itself, is not sufficient to describe the problem; to figure out the probabilities involved, we must also know what options were available to the host before opening Door C.

Indeed later two different study questions are posed:

Study question 1.3.5 (Monty Hall) Prove, using Bayes’ theorem, that switching doors improves your chances of winning the car in the Monty Hall problem.

Study question 1.5.4 Define the structural model that corresponds to the Monty Hall problem, and use it to describe the joint distribution of all variables.

The question 1.3.5 is solved above (even if the exercise number reported in the solution is erroneous)

More ahead (pag 41-43) the solution for the question 1.5.4 is outlined too

enter image description here

“ … Another example of colliders in action—one that may serve to further illuminate the difficulty that such configurations can present to statisticians—is the Monty Hall Problem, which we first encountered in Section 1.3. At its heart, the Monty Hall Problem reflects the presence of a collider. Your initial choice of door is one parent node; the door behind which the car is placed is the other parent node; and the door Monty opens to reveal a goat is the collision node, causally affected by both the other two variables. The causation here is clear: If you choose Door A, and if Door A has a goat behind it, Monty is forced to open whichever of the remaining doors that has a goat behind it. Your initial choice and the location of the car are independent; that’s why you initially have a 1/3 chance of choosing the door with the car behind it. However, as with the two independent coins, conditional on Monty’s choice of door, your initial choice and the placement of the prizes are dependent. Though the car may only be behind Door B in 1/3 of cases, it will be behind Door B in 2/3 of cases in which you choose Door A and Monty opened Door C.”

The figure 2.3 represent the DAG for the basic collider example; two independent causes with a common effect.

Now, the terms $P(Z=C|X=A,Y=y)$ reveals the dependencies, that are induced by the rules of the game. Indeed the three probabilities are different: $1/2$ if $Y=A$; $1$ if $Y=B$, $0$ if $Y=C$). The host cannot open the door chosen by the player, and the last two probabilities are true because the host cannot open the door that hides the car too; so his chosen is forced. The first probability is true because if $Y=A$ the host choice remain random ($C$ or $B$).

The same fact is expressed also in the term $P(Y=C|X=A,Z=C)=0$, that justify the advantage for the player to switch from the door $A$ to door $B$.

Therefore, about questions:

However this fact seems me in contrast with the main argument of causal inference. It is so? If yes, because it seems work? What is the solution proposed in causal inference?

There is no contradiction, the problem can be solved with standard statistical tools but causal tools can be used too. The point is that the rules of the game imply some dependencies, but the rules can be encoded in the SCM too, and in particular in the basic collider DAG above. Indeed the DAG imply some dependencies in the joint distribution $F(X,Y,Z)$. The SCM is not mandatory here but it can be used because causality play his role here. Moreover, as always, another workable point of view can help to understand what happen, and here the SCM does it.

As example, If the host don't know where the car is and, then, can exclude (more than open) the door with the car, we have that


then the advantage of switching disappear

$P(Y=A|X=A,Z=C)= P(Y=B|X=A,Z=C)=1/3$ (and $P(Y=C|X=A,Z=C)=1/3$ too)

because the previous dependencies disappear, and the DAG above not hold yet.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Sycorax
    Jan 5, 2021 at 14:43
  • $\begingroup$ Copy from the chat (which makes the latex unreadable): normally you have (for three doors) $$P(X=x, Y=y) = P(X=x) \cdot P(Y=y)=\frac{1}{9}$$ and $$P(X = Y) = \frac{1}{3}$$ but conditional on $Z$ you have $$P(X=x, Y=y | Z=z) = \begin{cases} \frac{2}{8} &\quad \text{if} \quad z \neq y \text{ and } z\neq x\text{ and } y\neq x \\ \frac{1}{8} &\quad \text{if} \quad z \neq y \text{ and } z\neq z \text{ and } y=x \\ 0 &\quad \text{else} \end{cases}$$ $\endgroup$ Jan 5, 2021 at 15:40
  • $\begingroup$ Relating to your latest edit. The labels $A,B,C$ seem to be just the different doors. Given $X=A$ you will either have $Z=B$ or $Z=C$ (with equal probability). Given that the contestant chooses door $A$ the door which the quizmaster opens is $B$ or $C$. $\endgroup$ Jan 6, 2021 at 16:30
  • $\begingroup$ "but the fact that it can involve causal theory and therefore can be made in SCM framework" You can apply the framework to it and draw DAGs, formulate SEMs, and apply $do$-calculus. But it is not something new and not something that classical statistics can not do (in fact in the end you need classical statistics to solve the problem)... $\endgroup$ Jan 6, 2021 at 16:38
  • $\begingroup$ ... In the Q the issue is stated as "standard statistical tools are adequate for passive observation but not for intervention". This example is false in two ways. 1 Classical statistical tools are not inadequate in the sense that they are wrong. In fact, they are needed for causal inference (but indeed 'intervention' is nicely formulated with graphs and causal models and classical statistical tools are inadequate in that sense). 2 The intervention by the quizmaster that occurs in the Monty Hall problem is not intervention as in causal inference and is not like the $do$-operator. $\endgroup$ Jan 6, 2021 at 16:40

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