Monty Hall Problem with a Fallible Monty

Question

Monty had perfect knowledge of whether the Door had a goat behind it (or was empty). This fact allows Player to Double his success rate over time by switching “guesses” to the other Door. What if Monty’s knowledge was less than perfect? What if sometimes the Prize truly WAS in the same Doorway as the Goat? But you could not see it until after you chose and opened YOUR door? Can you please help me to understand how to calculate IF— and by how much — Player can improve his success when Monty’s accuracy rate is less than 100%? For example: what if Monty is wrong — on Average-50% of the time? Can the Player STILL benefit from switching his Guess/Door? I imagine that if Monty has less than 33.3% chance of being correct that Prize is NOT behind the Door, then Player's best option is to NOT Switch his Door choice. Can you please provide me with a way to calculate the potential benefit of switching by inserting different Probabilities of Monty being Correct about the Prize NOT being behind the Door? I have nothing beyond High School math, and am 69 years old, so please be gentle.

---

Thanks for the insights and formulae provided. It appears to be that if "Fallible Monty" is only 66% accurate in predicting the absence of a Prize/Car that there is ZERO benefit to switching from your original choice of doors....because his 33% error rate is the default base rate for the Prize being behind ANY door. One assumes, though, that IF Monty gets better than 66% at predicting where there is NO PRIZE THEN switching derives greater Utility. I will be trying to apply this reasoning to a game where an "Expert" makes an "expert prediction" that one of three roughly equally probable options will be the correct one. I have little faith in the Expert being correct, and I am quite certain that his "hit rate" will be less than 33% - more like 15%. My conclusion from this will be that when the "Expert" chooses a different option from my choice, that it will be beneficial for me to switch my choice to the other option. And, when the Expert chooses the same option as me, I am probably wrong for sure, and should change to one of the other two! ;-)

Answer 1

Let's start with the regular Monty Hall problem. Three doors, behind one of which is a car. The other two have goats behind them. You pick door number 1 and Monty opens door number 2 to show you there is a goat behind that one. Should you switch your guess to door number 3? (Note that the numbers we use to refer to each door don't matter here. We could choose any order and the problem is the same, so to simplify things we can just use this numbering.)

The answer of course is yes, as you already know, but let's go through the calculations to see how they change later. Let $C$ be the index of the door with the car and $M$ denote the event that Monty revealed that door 2 has a goat. We need to calculate $p(C=3|M)$. If this is larger than $1/2$, we need to switch our guess to that door (since we only have two remaining options). This probability is given by: $$ p(C=3|M)=\frac{p(M|C=3)}{p(M|C=1)+p(M|C=2)+p(M|C=3)} $$ (This is just applying Bayes' rule with a flat prior on $C$.) $p(M|C=3)$ equals 1: if the car is behind door number 3 then Monty had no choice but to open door number 2 as he did. $p(M|C=1)$ equals $1/2$: if the car is behind door 1, then Monty had a choice of opening either one of the remaining doors, 2 or 3. $p(M|C=2)$ equals 0, because Monty never opens the door that he knows has the car. Filling in these numbers, we get: $$ p(C=3|M)=\frac{1}{0.5+0+1}=\frac{2}{3} $$ Which is the result we're familiar with.

Now let's consider the case where Monty doesn't have perfect knowledge of which door has the car. So, when he chooses his door (which we'll keep referring to as door number 2), he might accidentally choose the one with the car, because he thinks it has a goat. Let $C'$ be the door that Monty thinks has the car, and let $p(C'|C)$ be the probability that he thinks the car is in a certain place, conditional on its actual location. We'll assume that this is described by a single parameter $q$ that determines his accuracy, such that: $p(C'=x|C=x) = q = 1-p(C'\neq x|C=x)$. If $q$ equals 1, Monty is always right. If $q$ is 0, Monty is always wrong (which is still informative). If $q$ is $1/3$, Monty's information is no better than random guessing.

This means that we now have: $$p(M|C=3) = \sum_x p(M|C'=x)p(C'=x|C=3)$$ $$= p(M|C'=1)p(C'=1|C=3) + p(M|C'=2)p(C'=2|C=3) + p(M|C'=3)p(C'=3|C=3)$$ $$= \frac{1}{2} \times \frac{1}{2}(1-q) + 0\times \frac{1}{2}(1-q) + 1 \times q$$ $$= \frac{1}{4} - \frac{q}{4} + q = \frac{3}{4}q+\frac{1}{4}$$

That is, if the car was truly behind door 3, there were three possibilities that could have played out: (1) Monty thought it was behind 1, (2) Monty thought 2 or (3) Monty thought 3. The last option occurs with probability $q$ (how often he gets it right), the other two split the probability that he gets it wrong $(1-q)$ between them. Then, given each scenario, what's the probability that he would have chosen to point at door number 2, as he did? If he thought the car was behind 1, that probability was 1 in 2, as he could have chosen 2 or 3. If he thought it was behind 2, he would have never chosen to point at 2. If he thought it was behind 3, he would always have chosen 2.

We can similarly work out the remaining probabilities: $$p(M|C=1) = \sum_x p(M|C'=x)p(C'=x|C=1)$$ $$=\frac{1}{2}\times q + 1\times \frac{1}{2}(1-q)$$ $$=\frac{q}{2}+\frac{1}{2}-\frac{q}{2}=\frac{1}{2}$$

$$p(M|C=2) = \sum_x p(M|C'=x)p(C'=x|C=2)$$ $$=\frac{1}{2}\times\frac{1}{2}(1-q) + 1 \times\frac{1}{2}(1-q)$$ $$=\frac{3}{4}-\frac{3}{4}q$$

Filling this all in, we get: $$ p(C=3|M)=\frac{\frac{3}{4}q+\frac{1}{4}}{\frac{1}{2}+\frac{3}{4}-\frac{3}{4}q+\frac{3}{4}q+\frac{1}{4}} $$ $$ =\frac{0.75q+0.25}{1.5} $$ As a sanity check, when $q=1$, we can see that we get back our original answer of $\frac{1}{1.5}=\frac{2}{3}$.

So, when should we switch? I'll assume for simplicity that we're not allowed to switch to the door Monty pointed to. And in fact, as long as Monty is at least somewhat likely to be correct (more so than random guessing), the door he points to will always be less likely than the others to have the car, so this isn't a viable option for us anyway. So we only need to consider the probabilities of doors 1 and 3. But whereas it used to be impossible for the car to be behind door 2, this option now has non-zero probability, and so it's no longer the case that we should switch when $p(C=3|M)>0.5$, but rather we should switch when $p(C=3|M)>p(C=1|M)$ (which used to be the same thing). This probability is given by $p(C=1|M)=\frac{0.5}{1.5}=\frac{1}{3}$, same as in the original Monty Hall problem. (This makes sense since Monty can never point towards door 1, regardless of what's behind it, and so he cannot provide information about that door. Rather, when his accuracy drops below 100%, the effect is that some probability "leaks" towards door 2 actually having the car.) So, we need to find $q$ such that $p(C=3|M) > \frac{1}{3}$: $$\frac{0.75q+0.25}{1.5}>\frac{1}{3}$$ $$0.75q+0.25 > 0.5$$ $$0.75q > 0.25$$ $$q > \frac{1}{3}$$ So basically, this was a very long-winded way to find out that, as long as Monty's knowledge about the car's true location is better than a random guess, you should switch doors (which is actually kind of obvious, when you think about it). We can also calculate how much more likely we are to win when we switch, as a function of Monty's accuracy, as this is given by: $$\frac{p(C=3|M)}{p(C=1|M)}$$ $$=\frac{\frac{0.75q+0.25}{1.5}}{\frac{1}{3}}=1.5q+0.5$$ (Which, when $q=1$, gives an answer of 2, matching the fact that we double our chances of winning by switching doors in the original Monty Hall problem.)

Edit: People were asking about the scenario where we are allowed to switch to the door that Monty points to, which becomes advantageous when $q<\frac{1}{3}$, i.e. when Monty is a (somewhat) reliable "liar". In the most extreme scenario, when $q=0$, this means the door Monty thinks has the car actually for sure has a goat. Note, though, that the remaining two doors could still have either a car or a goat.

The benefit of switching to door 2 is given by: $$ \frac{p(C=2|M)}{p(C=1|M)} = \frac{ \frac{0.75 - 0.75q}{1.5} } { \frac{1}{3} } = 1.5-1.5q $$ Which is only larger than 1 (and thus worth switching to that door) if $1.5q<0.5$, i.e. if $q<\frac{1}{3}$, which we already established was the tipping point. Interestingly, the maximum possible benefit for switching to door 2, when $q=0$, is only 1.5, as compared to a doubling of your winning odds in the original Monty Hall problem (when $q=1$).

The general solution is given by combining these two switching strategies: when $q>\frac{1}{3}$, you always switch to door 3; otherwise, switch to door 2.

Answer 2

This should be a fairly simple variation of the problem (though I note your limited maths background, so I guess that is relative). I would suggest that you first try to determine the solution conditional on whether Monte is infallible, or fully fallible. The first case is just the ordinary Monte Hall problem, so no work required there. In the second case, you would treat the door he picks as being random over all the doors, including the door with the prize (i.e., he might still pick a door with no prize, but this is now random). If you can calculate the probability of a win in each of these cases then you an use the law of total probability to determine the relevant win probabilities in the case where Monte has some specified level of fallibility (specified by a probability that we is infallible versus fully fallible).

Answer 3

Based on the comments on Ben's answer, I am going to offer up two different interpretations of this variant of Monty Hall, differing to Ruben van Bergen's.

The first one I am going to call Liar Monty and the second one Unreliable Monty. In both versions the problem proceeds as follows:

(0) There are three doors, behind one of which is a car and behind the other two are goats, distributed randomly.

(1) Contestant chooses a door at random.

(2) Monty picks a door different to the contestant's door and claims a goat is behind it.

(3) Contestant is offered to switch to the third unpicked door, and the problem is "When should the contestant switch in order to maximise the probability of finding a car behind the door?"

In Liar Monty, at step (2), if the contestant has picked a door containing a goat, then Monty picks a door containing the car with some predefined probability (i.e. there is a chance between 0 and 100% that he will lie that a goat is behind some door). Note that in this variant, Monty never picks a door containing the car (i.e. cannot lie) if the contestant chose the car in step (1).

In Unreliable Monty, there is a predefined probability that the door Monty pick's in step (2) contains a car. I take from your comment on Ben's answer that this is the scenario you are interested in, and both of my versions differ from Ruben van Bergen's. Note that Unreliable Monty is not the same as Liar Monty; we will rigorously differentiate between these two cases later. But consider this, in this scenario, Monty's door can never contain the car more than $\frac{2}{3}$ of the time, since the contestant has a probability of choosing the car $\frac{1}{3}$ of the time.

To answer the problem, we are going to have to use some equations. I am going to try and phrase my answer so that it is accessible. The two things that I hope are not too confusing are algebraic manipulation of symbols, and conditional probability. For the former, we will use symbols to denote the following:

$$\begin{split} S &= \text{The car is behind the door the contestant can switch to.}\\ \bar{S} &= \text{The car is not behind the door the contestant can switch to.}\\ M &= \text{The car is behind the door Monty chose.}\\ \bar{M} &= \text{The car is not behind the door Monty chose.}\\ C &= \text{The car is behind the door the contestant chose in step (1).}\\ \bar{C} &= \text{The car is not behind the door the contestant chose in step (1).} \end{split} $$

We use $\Pr(*)$ to denote "the probability of $*$", so that, put together, something like $\Pr(\bar{M})$ means the probability that the car is not behind the door Monty chose. (I.e. wherever you see an expression involving the symbols, replace the symbols with the "English" equivalents.)

We will also require some rudimentary understanding of conditional probability, which is roughly the probability of something happening if you have knowledge of another related event. This probability will be represented here by expressions such as $\Pr(S|\bar{M})$. The vertical bar $|$ can be thought of as the expression "if you know", so that $\Pr(S|\bar{M})$ can be read as "the probability that the door the contestant can switch to has the car, if you know that the car is not behind Monty's door. In the original Monty Hall problem, $\Pr(S|\bar{M}) = \frac{2}{3}$, which is larger than $\Pr(S) = \frac{1}{3}$, which corresponds to the case when Monty has not given you any information.

I will now demonstrate that Unreliable Monty is equivalent to Liar Monty. In Liar Monty, we are given the quantity $\Pr(M|\bar{C})$, the probability that Monty will lie about his door, knowing that the contestant has not chosen the car. In Unreliable Monty, we are given the quantity $\Pr(M)$, the probability that Monty lies about his door. Using the definition of conditional probability $\Pr(M \text{ and } \bar{C}) = \Pr(\bar{C} | M) \Pr(M) = \Pr(M | \bar{C}) \Pr(\bar{C})$, and rearranging, we obtain:

$$ \begin{split} \Pr(M) &= \frac{\Pr(M | \bar{C}) \Pr(\bar{C})}{\Pr(\bar{C} | M)}\\ \frac{3}{2} \Pr(M) &= \Pr(M | \bar{C}), \end{split}$$ since $\Pr(\bar{C})$, the probability that the car is not behind the contestant's chosen door is $\frac{2}{3}$ and $\Pr(\bar{C} | M)$, the probability that the car is not behind the contestant's chosen door, if we know that it is behind Monty's door, is one.

Thus, we have shown the connection between Unreliable Monty (represented by LHS of the above equation) and Liar Monty (represented by the RHS). In the extreme case of Unreliable Monty, where Monty chooses a door that hides the car $\frac{2}{3}$ of the time, this is equivalent to Monty lying all the time in Liar Monty, if the contestant has picked a goat originally.

Having shown this, I will now provide enough information to answer the Liar version of the Monty Hall Problem. We want to calculate $\Pr(S)$. Using the law of total probability:

$$\begin{split} \Pr(S) &= \Pr(S|C)\Pr(C) + \Pr(S|\bar{C} \text{ and } M)\Pr(\bar{C} \text{ and } M) + \Pr(S|\bar{C} \text{ and } \bar{M})\Pr(\bar{C} \text{ and } \bar{M})\\ &= \Pr(\bar{C} \text{ and } \bar{M}) \end{split}$$ since $\Pr(S|C) = \Pr(S|\bar{C} \text{ and } M) = 0$ and $\Pr(S|\bar{C} \text{ and } \bar{M}) = 1$ (convince yourself of this!).

Continuing:

$$\begin{split} \Pr(S) &= \Pr(\bar{C} \text{ and } \bar{M})\\ &= \Pr(\bar{M} | \bar{C}) \Pr(\bar{C}) \\ &= \frac{2}{3} - \frac{2}{3}\Pr(M | \bar{C})) \end{split}$$

So you see, when Monty always lies (aka $\Pr(M | \bar{C})) = 1$) then you have a zero chance of winning if you always switch, and if he never lies then the probability the car is behind the door you can switch to, $\Pr(S)$, is $\frac{2}{3}$.

From this you can work out the optimal strategies for both Liar, and Unreliable Monty.

Addendum 1

In response to comment (emphasis mine):

"I added more details in my comment to @alex - Monty is never hostile nor devious, just FALLIBLE, as sometimes he can be wrong for whatever reasons, and never actually opens the door. Research shows that Monty is wrong roughly 33.3% of the time, and the car actually turns out to be there. That is a Posterior Probability of being correct 66.6% of the time, correct? Monty never chooses YOUR door, and you will never choose his. Do these assumptions change anything?"

This is as I understand, the Unreliable Monty Hall Problem introduced at the start of my answer.

Therefore, if Monty's door contains the car $\frac{1}{3}$ of the time, we have the probability of winning when you switch to the last unpicked door as:

$$ \begin{split} \Pr(S) &= \frac{2}{3} - \frac{2}{3}\Pr(M | \bar{C})\\ &= \frac{2}{3} - \frac{2}{3} \times \frac{3}{2}\Pr(M) \\ &= \frac{2}{3} - \frac{1}{3}\\ &= \frac{1}{3} \end{split}$$

Thus, there is no difference between switching, remaining with the original door or if allowed, switching to Monty's chosen door (in line with your intuition.)

Answer 4

For some reason, a moderator decided to delete my own answer to my own question, on the grounds that it contained "discussion." I don't really see HOW I can explain what is the Best Answer without discussing what makes it work for me, and how it can be applied in practice.

I appreciate the insights and formulae which were provided in the previous answers. It appears to be that IF "Fallible Monty" is only 66% accurate in predicting the absence of a Prize/Car THEN there is ZERO benefit to switching from your original choice of doors....because his 33% error rate is the default base rate for the Prize being behind ANY door. One assumes, though, that IF Monty gets better than 66% at predicting where there is NO PRIZE THEN switching derives greater Utility.