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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! ;-)

gung - Reinstate Monica
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Pseudoego
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    If Monty's accuracy is less than 100%, does that mean he sometimes opens the door with the prize behind it? If so, you should probably pick that door. – Fax Jul 25 '19 at 12:12

4 Answers4

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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.

Ruben van Bergen
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  • Wouldn't the expected value actually go back up when q < 1/3, because it isn't modeling how likely he is to be accurate, it is modeling how likely he is to be wrong? As it approached 0 it would mean he is always lying if he can, and your expected winnings would return to 2/3 – Cireo Jul 24 '19 at 07:13
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    @Cireo He wouldn't be lying, he would simply be wrong. Lying would imply him knowing his answer was wrong. I suspect the reason the expected value doesn't go back up is because the chance that he accidentally points to the door with the car behind it (ie, p(M|C=2) is going up) and you can't pick that door, no matter what). q=0 means he always misremembers where the car is, ie there's now a relatively high chance of him pointing to the door with the car behind it. – Buurman Jul 24 '19 at 08:04
  • I'd be interested to see how this changes if you can open the door that Monty points to. As @Cireo points out, for q<1/3, you should switch to the door that Monty points to, and an always-wrong Monty is as good as an always-right one, – Nuclear Hoagie Jul 24 '19 at 12:37
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    A more general solution (which this obviously needs) includes a "hostile" Monty; one who changes what he points at (or even if he points at something) depending on if you picked a goat or a car. – Yakk Jul 24 '19 at 13:22
  • @Cireo, Buurman & Nuclear Wang: see edits. – Ruben van Bergen Jul 24 '19 at 14:00
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    @Yakk: There are infinitely many more scenarios you can imagine that change the odds in infinitely many ways. It also all depends on if you know how Monty operates. If you know he's hostile, then actually he cannot decrease your odds below 1/3, because you would just decide to ignore whatever he does. If you don't know his decision process, then it totally depends on what you do assume and what he does do exactly, and there are many degrees of freedom there. – Ruben van Bergen Jul 24 '19 at 14:12
  • @RubenvanBergen Yes, you have to solve it analytically; you end up with something like a Monty field theory. ;) – Yakk Jul 24 '19 at 14:15
  • In your analysis of lying Monty, shouldn't he still be saying "I'm pretty sure it's a goat behind that door", and thus q=0 would imply 100% chance of the car being behind the indicated door? – Kalev Maricq Jul 24 '19 at 14:29
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    @KalevMaricq: I didn't really talk about lying Monty. The problem with that is the car could be behind the door you initially picked, which Monty isn't allowed to choose (otherwise, I would argue it's no longer the Monty Hall problem). So he may only have two goat-doors he can choose from, in which case he cannot lie by saying there's a goat behind one of them. So I don't think it's possible to construct a true "liar Monty" within the confines of the problem. What I went with instead (for $q=0$) is a Monty who always mistakes a goat-door for the car-door, but we don't know which goat-door. – Ruben van Bergen Jul 24 '19 at 14:54
  • @KalevMaricq: Also note that if Monty reliably points towards the door with the car and tells you it has a goat, the game becomes trivial and you can always win by choosing Monty's door (or always lose if you're not allowed to). So I stuck with a fallible Monty that seemed more interesting (and I think is what the OP was after). But as I said in an earlier comment, there are many more interesting variations you can think of. – Ruben van Bergen Jul 24 '19 at 15:00
  • Good point. I guess it would change the problem slightly to allow Monty to pick your door. So, is your setup that Monty randomly picks one of the two other doors and makes the true claim with probability q or the false claim with probability 1-q, about that door? – Kalev Maricq Jul 24 '19 at 15:06
  • @KalevMaricq: There may be different ways to think about it that agreee with the maths, but one way is to assume that at the beginning of a round in the game, a biased coin is flipped that determines whether Monty receives accurate information about where the car is located. With probability $q$, it lands on "accurate", and with $1-q$, it lands on "inaccurate". If the information is to be inaccurate, another (fair) coin is flipped that determines which goat-door Monty will mistakenly think has the car. Monty then makes his usual decisions, based on where he believes the car to be. – Ruben van Bergen Jul 24 '19 at 15:37
  • 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? – Pseudoego Jul 24 '19 at 16:23
  • @Pseudoego: Those are pretty much the assumptions I used in my calculations (esp. Morty being just fallible). However, note that for Morty to be wrong $\frac{1}{3}$ of the time when he says "this door has a goat", he would actually have to be just randomly guessing, because somebody who points randomly at a door has a 2 in 3 chance of pointing at a goat. So this corresponds to $q=\frac{1}{3}$ in my equations, in which case there is no benefit to switching. – Ruben van Bergen Jul 24 '19 at 17:46
  • @RubenvanBergen: So, even with q=0 there is a 50% chance he gives you accurate info? Because 1/3 of the time, you pick the car and he tells you correctly that one of the other doors is a goat, and 1/6 of the time you pick a goat, he believes you picked the car, and he accidentally picks the other goat to tell you is a goat. – Kalev Maricq Jul 24 '19 at 18:14
  • @KalevMaricq: Yes. In other words: when $q=0$, if we ask Monty to identify the door with the car, he will always pick the wrong door. But if we ask him to pick a door with a goat, he will still get it right 50% of the time. Since two doors have the same thing behind them, it is impossible for Morty to be misinformed about all of them. If one of the goats were replaced by (e.g.) a sheep, then we could define things such that when $q=0$, Morty would be mistaken about every item (e.g. mistake car for sheep, sheep for goat, and goat for car). – Ruben van Bergen Jul 24 '19 at 18:35
  • "If q is 0, Monty is always wrong" You've been talking to my wife? – Monty Harder Jul 24 '19 at 19:08
  • @RubenvanBergen If the hostile Monty knows that you know that Monty is hostile, his optimal strategy would include pointing at the actual goat with some probability. If Monty tells the truth (points at a random goat) with probability 0.5, and points at the car (if it's behind one of the unchosen doors) with probability 0.5, then he provides no additional information to the player, and staying, switching to the door Monty pointed at, and switching to the other door all have probability 1/3 of being the car. Any other probability of telling the truth permits a player strategy with higher payout. – Ray Jul 24 '19 at 19:24
  • Having trouble with your reasoning: "However, note that for Morty to be wrong 1/3 of the time when he says "this door has a goat", he would actually have to be just randomly guessing, because somebody who points randomly at a door has a 2 in 3 chance of pointing at a goat. " Your assertion that Monty will get it right 50% of the time is wrong. I specified that research shows that Monty is correct that there is NO CAR roughly 2/3 of the time. He does not actually KNOW where the car is any more. He is predicting where it is NOT. But I can see that the Monty Analogy is breaking down now. Sorry – Pseudoego Jul 24 '19 at 19:25
  • @Ray: Exactly. As long as the player knows Monty's decision policy, there is nothing Monty can do that will cause the player to win less than 1/3 of the time (on average). Monty's actions either provide information or not, and any information they provide increases the odds of winning. It's only when the player makes incorrect assumptions about Monty's behavior that Monty can (reliably) misdirect the player (but the space of possible incorrect assumptions is infinite, so I didn't go there in my answer). – Ruben van Bergen Jul 24 '19 at 19:43
  • @Pseudoego: Let me first clarify that I didn't say that Monty would get it right 50% of the time in your scenario (I mentioned a 50% figure in a different context, but not in reply to you). Then to explain my reasoning: you asked me to assume Monty is correct that there is no car 2/3 of the time. Consider what would happen if Monty just pointed at a random door and said that door had no car. Since 2 out of 3 doors have no car, he would be correct 2/3 of the time. So, correctly predicting "no car" 2/3 of the time is the same as random guessing, i.e. Monty having no actual information. – Ruben van Bergen Jul 24 '19 at 19:52
  • Well, YOU certainly have some actual information that I can see having utility, @Ruben van Bergen . If Monty is only going to be right about the door being empty 2/3 of the time, then that boils down to the original "state" of the game, with each door having only 1/3 P of being correct. So, yes, indeed, no point switching unless Monty can up his accuracy. The question THEN is : How accurate does he need to be in order to justify a switch? Anything above 66.6%, I assume. I asked for this information so that I could understand a different - but analogous - game. Thanks for helping out. – Pseudoego Jul 24 '19 at 20:09
  • @Pseudoego: Yes, if you quantify accuracy in those terms, then Monty needs to be better than 2/3 to justify a switch (this is equivalent to $q>1/3$ in my calculations, as I defined accuracy w.r.t. him knowing where the car is). – Ruben van Bergen Jul 24 '19 at 21:06
  • @RubenvanBergen I do not think that your version of the problem agrees with the OPs. As I understand it, your Monty will choose to reveal information about the contestant's chosen door sometimes; in a comment the OP has said that Monty and the contestant chooses different doors. – Alex Jul 25 '19 at 01:44
  • @Alex: No, my Monty will never reveal anything about the contestant's chosen door. What made you think that? Note that you can see in my calculations that $p(C=1|M)$ stays constant at $\frac{1}{3}$, indicating that Monty's actions do not influence the probability that the contestant's door has the car. Only the probabilities for the other doors change. – Ruben van Bergen Jul 25 '19 at 01:55
  • @RubenvanBergen you wrote: "so we'll change things up slightly and say that he doesn't actually open any door ..., he just points at one and tells you he's pretty sure that one has a goat behind it." – Alex Jul 25 '19 at 01:58
  • @Alex: I see, I'll change that sentence to be more clear. (What I meant with that is just that if Monty will sometimes be wrong, we don't want him to accidentally open the door with the car behind it, because that changes the game entirely. But he still only chooses between doors that the contestant didn't initially choose, and we still refer to the door that he chooses as door number 2.) – Ruben van Bergen Jul 25 '19 at 16:06
  • The question seems to have a parameter $p$ that is the probability that the door Monty points to has a goat. Your answer has a parameter $q$ where Monty has probability $q$ of correctly knowing where the car is and choosing a door based on that knowledge, and probability $1-q$ of thinking that one of the doors with a goat has a car, and having a 50% chance for each goat of thinking that that is a car, then choosing a door to point to based on that belief. – Acccumulation Jul 25 '19 at 19:31
  • I'm not clear on whether you misunderstood the question wording, or whether you felt that the original parameterization was unwieldy and deliberately decided to introduce a different parameterization, and I don't think that your answer clearly communicates that you're reparameterizing the question. – Acccumulation Jul 25 '19 at 19:31
  • Ok, I see that now, reading on, how does this line work? $p(M|C=3) = \sum_x p(M|C'=x)p(C'=x|C=3)$ – Alex Jul 26 '19 at 05:09
  • @Alex: This computing the probability of what we observed (Monty pointing towards door number 2), supposing that the car is located behind door number 3. To compute this, we have to sum over all possible beliefs that Monty could have when the car is behind door 3 (he could believe it to be behind 1, 2, or 3), and the probability of him pointing to door 2 given each of these beliefs. – Ruben van Bergen Jul 26 '19 at 16:26
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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).

Ben
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    I appreciate the response, but I was looking for something more specific. I'm specifying that Monty has picked a Door. I'm specifying that the probability of the Prize being behind that door could be anywhere from Zero to 100%. I was hoping for a formula which would allow me to simply enter the Probability that Monty is Right/Wrong and then working out the rest of the formula would provide a Numerical Estimate which indicates the Probability that Switching will result in a Win. Is that degree of assistance an unrealistic request? – Pseudoego Jul 24 '19 at 03:41
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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.)

Alex
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  • Alex and @Ruben van Bergen et al Thanks for the helpful details. Assume Monty's never hostile, just fallible and tells you "I am pretty sure that the the Car is NOT behind this door." but does not open the door. Let's assume Research shows he is WRONG only about 33.3% of the time, thus correct 66.6% (a Posterior Probability?). There is still some benefit to switching, but once his accuracy reaches only 33.3% then it would make NO sense to switch to either HIS door or the other one. Literally a case of "your guess is as good as mine." Does any of this change your Analyses or Formulae? – Pseudoego Jul 24 '19 at 16:04
  • No, this doesn't change my analysis. I added something which I hope clarifies the question in your comment. Btw, I wouldn't read too much into the words "hostile", "fallible", "monty lies". These don't really mean anything unless defined with precision as the (conditional) probability that Monty is wrong about a door containing a goat. – Alex Jul 25 '19 at 01:18
  • Pretty annoyed that my OWN answer to MY OWN question would be deleted with the only explanation given is that this site is not for "discussion" - when I am mainly explaining why I think the Answers given so far are Correct, and explaining how they will be useful. There was far more discussion in most of the other answers given. This seems myopic to me - at best - and moronic - at worst - to delete somebody's answer to their own question: how can you possibly explain WHY you are rating an Answer as the BEST without discussing it? Thanks to all who replied regardless. – Pseudoego Jul 25 '19 at 02:47
  • @Pseudoego your last comment is better post as a comment on your original question. I didn't see your answer, but it sounds like you want to discuss the existing answers, in which case you can modify your original question. – Alex Jul 25 '19 at 03:07
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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.

Pseudoego
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