Mdthbs

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.

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).

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. We don't want him to give away the game entirely though, so we'll change things up slightly and say that he doesn't actually open any door (because we don't want him to accidentally reveal the car - at least if I understand your question correctly), he just points at one and tells you he's pretty sure that one has a goat behind it. So, 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: $p(C'=x|C=x) = q = 1-p(C'=x|Cneq x)$. 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:
$$p(M|C=3) = sum_x p(M|C'=x)p(C'=x|C=3)$$
$$= p(M|C'=1)p(C'=1|C=3) + p(M|C'=2)p(C'=2|C=3) + p(M|C'=3)p(C'=3|C=3)$$
$$= frac{1}{2} times frac{1}{2}(1-q) + 0times frac{1}{2}(1-q) + 1 times q$$
$$= frac{1}{4} - frac{q}{4} + q = frac{3}{4}q+frac{1}{4}$$

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

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

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 Morty'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.)