I know that the Monty Hall problem is old news. However, I can't get this out of my head and I am looking forward to your thoughts about it.
If the rules of the game are known from the beginning, especially that the game host will guarantee to open a door containing a goat, isn't this information extremely relevant when calculating the initial probabilities of my chances to select the door with the car behind it? This aspect seems to be insufficiently considered in explanations stating that switching doors is statistically the better option - which of course I can accept as it has been discussed by far more intelligent people than me ;)
My line of thought is as follows: if I know from the start that the host will definitely open a door that contains a goat with 100% certainty, then I can ignore that door from the beginning (no matter which of the two remaining doors it ultimately is). Thus, from the start, I only have two doors to choose from and I end up with the seemingly incorrect 50:50 probability of winning the car.
The Wikipedia article uses the following pictures to explain the usual calculations where you initially have a 1/3 chance of selecting the car (I am not yet allowed to directly embed pictures):
Tree showing the probability of every possible outcome if the player initially picks Door 1.
From how I see it, this is not taking into account the information that it its guaranteed that the host will reveal the door with the second goat. So, from the two doors in the set of doors that I did not choose I can ignore one because the host knows that there is a goat behind it and it is guaranteed that it will be eliminated.
For a better understanding of the problem, the scenario is often expanded and I think my argument still holds: I choose from 1,000,000 doors instead of only 3. With the knowledge that, after my choice, 999,998 doors will be guaranteed to be opened, revealing a goat behind each, it is not really a choice among 1,000,000 doors from the start but only between 2 doors, isn't it?
[Edit after a round of comments] we want that car, so our first choice is between two sets: the first "set" with only one door and the second set of all other doors. What is the probability of the car being in one of two sets? Does the size of the sets matter? I'd say no because we don't care about the goats, they're all the same to us. We only care about the chance of the car being in our selected "set". Isn't that a 50:50 chance?
What do you think? :)
In the Comments you write:
You seem to be using some kind of Wagner Principle in your reasoning here. The Wagner Principle states that if an event can either happen or not, then there is a 50:50 chance of it happening or not happening.
The Wagner Principle is fallacious, because the probability of some event happening does not have to equal the probability of an event not happening. For example, the probability of rolling a $6$ with a die is not the same as the probability of not rolling a $6$.
You are saying: if you consider two sets, the first one being the set containing just the door that you picked, and the second one the set of all other doors, then there is a $50:50$ chance of the door with the car belonging to one set or the other.
This reasoning is just as fallacious. Just because I can split all possible outcomes into two sets does not mean that the probability of the outcome belonging to one set equals the probability of the outcome belonging to the other.
Yes, I can split the possible outcomes of rolling a die in two sets: the set of outcomes where I roll a $6$ versus the set of outcomes where you do not roll a $6$. But obviously the probability of the outcome of belonging to one set does not equal to the probability of belonging to the other.
Now, you say: But wait! Monty opens up a bunch of doors, revealing a goat behind all of them. Doesn't that make a difference as to the probability of the door that you initially picked to be the one with the car?
No, it does not. As you pointed out before, no matter whether you initially picked the door with the prize, Monty is certain to open a door with a goat that is a door other than the one you initially picked. So, the fact that he does this tells you absolutely nothing regarding your initial pick being the winner: the probability that your initial pick was the winner equals the probability that your initial pick was the winner given that Monty opened some other door and revealed that it had a goat.
If it helps to do this in terms of conditional probability: we know in general that
$$P(A|B)=\frac{P(A)}{P(B)}$$
Now, let's define $A$ to be the event of your initial pick being the winner, and $B$ be the event of Monty Hall opening a door that is other than initial pick and that has a goat.
Clearly, we have:
$$P(A) = \frac{1}{3}$$
And since Monty is certain to open a door that is other than initial pick and that has a goat, we have:
$$P(B)=1$$
and so we have:
$$P(A|B) = \frac{P(A)}{P(B)} = \frac{P(A)}{1} =P(A)=\frac{1}{3}$$
That is, given that the probability of your initial pick being the right one is 1 out of 3, it is still 1 out of 3 after Monty has opened his door.
OK, but why doesn't the same logic apply to the leftover door (the door that you didn't pick and that Monty didn't open either?) Why does its probability of containing the prize suddenly increase to 2 out of 3?
Well, that's because Monty could have opened that very door. That is, after you pick door 1, Monty could pick door 2 or door 3: both are still live options as far as your knowledge of the situation is concerned, because as far as you know, both could still be a goat or a car. But when we then see Monty pick door 3 and reveal it to have a goat, we now have learned something about both doors: the probability of door 3 containing the car has now gone down from 1 in 3 to 0 ... and for door 2: given that Monty decided to open up door 3 is suggestive of Monty intentionally skipping door 2: it is quite possible that Monty skipped door 2 exactly because door 2 had the prize.
Indeed, given that there was a 2 out of 3 chance that one of the doors other than your initial pick contained the prize, and given that Monty was certain to open a leftover door with a goat, there is still a 2 out of 3 chance that one of the leftover doors has the car, But given that there is now only one leftover door, that door now has that 2 out of 3 chance of being the winner.
Again, formally:
If $A$ is the event of one of the doors other than your initial pick to be the winner, and $B$ the event of Monty opening up a door other than your initial pick and that door containing a goat, we have:
$$P(A)=\frac{2}{3}$$
but again $B$ is certain to occur, and so we have:
$$P(B)=1$$
and so:
$$P(A|B)=\frac{P(A)}{P(B)} = \frac{P(A)}{1} = P(A) = \frac{2}{3}$$
But wait! If all these conditional (a posteriori) probabilities remain exactly the same as the initial (a priori), doesn't that mean that since door 2 had an initial probability of 1 out of 3 of being the winner, it would have to have an a posteriori probability of 1 out of 3 if being the winner as well?
Well, yes! After initially picking door 1, and Monty opening some door other than door 1 and revealing it to have a goat means that the probability of door 2 having the prize remains 1 out of 3:
If $A$ is the event of door 2 being the winner, and $B$ being the event of Monty opening up some door other than your initial pick (door 1) containing a goat (which is certain to happen), then yes, again we have:
$$P(A|B)=\frac{P(A)}{P(B)} =\frac{P(A)}{1}=P(A)=\frac{1}{3}$$
So note: the certainty that Monty is 100% guaranteed to reveal a door with a prize does matter … but only in so far as that means that various a priori probabilities do not change!
Indeed, knowing that Monty opens some door other than your initial pick with a goat tells us nothing. This would be like you picking door 1, then leaving the room, coming back, and being told that Monty in the meantime opened either door 2 or 3 and showed it had a goat ... but you are not told which one.
But in the case that you see him open door 3 to reveal a goat, we know the specific door that Monty opens that has a goat. And as I argued, that raises the probability that door 2 has the car.
In terms of conditional probability: If $A$ is the event of the car being behind door 2, and $B$ being the event of Monty opening up door 3 after your initial pick of door 1, we no longer have that $P(B)=1$, because we were not certain that Monty would specifically open door 3: maybe the prize is behind door 3, in which case Monty cannot open door 3, and even if door 3 does not have the car, Monty could still open up door 2, in case your initial door was the one with the car.
Indeed, given the symmetry of the situation between doors 2 and 3, after you picking door 1, Monty is just as likely to open up door 2 as door 3, meaning that:
$$P(B)=\frac{1}{2}$$
and therefore:
$$P(A|B)=\frac{P(A)}{P(B)}=\frac{P(A)}{\frac{1}{2}}=2 \times P(A) = 2 \times \frac{1}{3} = \frac{2}{3}$$
A final comment: let's go back to that scenario with $1,000,000$ doors. Now, if you picked door $1$, and then you see Monty opening doors $2$, $3$, $4$, .... all the way through $317,652$ ... but then he skips door $317,653$ ... and proceeds to open up doors $317,654$ through $1,000,000$ ... Isn't it totally intuitive now that most likely Monty skipped door $317,653$ exactly because that was the one with the prize? I mean, the only alternative would be that door 1 was the winner, and Monty just randomly had to skip one of the doors, but how likely is that?! So, yes, those are two possibilities, but in probability, I hope you see that they are wildly different. And indeed, if you do that math, you will find that there is (of course!) a $1$ in a million chance that you initially picked door was the winner immediately, but whatever door Monty skips has a $999,999$ our of $1,000,000$ chance of being the winner.