Okay, I'm very embarrassed that there are already 71 questions (based on search of "monty hall") and I'm going to post another one. I read the first 5 before succumbing to choice-overload. I'll try to keep this short and sweet.
A host and contestant stand before 3 doors. The host advises the contestant that behind 1 of the 3 is a car while the other doors each have a goat.
The host advises the contestant to choose 2 of the 3 doors to reveal if either has the car.
The contestant chooses door
The host advises that behind one of the doors chosen is a goat and asks if the contestant wants to keep doors
3or switch to only revealing door
Big Question: Is the probability of revealing the car higher if the contestant switches or stays with the original choice?
As I understand the original problem, the above has the same result, so the contestant should switch, but I can't wrap my head around the math and don't want to hurt my brain trying if I'm incorrect about the above fundamentally being the same scenario.
Also, if the above is the same, how is it any different from the contestant saying "3, no wait 2", since no matter which 2 doors are chosen (either by the contestant alone or with the help of the host, as in the original problem), we know that at least 1 door has a goat?
Last bit: If this is the same mathematical scenario, is it even less intuitive than the original or does it help clarify (to someone other than me) why the original works?
Original MH problem, simplified:
There are 3 marbles in a bag; 2 are boring and grey, 1 is green. The host asks you to reach in and pull 1 out but not look at it. After doing this, the host, who can look into the bag, pulls out 1 grey marble. He then asks if you want to keep the 1 in your clutched hand or take the 1 still in the bag.
My version, simplified:
There are 3 marbles in a bag; 2 are boring and grey, 1 is green. The host asks you to reach in and pull 2 out but not look at them. After doing this, the host asks if you want to keep the 2 in your clutched hand or take the 1 still in the bag.
In both scenarios, 2 marbles are removed from the bag and 1 of those 2 is definitely grey. If we accept (and we all should at this point!) that in the first scenario the probability of the remaining door or marble being the winning choice is 2/3, shouldn't that hold true in the second scenario? If not, please explain at what point it diverges? If we know 1 of the 2 "out of the bag" is grey or a goat in either scenario, it shouldn't matter if we see which of the 2 it is, right?
Thanks to Eric T for helping me get my head around this. With either of my modified scenarios, where my logic diverged was I allow the contestant to choose 2 doors and then keep both choices or switch, whereas in the original MH problem, the contestant is given a second "choice" with the host-reveal but still only keeps the original (or switches). One of my goals in creating this alternative was to eliminate the host variable which is a clear source of confusion (and trickery) in the original, leading to such misassumptions as
The host's knowledge of what is behind all three doors creates a mathematical bias since he won't ever choose the car at random. If MH only presents the option when the car wasn't selected (to throw the contestant off), this would not change the math when testing for when the contestant chooses the car first. If the host always chooses a goat, it's because he always has at least one goat to choose from and is supposed to reveal a goat, not choose a door at random.
Showing the contestant that one of the other 2 doors has a goat gives the contestant new information that affects the outcome. it is not eliminating the goat (seeing the goat) that makes switching more likely to reveal the car, it is eliminating the door.
If my variation has the pick-2 parameter but has the "only one door allowed" rule reinstated , switching is still the better option. Here is the final version:
A host and contestant stand before 3 doors. The host advises the contestant that 1 of the 3 doors hides a car while the other doors each hide a goat.
The host advises the contestant to choose 2 of the 3 doors to check for the car.
The contestant chooses door
The host asks the contestant to choose between opening
3or switch to opening door
In this scenario, the host has done nothing to interfere and the contestant knows that at least one of the two selected doors has a goat, but must still risk choosing the wrong door of his selected 2 or switching. While the odds may still seem 1/2 at first, the possibility that the contestant chose both goats and thus may have no chance with his current subset makes the better odds in switching clearer.
Last question: What would be the actual probability of choosing the car if we don't know if the car exists in the subset 1 or 0 times? Just a comment mentioning a concept or wiki page is fine. Just curious on the math and have no idea what search for.