In post 18, shaft.ed wrote:Might be the wrong way of thinking about it, but what about a simpler explanation.
When you picked your door it had a 1/3 chance of being correct. Yes Monty opens a goat door, but that goat door doesn't change your odds because he can always open a goat door after your pick. Thus that doors odds remain 1 in 3.
However, for the switch door, you have new information. The goat door being opened does impact that door's odds since you are selecting it after the reveal. So it is a 1 in 2 chance of being the correct door.
We--eeell, this is the potential statistical artefact I'm thinking about.
I'm just trying to figure out if there's any result that changes based on the reveal - and it sort of seems that it does. Because from my own perspective I don't know if my pick fell into the 33% or the 66% bracket. Monty revealing a door reveals exactly one thing: of the two doors Monty could have picked, he picked -that door-. If he had both doors to choose from, then he took a 50/50 himself, otherwise he was forced. And he's forced 33% of the time (because 33% of the time, I picked correctly). That leaves a 66% chance that Monty picked something inconsequential. Granted, I no longer get to choose the option that Monty picked after he reveals - but then, he couldn't choose out of 3 either if I picked first. (So Monty gets information about the door I picked, although he always has the option to open a door with a goat because he knows which one not to open, if it exists between his choices.) Monty's expected choices are more logic-gate-like, as in 0|0, 0|1, 1|0. Perhaps there's some symmetry there?
I was just thinking about it in a different way - in terms of order of operations.
Let's say you are told that there are 3 options you will be able to choose from, but Monty reveals a goat *first*.
Monty can only pick from 2 of the 3, because one is a car. Monty is *forced* into 50/50 decision which is reduced by the fact he can only select from 66% of the options if he selects first.
This is definitely information because Monty *cannot* pick the car door. However, it doesn't give information on the remaining choices' probability for success because either of the remaining options has nothing qualifying it.
The information revealed amounts to "Monty picked door [x]" which you already know and provides nothing extra.
So...
You ignore that information and pick randomly of the 3 doors.
Then you look at whether you picked Monty's goat (which is no longer an option) and pick out of the remaining two doors if you did. If you did pick Monty's door, it seems like the other two options have the same probability of success because Monty was constrained similarly to you and your own decision is null. If you didn't - you're left with only one option for a switch (literally the "switch"), one of which Monty didn't pick.
You know that there are 3 states, some of which are already collapsed because you know what Monty picked, and you know if you picked the same.
You did not pick what Monty picked and Monty picked (A) door (goat) | can switch
You did not pick what Monty picked and Monty picked (B) door (goat) | can switch
You picked what Monty picked and it was a goat | must switch (to A or B)
Yeah I dunno, I'm searching for the extra information here and I'm not finding it. Something either mechanical or concrete that links Monty's choice to my own choice without resorting to analogies like "Imagine Monty had 999 doors and told you that 998 of them were wrong".
It reminds me of the whole "where did the extra dollar come from" puzzle that I read years ago:
https://math.stackexchange.com/questions/15524/riddle-simple-arithmetic-problem-illusion