If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.
You are taking an empty result and replacing it with a strict win.
That said, though, if you're trying to convince someone of the correctness of the Monty Hall Problem, and they're stuck on that misconception, saying "well you may as well switch in case I'm right" isn't a great argument.
Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".
What happens?
123
ABC - player was right at first, so they lose if they switch
ACB - player was right at first, so they lose if they switch
BAC - player was wrong, Monty chooses 2 which is the car, so the game is ruined
BCA - player was wrong, Monty finds a goat, player wins if they switch.
CAB - player was wrong, Monty chooses 2 which is the car, so the game is ruined
CBA - player was wrong, Monty finds a goat, player wins if they switch.
I believed you were wrong, but now I believe you.
1/3: you picked car, switch fails
1/3: you picked goat, monty ruins. Not included
1/3: you picked goat, monty open goat. Switch wins.
Of the two non ruined scenarios, half win
We can also demonstrate by using the 100 door example. Monty opens 98 of the 99 doors you didn’t choose. 98 times out of 99, monty ruins the game.
So random choice dramatically changes things.
If Monty chooses randomly both strategies fail 2/3 of the time.
Whether Monty knows for sure or is just picking at random only affects whether games are ruined, not the probability that the door the contestant picked is a winner.
If there are N doors and the player picks door 1, there is a 1/N chance he will be correct. If he sticks to that door, there is nothing that will ever change that 1/N odds that it was the winner.
Downthread someone enumerated the cases, but again, they discard the 1/3 of the cases where Monty ruined it. What that does is throw out half of the cases where contestant picked the wrong door and had the wrong strategy. Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.
By that logic, so does the remaining door. 1/3 vs 1/3.
Its like saying soccer players could score more if they picked up the ball and ran with it. While true, you are no longer describing the game of soccer.
So I'm not getting the point you're making, unless you're saying that would be a more grokable way to state the problem?
Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.
Marble 1 2
Bag
2R R R
BR B R
If, on the other hand, you use a red-marble-picking robot, you get:
Marble 1 2
Bag
2R R R
BR R R
So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.
tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.
EDIT: formatting.
Alright just working through your logic here. Let's say I pick the first door always. I'll write out all the possible configurationss with both possibilities for Monty.
CGG > Monty reveals mid (Goat)
CGG > Monty reveals right (Goat)
----
GCG > Monty reveals mid (Car)
GCG > Monty reveals right (Goat)
---
GGC > Monty reveals mid (Goat)
GGC > Monty reveals right (Car)
So there's 4 worlds where Monty reveals a goat. In two of them, I picked a car initially. In the other two, I get a car by switching.
My prior post is mistaken -- thank you!
(C)GG - switch and lose
(G)CG - switch and win
(G)GC - switch and win
There are other situations where the game is over before you get a decision.
Both GCG and GGC are eliminated half the time before you get to make a decision. But CGG is never eliminated before you get to make a decision.
Once you're making a decision and the game isn't already over, you can use this information -- that the game didn't end -- to assign a 50% chance of being in CGG, and 25% of each of the other two.
I wrote a simulation to demonstrate how right I was.
I was wrong and you are too.
It's actually an interesting, possibly general story, that kind of makes me think. Perhaps for certain personality types, misplaced conviction will put you on a higher velocity trajectory toward truth than honest confusion.
The key part of this sentence that makes it true, is the "for certain personality types".
I would wager the majority of personality types however, it would put you on a higher velocity trajectory to willful ignorance.
Therefore, no games are removed as moot when you have initially selected a car, and 50% of games are removed as moot when you have initially selected a goat.
So it ends in a coin toss whether you switch or not.
The 2/3 to 1/3 split only happens when Monty removes a goat using privileged information.