I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.
If he opens a door and tells you he knows it's a goat, you double your likelihood of winning by switching to the remaining door.
If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others?
In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"...
Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?
He chooses randomly from the doors you did not choose, and draws a goat. When offered the choice to switch to the third door, it offers you no benefit. Both are equally likely to contain the car.
Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.
Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch.
Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch.
The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2)
exactly 50/50.
In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.
- you choose door 1, goat
-- host chooses door 1, goat
--- switch: 1/2 chance of being right
--- stay: you're wrong
-- host chooses door 2, goat
--- switch: you are right
--- stay: you are wrong
-- host chooses door 3, car
--- switch: you are right
--- stay: you are wrong
- you choose door 2, goat
-- host chooses door 1, goat
--- switch: you are right
--- stay: you're wrong
-- host chooses door 2, goat
--- switch: 1/2 chance of being right
--- stay: you are wrong
-- host chooses door 3, car
--- switch: you are right
--- stay: you are wrong
- you choose door 3, car
-- host chooses door 1, goat
--- switch: you are wrong
--- stay: you are right
-- host chooses door 2, goat
--- switch: you are wrong
--- stay: you are right
-- host chooses door 3, car
--- switch: you are wrong
--- stay: you are right
so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times.
there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too.
this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.
No. He picks goat 100% of the time. Choosing not goat is outside the bounds/parameters of the initial constraints.
At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked.
The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck.
>Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.
I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.
Hey lets put a car behind three doors, have people choose, open one of the other two doors, then ask them to switch.
Sounds good Bob, but wait, won't the show end prematurely 1/3 of the time because you will randomly open a car?
Huh, good point. Um, let's sneak a peek before opening so we never open a car?
Sounds good Bob.
That’s extremely obvious! The entire reason to phrase the problem as a game show with a game show host is to make it abundantly clear that the host “knows the answers.”
Reordering the sentence makes it clearer. "In all games where Monty chooses goat...". The premise creates a subset.
Monty’s knowledge is irrelevant. The original problem can be modified with no change in probabilities to:
Pick one of three doors. You can have whatever is behind that door, or you can change your pick to both of the other doors and win whatever is behind both of them. Should you switch? Obviously.
Step 2: Divide it into a pair of cards, and a single card.
Step 3: Reveal one of the pairs of cards. If you reveal and ace, return to step 1 since this history is eliminated from the story we've been told: we know this wasn't a possible path to our endgame. Otherwise continue to step 4.
Step 4: Which of the remaining two cards is more likely to be an ace?
Step 5: Realize that they're just as likely as each other to be the ace. Step 2 didn't magically imbue the card remaining in the pair with extra probability juice.
Long story short, you definitely need Monty to make an intelligent selection, so Monty's knowledge is far from irrelevant. It matters whether he revealed a goat by luck or by knowledge, because he's 2x as likely to get "lucky" in the case where there are 2 goats behind the doors you didn't choose in your original guess.
https://news.ycombinator.com/item?id=24713352
When he shows (randomly from the doors you didn’t pick) the car, switching to open door (where the car is) and the other door (which hides a goat) you increase your chances from 0% to 100%.
When he shows a goat, switching to the open door (where there is a goat) and the other door (which may or may not hide the car) your chances stay at 50%.
Switching is the best ex-ante strategy, not necessarily so after a door has been opened.