I think I’m going to have to run my own sim to make this sink in. I even watched Adam Savage run his own analog sim, but that just wasn’t enough.
One of the big challenges to science, I think, is that you have to learn to trust the science over your intuition, which is hard.
I was born in the 70’s, but not enough of it apparently.
There’s an even weirder way to state the Monty Hall problem that only involves one choice:
The host presents three doors, each with an equal chance of containing a prize. Then the host opens either door 2 or door 3 (never door 1 and never the door with the prize). Now choose. Hint: Don’t pick door 1.
The ‘real’ problem (being Monty Hall) was always there. Always. Therefore, the real problem is with those who coined the term “The Monty Hall Problem” and – when having done so – NOT considering the effect of Monty Hall in their definition of “The Monty Hall Problem”. 
With enough curry, you’d never notice the difference. 
No… I get that revealed door if I decide to switch or not. Actually, the revealed door is no longer an option. I can choose between my original choice, or the 1 other door that hasn’t been revealed.
With all due respect, I’m not sure you’re following what the Monty Hall Problem is. Your options are to remain with your original choice, thus giving yourself a 1/3 probability of winning, or switching to the remaining door, thus giving yourself a 2/3 probability of winning.
My problem is that I keep trying to divorce the final problem (“Here are two doors, there’s a 50/50 chance of picking the right one.”) from the total context, as if it were just a matter of probability. The problem is, it isn’t a matter of probability at all, really - the second door has been artificially selected to change the odds.
It’s no more about probability than playing a card game where you randomly pick a card that you don’t get to see, but your opponent does - and then they rifle through the pack to find a card to counter your own. You would know who to bet on instantly in that situation, but with the doors, there’s the illusion that it’s a matter of probability because the selection is hidden, essentially. Especially with only three doors - the opening of the third door distracts from the fact that it’s about knowledge of the (first and) second door.
I really don’t understand this. Why is switching giving you a 2/3 probability but staying not?? Let’s say I choose door 1. Door 3 is shown as not being correct. I only have door 1 or door 2 (3 is known to not be correct). So the options are :
I have 2 options, how is that not 1 in 2?
There are three possible places for the prize, and three possible initial picks.
prize.
A/A
A/B
A/C
B/A
B/B
B/C
C/A
C/B
C/C
In three of them, you have picked the winning door and in 6 you haven’t Since we do not know which door the prize is behind, we assign each possibility an equal probability. In 1/3 of the occurences, you have picked the winning door.
We know that the prize was not revealed when Monty opens a door. It actually doesn’t matter whether that was on purpose, or by chance, but we know THAT DIDN’T HAPPEN, And he can’t pick the door that you have picked
When you have picked the correct door, he has two options, but when you have picked the incorrect door, he only has one choice. This leaves us with
prize/pick/Monty
A/A/B
A/A/C
A/B/C
A/C/B
B/A/C
B/B/A
B/B/C
B/C/A
C/A/B
C/B/A
C/C/A
C/C/B For 12 total possibilities. But for any particular combination of our pick, and Monty’s reveal, there are only two possibilities: The odds are now 50/50.Keep in mind that we decided in the beginning that A/A had the same probability as B/A. and as C/A. So if Monty opens C, the two possibilities left are A/A and B/A, which we have already assigned equal probabilities to.
It took me a while, but I’m getting it, and where I went wrong. I was just jumping to the (mathematical) assumption that the prize re-randomizes when MH opens one door. Which they hypothetically could have done in the show, but it would be, I assume, unethical. Without redistribution of the prize, the denominator stays the same as the original randomization.
Thanks!
The thing that drives me crazy about the whole Monty Hall thing is that it is not usually stated whether Monty is picking a door at random or, as you say, based on prior knowledge.
Perhaps the problem would be less famously obtuse if it were framed as clearly as you just did.
Pretend they never open a door. You get to choose one and open it. You choose door 1. What is the chance that you are correct? It’s 1 in 3. That never changes.
So you have a 33% chance of being right. Then they show you an empty door. The chance you picked the correct door in the first place doesn’t change. And if you have a 33% chance of being right, you have a 67% chance of being wrong. And there’s only one other door.
As near as I can tell, this is where your reasoning goes wrong. He knows which of these cases is true, but crucially, YOU DON’T.
Just run the simulation that Humbabella helpfully posted. Here it is again: Monty Hall Problem Online. Run the monty hall game and simulation , over and over to understand the probability of this problem.. You can see a demonstration of the correct resolution of the problem for yourself.
The problem first appeared in this form, complete with solution, in the American Statistician in 1975; I reproduce it below. It is just a ‘sexing up’ of a much older, mathematically identical problem, the three prisoners problem. I think it was meant as an exercise people like me could use when teaching very basic probability classes, and is not difficult. (There is a similar problem, the “boy-or-girl” problem, which is quite difficult, mainly because the problem is less well-formed than it appears.)
Anyway, here is the original:
Because Monty would never open the door with the good prize before giving you the option to switch doors (otherwise the game would end right there and then). Once Monty opened a door with a bad prize, there’s still a 1 in 3 chance your first pick was right all along since those were the odds you started with, and a 2 in 3 chance you guessed wrong… But now you have the added bonus that now you know what’s behind a door without the good prize… So the door that’s left you didn’t pick has a 2 in 3 chance of being right.
After you’ve made your first choice, the doors are divided into two groups - the doors you initially chose (probability 1/3) and the doors you didn’t initially choose (probability 2/3).
If Monty were to put on a stupid hat (or some other arbitrary action), the probability of the car being in the group “cars you didn’t initially choose” is still 2/3.
Monty’s action of opening a door is similarly arbitrary woo; opening a door doesn’t change the odds of the group “doors you didn’t choose” from 2/3. All he’s asking you to do is choose between the door the group “doors you initially chose” (1/3 odds) and the group “doors you initially didn’t choose” (2/3 odds). That these groups are represented by hinged bits of wood is neither here nor there.
Actually, that isn’t quite right. If Monty opened a door at random, and it could be the good prize, but it wasn’t, thus leaving only two doors to choose from… In that case, one door is as good as the other, and it would be a 50-50 chance either way. This is the intuitive solution, and it would be correct in this case.
The fact that Monty would never reveal the good prize before giving you the option to switch is what preserves the initial 1 in 3 odds of the selected door and the 2 in 3 chance you were incorrect even when there’s only two doors left.
Because two options doesn’t automatically mean 1 in 2. Suppose your choice was that the car was either behind door number 1, or somewhere behind doors 2 and 3 combined. Clearly the first option has a 1 in a 3 chance of winning and the other has a 2 in 3 chance of winning even though you have two choices.
On the game show you start with 1/3 chance of being right. It’s always stays 1/3 chance of being right. What Monty is doing is giving you the option to switch to the 2/3rds of the chance your original choice was wrong.
I think you have the secret for those who cannot do statistics and probability. I know I do not trust my guts with questions of probability, so I always try to go back to what I know. The maths is then usually trivial.
Once you have mastered that, try quantum physics. No, really. It is wonderfully powerful, but it so counter-intuitive that your brain wants to throw up. Science is not all about conquering instinct, but sometimes the two don’t get along at all.
