Forgive me if this explanation has been used before and I haven’t seen it.
So I understood how the math supports the Monty Hall Problem, but couldn’t wrap my head around WHY it works.
Then i thought:
Don’t make it 3 doors. Make 100 doors. You pick one out of 100 doors, and Monty narrows it down to your door and one other door. Which do you choose? Do you REALLY think you picked the right one when you only had a 1% chance of doing so??? Take the other door fool!
So…that’s it.
The studio isn’t that big.
I think the best way to explain it intuitively is as follows. This is the two-goats-and-a-car format.
You make your initial choice, say door A.
Would you now switch away from A if you were offered the better of doors B and C, if their contents are different? Of course you would - that gets you the car if it’s behind either of B or C, doubling your chance of getting the car.
When Monty opens one of B and C (showing you a goat) and gives you the opportunity to switch to the other one, he is effectively offering you just that - the better of B and C, if they are different.
I’m pretty sure the “100 doors” extension has been discussed before. It is a good way to make the solution more intuitive.
It has been proposed before, and the correct action in the 100-door scenario is obvious. What’s I don’t think is obvious is how it relates to the 3-door scenario.
But what if you want a goat instead of a car? If you choose a door with a goat behind it do you actually get to take the goat home with you?
I recall vos Savant herself having used this argument before, but she made it even bigger; something like 7,000 doors.
The key that made me understand it all was the fact that Monty knows which door has the car, and will always eliminate a goat door.
I’ve heard a slight variant of this:
Monty doesn’t open any doors. The rule is that having chosen A, you can elect to stick with that, or switch to both B and C. But you get to keep only one of the two prizes found behind those two doors.
Right.
He is giving you information about the “system.”
That’s the piece that people who don’t understand why you should switch miss.
What if you want a goat more than a car? Goats are cheaper to own, and they do landscaping.
I suppose you are trying to reword the problem, but an empty (or goat) door will be opened in every game, and that tells you which of the two other doors would have the prize behind it two out of three times.
Yeah, in a slightly different system, where he just opens a random door after you choose yours, then he gives you no information in 2 out of 3 scenarios, and in the third, he gives you the information of exactly which door you should have chosen, but does so in a way that eliminates it.
It is the fact that he knows where it is, and so will eliminate one of your false choices for you that give you the advantage to switch.
Yes, exactly this.
And it’s the fact that his knowledge is implicit in the problem explanation, not explicit, that gives people so much trouble. A strict reading of the usual explanation of the problem doesn’t explain that a) Monty always knows where the car is, and b) always deliberately chooses to show you a goat.
It’s those two implicit "always"s that trip people up who are stuck on “It’s all random; how could the odds change after the fact?”
I remember positing the 100 doors variation on the AOL SDMB back in 1996 or so.
The key is driving home the fact that the host isn’t eliminating a random door. He’s imparting additional information and that changes the odds in your favor.
Yes, it’s critical to specify that the rules of the game are:
Monty knows what’s behind the doors. Monty has no discretion, he always mechanically follows the same procedure: he always reveals a goat, and he always gives the opportunity to swap.
If these rules do not apply, the classic solution is not valid.
The casual observer rarely understands the problem. Tell someone there are 3 doors, 1 with a prize behind it and nothing behind the other two, and then have them pick one. Now tell them that you are betting $100 that the prize is behind one of the two doors they didn’t pick, do you think they’ll take up that bet? Of course not, they understand right away the odds are against them. Tell them you’ll pick the specific door out of the other two with the prize they may start to consider the bet, but give them the final piece of information, that one of the two other doors is empty and that one will be revealed before you pick your door and they won’t take the bet, and they’ll probably understand the problem.
I think it gets confusing by talking about odds changing or anything like that. The odds always favor the prize being behind one of the two doors that wasn’t selected. The odds don’t change, you are just given more information about which of the two other doors has the prize two out of three times.
Well in any event…I think if Amy had simply used the ‘100 door’ ( or 7000 ) explanation for Captain Holt, he would have figured it out a lot easier and been “BONNNNNNNEDDDD!!!” a lot sooner.
Take Monty Hall out of the Monty Hall problem. Don’t open any doors.
You have three doors. There’s a prize behind one of them. There’s nothing behind the other two.
You pick one door. You’re now offered a choice. You can have what’s behind the door you picked. Or you can have everything that’s behind the other two doors. Do you stay with the door you picked or switch to the other two doors?
Here’s the way I look at it.
Let’s say I have really bad luck. I mean, really bad luck. No matter what, I always make the wrong choice in everything I do for some unknown reason.
Got that? O.K.
Now let’s say I am playing the Monty Hall Game. Because I have such terrible luck, I always choose a door with no prize. (At least initially.)
Given the rules of the game - Monty opens another door (without a prize) and gives me the option to switch - I will always win if I switch.
Do you agree?
O.K., now… what are the odds I initially choose a door without a prize? 2/3.
If you’re still confused, keep rereading the above until it sinks in.
An abbreviated explanation:
If you initially choose a door without a prize, and you switch (after Monty opens another door), you will win. Agree?
So what are the odds of initially choosing a door without a prize? 2/3.