Of course plants don’t create the oxygen in the air. They just debond it from carbon. What do you think plants are, fusion reactors?
Guys: Don’t call people stupid or idiots. This is a difficult problem and has fooled many statisticians and mathematicians. You especially don’t call someone an idiot when they’re right and you’re wrong.
First of all, draw out a table of probabilities. After all, there are only 18 rows in the table (3 doors you choose x 3 doors where the car is located x 2 doors Monty can choose).
Let’s consider the Original Monty Hall problem:
There are 6:18 rows in the probability table where you picked the car. That means, there are 12:18 rows where you didn’t pick the car. Monty picks the car ZERO times in those remaining 9 rows. Since there is only one remaining door, picking it will give you the car 12:18 times.
Let’s look at the Monty Fall problem:
Same initial condition: There are 6:18 rows in the probability table where you picked the car. That leaves 12:18 rows where you didn’t pick the car. Now, Monty picks the car in six of those remaining rows (That is, there are two remaining doors, and he has a 1:2 chance of picking the door with the car).
Since the game states it isn’t considering those 6:18 possibilities, we’ll toss them out of our probability table. Now, there is a 6:12 rows that you initially picked the car, and a 6:12 rows that you didn’t picked the car. In this case, switching doors does not improve your chances of getting the car.
BTW, it is still a 6:18 initial chance of picking the car, we simply tossed the six rows where you picked a door, Monty picks a door, and the car is behind it. Obviously, switching in those six instances doesn’t help you in that situation unless Monty lets you pick the same door he just did.
===
Let’s look at the Monty Crawl problem. Initially, it seemed to me that it still should be the same 12:18 times that switching will get you the car. What concerned me is that I could think of a few situations (where Monty picked Door #3) where you’ll always win the car by switching.
Drawing out the probability table reveals that switching will get you the car 12:18 possibilities. However, if Monty picks Door #3 (which he’ll do 4:18 times), you will 100% of the time by switching. If Monty picks Door #2 (which he’ll do 6:18 times), you’ll will 2:3 times by switching. But, if Monty picks Door #1 (which he does 8:18 times), switching doesn’t give you an advantage. You lose 4 of those times and win 4 of those times.
====
Let’s take the two kids where at least one is a boy. Let’s give the kids names: One is Chris and one is Sandy.
There are three possibilities:
1). Chris is a girl and Sandy is a boy.
2). Chris is a boy, and Sandy is a girl.
3). Both Chris and Sandy are boys.
Yes, the odds are one out of three that they’re both boys. Isn’t possibility #1 and possibility #2 the same? If you were either Chris or Sandy, and one of you is a girl and one of you is a boy, would it make a difference to you which is which when you got dressed in the morning?
Still not convinced. Let’s play a game where we flip two coins. I pay you if they both land on heads. Otherwise, you pay me. Now, what are the odds that both coins will land on heads. If you think it’s still 2:1, I’d like to introduce you to a few investment opportunities.
No: The possibility of each coin landing on heads is 1:2, but when there are two coins, the possibility is 1:4.
1). Heads Heads
2). Tails Heads
3). Heads Tails
4). Tails Tails
If it is any help. Imagine the first coin is a dime, and the second coin is a nickel.
Let’s change the game. I still pay out if both coins land on heads. You pay otherwise, but we’ll consider it a do over if both coins land on tails. That makes three rows in our probability table:
1). Heads Heads
2). Tails Heads
3). Heads Tails
What is the probability that both coins land on heads? It’s one out of three! Now, let Heads = Boy and Tails= Girl. You can see it’s the same as before.
====
One more quiz:
There’s a game called Chuck-a-Luck (aka Crown and Anchor) that’s played on midways. The rules are simple. There are three dice kept in an hour glass cage. The cage is turned over and the three dice are rolled.
You bet on a number 1 to 6, and if that number comes up, you get paid even odds. That is, if you bet a dollar, you win a dollar. (You get two dollars back: The dollar you won, and your initial dollar you bet). That sounds very fair. The possibility of a number coming up on a die is 1:6 times. Since there are three numbers, the possibility must be 3 x 1/6 or 1/2 time. Even odds. Right?
However, the dealer gives a bit of a bonus: If two of the dice land on your number, you get double your money. If all three dice land on your number, you get triple your money! How can you lose?
But, lose you will. What is the actual odds of winning in this game?
BA you are a troll, or an idiot, or both.
Have a nice day 
I am microwaving a bowl of milk right now.
This even makes an appearance in the movie “21” doing one of the maths lectures, though it’s refered to as “Variable Change” rather than the Monty Hall problem
This problem nearly destroyed my relationship. Really. My fiance and I got into the biggest argument over this. I tried explaining the solution, my fiance called me stupid, and well, it erupted from there. I’m still a little angry about it quite frankly.
All I can imagine is the Benny Hill chase scenes between doors.
Jeff,
Have you ever said whether you got it right the first time? With such a big brain, I’d be surprised if you didn’t 
I know I didn’t. But by changing the story to 1,000,000 doors, where the contestant selects one door and then the host opens 999,998 other doors, leaving one un-opened, made it clear to me.
The psychological side of this question is far more fascinating than the question itself.
It’s no wonder casinos are so profitable if (as someone said) 92% of the population get caught up on this type of question.
I’ve never played poker/blackjack etc for money. But it seems obvious that if you’re ever going to be any good at it, you need a very clear understanding of exactly this type of unintuitive answer.
I would also suggest that many good players have an innate understanding of this, ie have never been taught.
@Simon: actually, it makes no difference whether Monty opens the door accidentally or not. The bottom line is you have a 1/3 chance of your initial choice having been correct. If an incorrect door then opens, there is a 2/3 chance that the OTHER door is correct, therefore you should still switch.
I don’t like the construction of the “Monty Fall” problem. What happens if the host slips and reveals the prize? Is it possible for the host to slip and re-open the contestant’s door or only one of the non-selected doors? Without knowing if these probabilities are possible, I can’t address it.
Monty Crawl, though, is neat.
The best way to explain it is actually change the problem slightly.
Say you have a million doors. Only 1 has the car behind it. the 999 999 others have goat.
So you pick a door.
Similarly, the host then opens all the other doors but one.
Do you switch?
Before the odds were 1 in a million to pick the car. But after the host opens the doors, you know that one of the doors has a goat, and the other is the car. In this variant, you know, that switching is better. Because you’ve changed the odds of getting the car from 1 in a million, to 1 in 2. Shrinking the problem you can always reason the same way, that switching is better, and so on down to the usual case.
The moment I was informed of the variant in this way, I beleived.
Wow, this is really non-intuitive but fun to think about. I thought it would make no difference to switch at first, but after reading the replies from Marilyn, it does make sense to switch.
Our minds truly suck at statistics and probabilities.
What also frightens me is that plenty of mathematicians were against her, they’re supposed to understand this kind of thing.
@Gabriel Ross Actually it makes all the difference in the world. Read the link that Jeff provided (the pdf link)
Wait, I’ll admit I don’t understand, but after the first door is opened, she has a .5 probability of getting a car, substantially better than she had a second ago. Switching doors isn’t going to change those odds one bit. What am I missing?
On a side note, and it’s been decades since I saw the show, but I remember Monty pulling a $100 bill out of his pocket to offer the contestant a new choice altogether: Take the money and sit down, or proceed with the chance of getting a new car. Different decision, that.
Robert B: No, the odds are always better than 1 in 2 (and you don’t need to use a million doors in an example to show that)
The important thing to realize is that the host always brings it down to a 2-door choice where one door has the prize. This is not a 1-in-2 choice.
If you initially picked the right door, switching will lose you the prize.
If you initially picked the wrong door, switching will gain you the prize.
So, the odds of winning are precisely 1 - (odds of you guessing from the original set of doors)
I mean to say, the odds of winning if you switch doors
My problem is that “switching” means “taking the other one” and intuitively, it’s simply turning a 1-in-3 chance into a 1-in-2 chance, so there’s no reason why switching should make any difference at all.
I’m with Erdos on this one. I only believe it because the statistics say it’s correct.
I still don’t understand how it’s not a 50/50 chance, no matter how many times people tell me the one door being wrong increases your knowledge about the validity of your initial choice (so it’s a 33/33/33 choice and you drop one – how does that not make it a 50/50 choice?).
A very simple way to explain…
Your first choice is usually wrong (66%) and thus 66% of the time the host is essentially showing you the correct door.
it’s a 33/33/33 choice and you drop one – how does that not make it a 50/50 choice?
There was a 0.33 chance you picked the right door in the first place. Probabilities have to add up to 1, so there’s a 0.66 chance that you win if you switch.