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[pikachu17]

It isn't a symmetrical situation because of the stuff that happened before. Make a graph of all the possible situations, it might help.

I already did so.

[Felissan]

OK then, here's an easier problem to contextualize. You're given 2 coins: one of them is normal, the other is fake and has 2 heads. You flip a coin and it lands on heads. What are the odds that it's the fake one?

[ChairmanPoo]

But this isn't the original choice, this is a choice between two doors.

No it isnt. Once again, the original setup matters a lot, because the odds work based on the dealer's choice. There is a 66% chance that the car is behind one of his doors, and he just showed you behind which one it isn't. There is only a 33% chance that you actually chose the good door on the first try. Its not a 50% at all. In order to understand the problem you have to look at the whole process not just one bit.

Think about it like this: would you rather be able to open one door from the onset or two?

It isn't a symmetrical situation because of the stuff that happened before. Make a graph of all the possible situations, it might help.

this

[ChairmanPoo]

But this isn't the original choice, this is a choice between two doors.
Please note the possibilies of each door being the one with car. Either door one, or door two has the car. Door three is not a possible container of the car.

the door choice matters. both yours and the dealers.

 If you didnt know behind which door was the car, but the dealer gave you the choice to open the other two doors and keep the car if it was there, would you do it?  would you say there's a 50% chance between it behind your door and it being behind one of the other (closed) doors?

[pikachu17]

Okay.
Either door OG is the car, or door SD is the car.
Two possibilities.
OG has the car in one possibility.
SD has the car in one possibility.
1 in 2 for OG.
1 in 2 for SD.

See? One in two chance.

The door with the shown goat can not possibly have the car, so it is not one of the possibilities.

[TamerVirus]

What if we scale up the problem then, because 'stuff happened before doesn't matter'

Instead of 3 doors, there is 1,000,000 doors. Only 1 car

You pick a door, and Monty eliminates 999,998 goats .
Now there are 2 doors. Do you have a 50/50 chance of winning?

[ChairmanPoo]

Okay.
Either door OG is the car, or door SD is the car.
Two possibilities.
OG has the car in one possibility.
SD has the car in one possibility.
1 in 2 for OG.
1 in 2 for SD.

See? One in two chance.

The door with the shown goat can not possibly have the car, so it is not one of the possibilities.

you're really stuck in a loop. consider the problem like I told you:
you have three doors. would you rather open ONE door or TWO doors?

[pikachu17]

What if we scale up the problem then, because 'stuff happened before doesn't matter'

Instead of 3 doors, there is 1,000,000 doors. Only 1 car

You pick a door, and Monty eliminates 999,998 goats .
Now there are 2 doors. Do you have a 50/50 chance of winning?

Yes. Because there are only two possibilities of which door has the car.

EDIT:
How are we opening one door in one case, and two doors in the other? Either case, we opening the final door we choose, and the Goat Door the host showed us.

[Arcvasti]

The key to understanding the Monty Hall problem is knowing that humans are intrinsically awful at conditional probability. That's really all there is to it.

[TamerVirus]

[Felissan]

OK then, here's an easier problem to contextualize. You're given 2 coins: one of them is normal, the other is fake and has 2 heads. You flip a coin and it lands on heads. What are the odds that it's the fake one?

[Biowraith]

What if we scale up the problem then, because 'stuff happened before doesn't matter'

Instead of 3 doors, there is 1,000,000 doors. Only 1 car

You pick a door, and Monty eliminates 999,998 goats .
Now there are 2 doors. Do you have a 50/50 chance of winning?

Yes.

The host will never reveal a car.  So if you picked the car initially (1/1,000,000 chance) the other door will *always* be a goat, and if you picked a goat initially (999,999/1,000,000 chance) the other door will *always* be a car.   So switching has a 999,999/1,000,000 chance of paying off.  The fact there's only two choices left doesn't mean they're equally weighted towards the car - what came before affects that weighting.

[ChairmanPoo]

What if we scale up the problem then, because 'stuff happened before doesn't matter'

Instead of 3 doors, there is 1,000,000 doors. Only 1 car

You pick a door, and Monty eliminates 999,998 goats .
Now there are 2 doors. Do you have a 50/50 chance of winning?

Yes.

EDIT: Chairman Poo, clearly neither of us is going to change the other's mind.

I will keep trying to teach you because I want to see what you post once you reach enlightement.

I promise I'm not angry or annoyed. Its a difficult concept and clearly noone has taken the time to explain it to you. I'm going to educate you.
 

Please answer: would you rather choose one door or two doors?

[pikachu17]

Please explain to me where there is a different number of doors I could get in the Monty Haul problem.

The probability we want to know is not the initial choice, but the current one.
The current choice has two possible ends.
The door you chose has the car, or
The door you chose has the second goat.
That is one of two for the choice of two doors, one of which is a goat, and one of which is a car.
That we saw another goat beforehand does not change that.

[Felissan]

OK then, here's an easier problem to contextualize. You're given 2 coins: one of them is normal, the other is fake and has 2 heads. You flip a coin and it lands on heads. What are the odds that it's the fake one?