The role of probability in solving the Monty Hall problem

[ErGingerbreadMandude]

Whether it is beneficial to switch or not depends entirely on whether you choose a goat first.
And after you choose a door the host opens up another door which has a goat behind it. Now your probability of having a goat behind the door you first chose is 50-50.

So you have a 50-50 chance of you getting a car if you switch when the host asks you to.

[robvalue]

Wrong.

You are again missing the fact that the host makes a forced choice.

I just laid it all out in detail. You're just saying I'm wrong. I'm right, though.

[Mr.wizard]

It's really as simple as my chances initially of winning are 33% and losing are 67%, so when you get to the second round do you stick with your 33% or switch to the 67% chance, I know what I'm doing.

[ErGingerbreadMandude]

Why am I wrong. Read my above post carefully before you reply.

[ErGingerbreadMandude]

The chances of you getting a goat behind the door you initially chose is not 50-50 given that you don't know what's behind all the three doors. However after you know that behind one of a door there isn't a car, your chance of having chosen a car or goat initially essentially becomes 50-50

[robvalue]

I give up. I've explained it every way I can think of.

[Excited Penguin]

It all comes down to whether you pick the car on that first try or not. Forget about everything else. This is all that matters. I'm arguing that it's not 1/3 that you pick a car on your first try because there's nothing to say that whatever "random" process makes you choose your first door nothing guarantees that you won't choose the car 10 times out of 10.

[ErGingerbreadMandude]

Oh shut up penguin. It's not about picking a car in your first try that matters whether you win or not when you switch when the host asks you to,it is about picking a goat first.

Too bad we only have a 50-50 chance to pick a goat the first try else they'd be right.

[Mr.wizard]

The chances of you getting a goat behind the door you initially chose is not 50-50 given that you don't know what's behind all the three doors. However after you know that behind one of a door there isn't a car, your chance of having chosen a car or goat initially essentially becomes 50-50

Wow I don't think you understand how this game works at all. He is never going to open a car, he will always open a goat because he knows whats behind the doors.

[Excited Penguin]

Oh shut up penguin. It's not about picking a car in your first try that matters whether you win or not when you switch when the host asks you to is about picking a goat first.

No, it's not. Probability is irrelevant in this puzzle. Particularly because it doesn't work(at least here) as you seem to think it does. You can only talk of chance and probability once you know for sure that each door will be picked the same amount of times throughout a certain number of iterations of the same game.