Can Yas Forums into probability? Part 2

That was 2/3rds you nigger it was explained repeatedly

I don't switch so I have a better probability of getting my own goat

Lessons in critical thinking 101. Reminds me of the opening lines of one of my first lecture as a freshman by the shiftiest most incompetent but somewhat hot lecturer.

Then again I think I will take the goat. its built like a horse but can buck harder.

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I didn't get this problem until right now, because I saw the other one with the green/red balls. For years I read this problem as there's only ONE goat, two doors safe. Durrrrr

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Work out roughly the odds/probability of this accidently happening and I will maybe send you some Zenon if you are close to being right.

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Actually, wait. There IS one goat left once a door is opened.

DOOR 1: GOAT
DOOR 2: NO GOAT (PICK)
DOOR 3: GOAT

Door 3 revealed.

DOOR 1: GOAT
DOOR 2: NO GOAT (PICK)

Why should I switch? If he is going to reveal a door anyway, the odds were 50/50 from the start.

Intuitively seems like it should be 50/50, but if you switch you have a 2/3rds chance of winning the car.

83% of PhDs got this wrong when it was published

The Monty hall problem
You should always switch
the host is giving you the right answer two thirds of the time because you are wrong one third of the times
youtube.com/watch?v=TVq2ivVpZgQ

>but if you switch you have a 2/3rds chance of winning the car.
But why? He would open a door with a goat, regardless of what you pick.

Yes in the future this may be argued similarly to
As the car is in a quantum super position until it is revealed.

When you picked a door, you had a 1/3rd chance of being right (1 of 3 doors has a car).

That means those 2 remaining doors, combined, have a 2/3rds chance at containing the car. So imagine drawing a circle around the 2 doors you didn’t pick. That circle has a 2/3rds chance of having the car.

Once the goat door has been opened, there is now only one door in that circle. You can switch to be in the circle.

oops i meant u are right 1 third of the time

*correction - simmilar to But under current theorem:
is correct.

this.

Literally none of this applies to trading. As a matter of fact, that question doesn't even apply to card counting like in that shitty movie.

>Literally none of this applies to trading.

Not directly. But training your mind to overcome the faults of instinctive human reasoning is useful for many complex pursuits.

>He would open a door with a goat, regardless of what you pick.
That's the answer to your question. There's three scenarios:
1. You pick goat A, the host opens the door to goat B. You switch and win the car.
2.You pick goat B, the host opens the door to goat A. You switch and win the car.
3. You pick the car, the host opens any door. You switch and lose.

Thanks for taking the time to explaining the theory, but I don't see it. Why does it matter what the chances were initially? A chance won't drive you around. Whether you win or lose is determined at the end, when you have a choice between picking a goat or a car, which is 50/50.

But that's just not true. Yes, prior to opening the goat door we would obviously want to open two doors instead of one, because yes that would be 2/3 chance. However, we've already eliminated one of the possibilities, one of the doors in this "circle" has been proven to be the non-car one. How are we left with anything, then, but a choice between one unopened door and another unopened door with equal chance of being the car?

No. I hope you're trolling, but here goes:

Imagine if there were 100 doors. You pick a door, and the host opens up all the doors except for one other door, revealing 98 goats.

There are now only two doors left, including the one you picked. Do you switch?

Think it through again. The door you picked has a 1/3rd chance of winning. So the circle you draw around the other 2 doors has a 2/3rd chance of winning. If you’re within that circle, you have a 2/3rd chance of winning. The car doesn’t move so that can’t change.

This runs through the options: You can make a quick chart and see the results yourself. It’s definitely not immediately intuitive at all for most people, myself included.

>There are now only two doors left, including the one you picked. Do you switch?
I'm not trolling. Why should I switch if we know from the start we're eliminating 98 bogus doors? The only one that will remain at the final decision is the one that I picked and a potential bogus one. I cannot lose with my initial pick, because the part where he's going to take out 98 doors has yet to come. Then I get to make a decision between two doors.

I can understand these problems only by doing this, is there another method one can learn or the only other option is to memorize the general rules?

When you picked, you had a 99% chance of being wrong.

When you know you have a 99% chance of being wrong, switching when 98% of the other options have been eliminated is definitely the right choice.

I encourage you to draw out the possibilities and look at them like has done.

The main point here is that the gameshow guy would never open a door with a car behind it. The gameshow guy is biased into opening doors that do not contain cars.
Let me try to explain my example again:

Since there were 100 doors, there is only a 1% chance that you choose the car on your first pick. The host opens 98 doors, but doesn't open the door with a car behind it. Do you think your initial pick was correct (1% chance), or do you think the door that wasn't opened contains a car?

Actually, fren, I get it now. Thanks. It only makes sense in my head when I visualize a million doors.