Can someone please explain this shit to me?

She is absolute god desu. I wonder what it's like to be a super-IQ god literally schooling phd plebs

Your first pick has a chance of being right 1 of every 3 times. Thus whatever you picked at first, it's most likely wrong.

Take away another wrong door, and now the remaining door has one in half chance of being the good one.

So the question is, was your first choice the good one? Or would you rather try again?

this.

Philosophy major here to explain this to you brainlets.

The factor you aren't considering is that the host always picks a door with a goat on purpose. If the car is behind door 1, he will open 2 or 3. If it's behind 2, he will open 1 or 3, and so on. Therefore, when you pick 1 and he opens 3, which has a goat, and he offers the chance for you to switch your choice, considering all the possibilities, you're better off accepting the offer.

It goes like this. Pick door 1 and assume he either opens that one or another with a goat, and calculate if it would pay to switch your final answer or not. You'll find that you're 2/3 times more likely to get the car if you switch your final answer.

1 2 3 (declining to switch)
G C G lose
C G G win
G G C lose

1 2 3 (accepting to switch)
G G C win
G C G win
C G G lost

It really comes down to "Do you understand what conditional probability is?" or "Do you know how probability changes when new information is presented?".

There are two important facts given from the door reveal:
1) The probability of the car being behind door 3 has changed, once you know it is not there
2) You know that the host couldn't reveal the car OR the door you decided on

Step by step:
>No information about the probability weighting for 3 doors, so we assume the weighting is equal like a fair die, 1/3 probability of getting the car.
>Door 3 is revealed to have goat
>Probability of car behind door 3 is now 0, but probabilities need to sum to 1, need to update
>Game show host could not reveal the door you chose AND he had to reveal a goat.
>Keeping all of this in mind we can rebuild our probability space
>From this first selection you picked either a car, goat A, or goat B
>If you picked the car, the host reveals one of the goats, you switch to the other goat and lose
>If you picked goat A, the host HAS TO reveal goat B, you switch, you get the car, you win
>If you picked goat B, the host HAS TO reveal goat A, you switch, you get the car, you win
>3 possibly scenarios with switching, 2 out of 3 get you the car. Hence, 2/3 chance of car.
>It should be easy to see by negating the switching statement in the above three scenarios, you'll end up with the opposite outcomes, leaving a 1/3 chance of getting the car when not switching

> Pick door 1 and assume he either opens that one or another with a goat, and calculate if it would pay to switch your final answer or not. You'll find that you're 2/3 times more likely to get the car if you switch your final answer.

No, you're not. The chance of getting a car or a goat is 50/50 because you have NO information about what's behind those doors.
>If you picked the car, the host reveals one of the goats, you switch to the other goat and lose
>If you picked goat A, the host HAS TO reveal goat B, you switch, you get the car, you win
Yeah but you don't know which door you picked. Hence the chance is still 50/50.

>you don't know which door you picked
We're not considering this problem in the case of dementia patients or Guy Pearce.

I mean you don't know what's behind the door you picked, so it doesn't matter than Monty Hall shows you a goat. You still have a perfect 50-50 chance irrelevant of door switching meme shit

It makes a little more sense if you think about it with more doors. If you had 10 doors to choose from, and he revealed 8, would you still want to keep your door even though there was originally a 1/10 chance of being correct? 1000 doors? If you just look at the end, it's 50/50, but taken with additional context, it makes more sense to switch.

fuck this shit