Monty Hall/Three Prisoners Problem

Let’s talk about probability. This is a continuation of the “green ball/red ball” themes from the earlier threads.

(Beloved, honored jannies: this is related to politics because it exhibits the way human reasoning and perception works and how it can conflict with rational statistical analysis; one of the primary sources of political conflict is discrepancies between perceptions of what actions have the greatest probability of bringing about certain outcomes in a society).

In 1959, Martin Gardner introduced the Three Prisoners Problem. The same principal was later rephrased in the Monty Hall Problem by Marylin vos Savant in 1990. The mathematical question concerns the likelihood of a certain outcome being true for, initially, one of three choices. I’ll use the Monty Hall version here as it seems to be the more well known version, but feel free to use either for discussing the correct answer.
—-The problem is as follows:—-

You are on a game show, and in front of you are 3 doors. Behind one door is a car; behind the other 2 are goats. You want the car, and you must pick one door to open. You pick a door (say, door #1). Before opening it, the host of the show, who knows what’s behind each door, then decides to open a door with a goat behind from among the two doors you didn’t pick (say, door #3, now open and showing a goat). He then asks if you’d like to switch your choice and open door #2, or stick with your original choice and open door #1.

What should you do to get the best chance of winning the car? Should you switch to opening door #2, stick with opening door #1, or does it not matter at all?

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en.m.wikipedia.org/wiki/Bertrand's_box_paradox
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The correct answer is that you should switch doors. Switching gives you the best chance of winning the car.

Stick with number 1. Look at the goat. Its body would show in door number 2. So just logically speaking, if they wanted a car to not be visible they would have to put it in door number 1 because it is just 1 room.

You have disrespected the concept of a thought experiment and should be ashamed

i already have a car, but no goat. i’ll take the goat, could be useful for milk and eating overgrown grass

You want the car you doublenigger

The car comes with 5 goats in it how about that

its bullshit. your odds don't get better the longer you play. the fault lies in the assumptions.

If you stay with your initial choice you only win 1/3 of the time. Swapping is the superior option.

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How long you play has nothing to do with it. Your odds get better if you switch to the other door.

If you switch doors, you have a 2/3rds chance of winning the car. If you keep your original choice, 1/3rd chance.

What if I wanna fuck the goat tho?

The answer to the ball problem was 50% and by god, anyone who disagrees is a leaf loving faggot who couldn't see through turk sophistry

The fact he didn't open that other door contains information, iirc switching doubles the chance you win to 2/3.

kek classic sand nigger

This has always and will always be the truth. Don't listen to the game show jew.

The game show knows where the car and goat are... they never show the car.
This means the probability of being shown a car is zero.
If you haven't accounted for this, your model is based on a basic understanding of "intro to statistics" class and you should be embarrassed.

The answer to the ball problem is 2/3rds

en.m.wikipedia.org/wiki/Bertrand's_box_paradox

Fully correct and based

Yes, the host will always show a goat.

Your odds get better if you switch doors.

If I flip a coin 99 times and it lands on heads 99 times, its still a 50/50 chance of being tails the next time. This is retarded. The chances don't change the longer you play. This is jewish magic to trick people.

I'm not judging, there's 2 goats in total and always room to share with my paki bro

I don’t know what you mean by “the longer you play”, but if you switch doors, yes, your chances do go up.

I had no idea people were this dumb about probability. Both this and the box problem are easy to understand, but people commit the same flaws in reasoning and think it's 50/50.

Thanks for making the ignorance of others clear, OP.

What's the ploblem?

But what would happen if it lands on tails 99 times?

Heres the thing with probability you should always change your choice with the Monty Hall problem HOWEVER I still maintain that the equation changes and thus needs a recalculation to 50/50.

OI! DO YOU HAVE YOUR PROBLEM LOICENSE?

50/50 would suggest that you don’t stand a better chance of getting the car by switching. But you do. You will get the car 2/3rds of the time if you switch.

Should you open the door you originally chose or switch to the other one

A lot of people get the wrong answer because people are retards.

Something like 83% of people with PhDs responding to the problem got it wrong. It’s very unintuitive.

The supposedly "correct" method of 2/3 is obtained by the following method:

1. Originally, all six balls equally as likely to be selected.

2. The ball selected can not have come from drawer R in box GR or from either drawer of box RR.

3. Therefore, it must have been drawn from the G drawer of GR, or either drawer of box GG.

4. The three remaining possibilities are equally likely, so the possibility from a field of RGG is 2/3.

However, in the question posed by the Turk we can reformulate the above as follows:

1. Ball G has already been selected, we need not account for the probability of it being selected.

2. Therefore, we can cast aside box RR, and concern ourselves only with box GR and GG.

3. This leaves us with selection of balls described above in (4): RGG.

4. The RGG selection does not mean probability of selecting a green ball is 2/3.

5. We do not know from which box we selected the first G ball, it is possible it was selected from the RG box or GG box.

6. We are asked to again select from the box we drew the initial G ball from, we concieve of the remaining SEPARATE boxes as Box R and Box G.

7. Given the field RG, what is the possibility we select G? Its 50%.

Baiting aside, I truly believe this is the correct answer.

I agree. But this is a fate vs free will argument. People either consider the entire system and say you could have picked the box that is now made irrelevant, but you didn't and thus you must continue to use it in your calculation, whereas those who argue fate would say you are born in a system where you were fated to choose only from the two relevant boxes and therefore fate made it 50/50.

Interesting take, had not thought of it from fate vs freewill ... thanks ameribro

Opening the goat door changes nothing(well, almost). The chances were always 2/3 for picking the wrong (goat)door and 1/3 for picking the right (car)door. Since you've had 2/3 chance for picking the wrong door, and since the host already eliminated one bad choice, you have a greater chance of winning if you switch because you were ALWAYS more likely to pick the wrong door at the beginning. At least I remember some user explaining it to me like that long time ago on Yas Forums.

What matters is my first choice

you change the door because you have a higher percentage of being correct than if you stay with your original decision

Five goats, huh? That would get me half of what I need...

There is is only two possible outcomes, green or red. 50/50. That is it. Don't over think it.

The door is a lie

You're fucking wrong. I'll prove it using a brute force method.

Let's label the first green ball G1, the second green ball G2, and the third green ball G3. And let's do the same for the red balls (this time from right to left. So the boxes look like this:

Box 1___Box 2__Box 3
G1 G2 | G3 R3 | R2 R1 |

Now let's go through all the combinations.
There is an equal probability to choose any of the balls. You are just as likely to pick up R1 as you are to pick up G2 on your first draw. Running through every single combination for the first, and second draw:

G1, G2
G2, G1
G3, R3
R3, G3
R2, R1
R1, R2

There are 6 balls, so there are 6 different ways to choose the first ball.
We have stated that you choose a green ball first. Therefore you simply cross out every single combination that starts with a red ball (those are not considered anymore)

G1, G2
G2, G1
G3, R3

Therefore, two times out of three you drew a green ball second.
This is not negotiable, the answer to this problem is well understood

>5. We do not know from which box we selected the first G ball, it is possible it was selected from the RG box or GG box.

Of course we don’t know, that’s the whole point of probability. The fact that we got a green ball means it’s more likely that we drew from a box with 100% chance of giving green balls, as opposed to the one with a 50% chance of giving green balls.

Problems like this tend to be less confusing when expanded to larger numbers:

Imagine two boxes each containing 500 trillion balls.

Box 1: All 500 trillion balls are green
Box 2: Only one ball is green, all the rest of the 500 trillion are red

You pick a box at random, take out a ball, and see that it’s green.

Of course it’s more likely that you have box 1.

Wrong.

Not particularly. Your first choice always has a 1/3rd chance of picking the car, and switching always gives you a 2/3rd chance of picking the car.

variable change makes it so changing gives you more of a chance of picking the car so you swap doors

He’s completely wrong

Correct, but not about the opened door not changing anything- it changes everything. Now you can switch to the “2/3rds block of doors” because it only contains one closed door.