/sci/ - Science & Math

is this the matter of fundamental perception of reality?
it appears there are two schools of thought
is this shit quantum?

>>15548965
You fully understand the Monty hall problem. Most people fail (Though they may say they understand, asking a few questions reveals they don't REALLY truly fathom it) and get filtered. If you succeed in understanding it to the FULLEST extent, not a shallow understanding, you already understand most of probability.

The monty hall problem is the best midwit filter.
I have never seen a person of subpar intelligence be able to comprehend why the Monty hall problem is true. No matter how much you explain it to them, they won't get it. This problem should be used as an offhand IQ test for schools and jobs.

>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Do you get it? or is it over for you?

It's so simple! All I have to do is divine from what I know of you. Are you the sort of man who would put the goat behind his own door, or the enemies? Now, a clever man would put the goat behind his own door, because he knows only a great fool would not switch when he was given the chance. I am not a great fool, so clearly I cannot choose the door in front of you. However, you would have known I am not a great fool, you would have counted on it, so I can clearly not choose the door in front of me.

And goats come from Australia, and Australia as everyone knows, is peopled entirely by criminals who are used to not having people trust them, as you are not trusted by me, so I clearly can not choose the door in front of you.

WAIT TIL I GET GOING. Where was I?

Yes, Australia. You would have suspected that I knew the goat's origin, so I can clearly not choose the door in front of me.

You've opened the other goat door which shows you're exceedingly strong, so you may have put the goat behind your own door counting on your strength to save you, so clearly I cannot choose the door in front of you.

>its 2/3 because...because it just is ok!?

Let's assume that instead of 3 there are 100 doors to choose. You pick 1 as normal but then 98 out of the 99 doors are rejected.
It's pretty obvious that it's not "50 - 50" anymore because initially you picked 1 out of 100 but now you are told 98 were duds.
i.e. On the original problem you were 1/3 and then swapping takes you to 2/3 and on the 100-door variation 1/100->99/100.

The solution with this is that math is a shortcut. Ultimately, the percentage calculations mean nothing, there are only 2 outcomes, either you are correct or you arent, and if we had knowledge of all the factors, we could always make the correct decision.

The math merely provides a chance. ITs essentially meaningless. Your chance of winning the loterally is 0.0003%. That doesnt mean anything if you know all the factors.

A diceroll is random. Its not if you can calculate the throw, gravity, wind resistance, the material of the surface it lands on and what the dice is made of.

>le probability
What a laugh. People lack common sense and use their flawed "logic".
Just put the best supercomputer to test this "probability" and you will see that it is bullshit.

https://youtu.be/dOQowCeAnRs

I am not denying that the woman is smart, but this is a classical mistake of the so called "smart" people.

The probability of winning will not improve if you switch doors. This is fucking obvious.

>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Bayes' theorem states P(A|B)=P(B|A)P(A)/P(B)
A=the player selects the car door
B=the host selects a goat door
P(B|A)=1
P(A)=1/3
P(B)=2/3
1*(1/3)/(2/3)=1/2
P(A|B)=The probability the player selects the car door given the host selects a goat door is 1/2. It is not advantageous to switch, given advantageous is presumably defined as increasing your likelihood to get the car.

Why is this so difficult for people to understand? I'm trying to explain this to someone and they aren't getting it.

alright fellas

>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

So switch or not, or does it not matter?

The people who say 50% probably don't know what the problem is even asking them.

Whenever you make a decision you should think how to get a bad outcome then change your original decision.

Two questions for /sci/:
1. I write a program that randomly picks a number C in {1,2,3}, the position of the Car, X in {1,2,3}, called the chosen position, and randomly Y in {1,2,3}\{X}, i.e. any number in {1,2,3} except X. Not counting the cases where Y=G, what does the ratio of cases where X=C to the total number of cases tend to?
2. I write a program that randomly picks a number C in {1,2,3}, the position of the Car, X in {1,2,3}, called the chosen position, and randomly Y in {1,2,3}\{X, C}, i.e. any number in {1,2,3} except X or C. What does the ratio of cases where X=C to the total number of cases tend to?

>>10636184

In the normal game, Monty knows which door has the goat and which has the prize.
And it's assumed that you know that he knows.

But what if we remove this assumption, so that now, you don't know that he knew where the goat was, when he opened one of the doors.

Would you switch in this case?

Both of the remaining doors have a 2/3 chance of giving you a car the moment you open one of them

Explain this to me - in the Monty Hall problem, you assign the initial probability of winning to be 1/3 because you have 3 doors and no info/bias about what's behind.

This is divided into p(door 1) = 1/3 and p(door 2 U door 3) = 2/3

Since it's impossible to win on both door 2 and 3 (mutually exclusive), you can write

p(door2 U door3) = p(door 2) + p(door 3) = 2/3

When one of the doors is opened, people say that you should switch because now the other choice has the probability of 2/3.

But how is that true? One of the doors now has probability 0 of being the winner.

Let's say you pick door 1.

p(door 1) = 1/3
p(door 2 U door3) = 2/3

Door 3 is opened, it contains a goat. i.e p(door 3) = 0

Which means that p(door 2 U door 3) = p(door 2) + p(door 3) = 1/3 + 0
p(door 1) = 1/3

But the total is not 1, so we need to change the probabilities for door 1 and 2. There is no reason to believe that door 2 has a higher probability of winning, so p(door1) = p(door2) = 0.5

The game show host opened all the doors that had goats behind them. Before he opened the doors, there was a 1/3rd chance.

After he opens the doors, let us imagine that the contestant is swapped out for a new contestant who has no knowledge of what occurred previously. This new contestant now has to choose between the two doors. The probability of winning is 50/50.

Probability of an event is decided by the internal structure of the event, not by subjective knowledge of outside factors. You can't smuggle in the influence and knowledge of a person playing a game into the mathematical probability of an event occurring, that's pure nonsense.

Fake science/math thread?

Pic related.

Is there more stuff like this?

The goat either IS or ISN'T behind the door. If you take out the choosing door part, it's clearly 50% chance. You can't refute this.

If you flip a coin, it's heads or tails.
If you pick a card it's red or blue.

Get your shit straight!

>>9328379
>>the greatest statistical problem of all time
*blocks your path*