Is Jow Forums intelligent?

I remember last time this was here many of you showed a lack of intelligence with your answers. I hope that's not the case anymore...

Attached: problem solving.png (1272x2096, 1.04M)

Other urls found in this thread:

martingalemeasure.wordpress.com/2014/02/02/monkey-typing-abracadabra-14/
en.m.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/NSFWRedditGif

I don't think the issue is people's intelligence so much as the notions of probability and randomness are epistemological but never explained as such bc statistics and probability courses are taught kind of shittily.

>unmarked coins
>the qeustion is about balls
>the boxes aren't marked either
Whoever made this picture is a retard.

This. Statistics don’t come intuitively to the majority of people, even if they have a background in other math like trig and calculus. It’s also taught horribly or not at all, at least in the US of A.

Well you're wrong, Monty Hall wins, nigger

Wait, there are actual retards out there who thought it's 2/3? Those people live among us? Jesus christ, how horrifying.

>t. 66% brainlet

Well, first you have a 33% chance of picking the right box, then you have a 50% chance of having the box with two gold balls, 50 + 33 = 83, so you have an 83% chance of getting two gold balls.

>83%
wew

he's right you know

A = Event that the second ball pulled is gold
B = Event that the first ball pulled is gold

P(A | B) = P(B | A) * P(A) / P(B)

P(B | A) = I will say this is 50%, you are handed one of two boxes and it is random which of the two you get.

P(A) = I will say this is 75%. Each of the 4 balls is equally likely to be chosen.

P(B) = Same as P(A), each of the 4 balls is equally likely to be chosen.

So, (0.5 * 0.75) / 0.75 = 75% chance.

You don't need any formal statistics education you just need to be non retarded.
If you're not just pretending to be retarded please do not reproduce.

Attached: 1536895596057.png (324x362, 45K)

read the question sweetie

Attached: 1513583451520.png (403x448, 53K)

P(B) is 1 your nword

>brainlets calling people with the correct answer brainlets
epic

How about this one. A monkey sits at a typewriter typing letters a-z at random until it types ABCDEFGHIJK. Is the expected time taken shorter, longer or the same for the monkey to type ABRACADABRA instead?

>75%

ok so we had a 83% and a 75%.
we're getting closer, just need to pass the 66% bridge and then we will be on track to the correct and one and only answer: 50%.

You can do it darlings.

Attached: 1515348341390.png (488x463, 28K)

No fucking shit the second is easier. How is this not obvious to anyone?

depends.
can the monkey touch type?

Show his "random" keyboard heatmap

it's a 50% chance you picked the box with 1 gold and 1 silver ball, but if you did the chance to pick up another gold ball is 0%. So the answer is 50% * 0% = 0%.

If you truly mean random, then neither possibility has an edge over the other in terms of which comes first. So the answer is "same"

It's not option 1 twice, they are two separate things.

There are three gold balls, so if you pick one, you are in one of three states, and two of those (picking the ball from the two gold box) leads to the next ball picked being gold.

Do you actually believe the answer is 50% or is this just a troll

ABCDEFGHIJK is faster, because.

same. both strings are of same length. in fact, he could type 11 "A"s and still have that same probability, for example.

No they're not. If you pick from the box with two gold balls the end result will be the same, another gold ball. You would have a point if it was a gold ball and a gold cube.

And I thought this board wasn't full of brainlets. Though I should have realized that from the fact that people truly think the answer to OP is 50%.

Can you actually not understand the question or are you just a troll?

that's not how statistics work
that's like saying if there's 99 red choices and 1 blue choice then there's a 50/50 chance of each because all red choices are the same

>answer to OP isn't 50%
>calls others brainlet

epic.

Attached: 1523894679170.jpg (600x600, 57K)

The precondition represents 3 possibilities, not 2

No. Read the question. Understand that there is a box.

Of course OP would post a problem about grabbing balls. I'm not gay and I'm not falling for this trap.

The fact that there is a box is irrelevant. There's 3 possible outcomes you can arrive at before taking out the second ball

The precondition says that you already picked a box that has one gold ball in it. The question is what are the chances that if you pull out another ball from the same box, it will be gold too? Is everyone Indian in this thread?

Attached: 1515598922348.jpg (465x509, 41K)

What an awful fucking question. If the thought of writing a book with exercises in probabilty ever crosses your mind, please do the world a favour and shoot yourself.

>There's 3 possible outcomes you can arrive at before taking out the second ball
What's is the third?

Attached: ur_a_niger.gif (320x240, 2.36M)

Yes, and there's three possible ways of reaching that outcome
Box 1, Ball 1
Box 1, Ball 2
Box 2, Ball 1
Not sure why you think you can fold the first two into one just because it's the same box

>Forced to take from same box
>Box is irrelevant

Attached: brainlet.jpg (645x729, 48K)

If you found a gold ball already wouldn't it be 50/50 since only 2 boxes have a gold ball and one has 2 gold balls and one is silver.

It's 2/3. The possible outcomes for the experiment are GG, GG, GS, SG, SS, and SS. You can eliminate SG, SS, and SS because you know the first ball was gold, leaving you with GG, GG, and GS; out of these three, GG and GG satisfy the problem's success criteria, so the probability is 2/3.

Is this a trick question in which the monkey is only typing lowercase letters so it would never type the given strings? If so, then the answer is "same"

No.
Outcome 1: Gold Ball 1 in hand from Box A
Outcome 2: Gold Ball 2 in hand from Box A
Outcome 3: Gold Ball 3 in hand from Box B

The question asks what is the probability that the next ball will be gold from the same box.
Outcome 1 and 2 is the fucking same, because you would get a gold ball from either, the question isn't assigning numbers to the balls like I am or you, and it isn't asking for the probability of that particular gold ball.
The question is asking the probability of you taking another gold ball from the same box.
Ball 1 or 2, is Box A
Ball 3 is Box B
Box A or B, your hand went into one of those, doesn't matter what number the gold ball was. So either you are in Box A and will take another gold, or Box B and take a silver.

/thread

Oh it says the anwser is at the bottom of the pic. Well at least I got it right. Next time op just put the question and no anwser, it's more fun that way.

martingalemeasure.wordpress.com/2014/02/02/monkey-typing-abracadabra-14/

The boxes are irrelevant you retard. Imagine if there are no boxes, just 3 gold balls and 3 silver balls, each labeled 1-3, all in one box.

>The boxes are irrelevant you retard.
>The boxes are irrelevant you retard.
>The boxes are irrelevant you retard.

Attached: e02.jpg (645x588, 33K)

It's threads like these when I realize Jow Forums is full of retards and I should never ever take any advice on here seriously.

Attached: 1536712475356.png (640x480, 397K)

Even of you just chose a random box for a golden ball without chosing one before it would still be 50/50. (3/6 which is the same thing. 6 outcomes and 3 good ones) Why are people saying 66% it's retarded.

People saying that it's 2/3 are missing that the problem states that you pick a BOX at random, not a BALL at random. The third box is eliminated from the get go (it's a red herring), and you don't have a higher probability of picking the first box over the second box, since it's the box that you're picking and not a gold ball.

DING DING DING DING
we have a winner

Good job for actually being able to comprehend English language unlike the many brainlets in this thread.

You flip a coin. It's heads.
You flip it again. What are the odds of getting heads again?
Another student in uni asked me that. I answered 50%. He laughed and said 25%. I had to spend the rest of the break to explain to him why he's wrong. THIS WAS IN UNI. Not surprised this place is full of brainlets too. The world is doomed.

First you have a 33% chance of getting the right box, then you have a 66% chance of having chosen the box with the two gold balls as soon as you know the box has at least one gold ball in it, 33 + 66 = 99, so 99%.

>Outcome 1 and 2 is the fucking same
they aren't, because they weight the probability
If you find you have a gold ball, you are more likely to be in Box A than Box B because it has 2 gold balls instead of 1
You can write a program to demonstrate this

How do you know the coin isn't heads on both sides? Retard.

The question says you pick a box at random then a ball at random

Anyone who uses these type of photos and claim others are not intelligent is pretty embarrassing

Because the student wouldn't have said 25% you cretin.
It's the fucking box.

So if the first box had 999 gold balls, and the second box had 1 gold ball and 998 silver balls, the answer would still be 50%?

>It's the fucking box.
what?

No, you're supposed to start off with the assumption that you have already taken a gold ball out whichever box you've picked.

So you chose a box already and you got a golden ball. There are only 2 boxes with golden balls. Box one has 2 golden balls and box 2 has one gold and one Gray. Since you already chosen a box and got a golden ball that means you have 50/50 of getting another golden ball. Box 3 or 2 silver balls is not chosen since it has no golden balls.

no the answer would be a very high % because it's much more likely you're in box A

Yes

You pick the fucking box. It doesn't fucking matter how many balls or dicks or vaginas there are in the box, you choose the box. the box chosen had a gold, HAD IT BY CHANCE, played no part in the probability of that box being picked, because you're picking the box not the fucking balls inside.

no, you have a 66% chance of getting a gold ball because if you got a gold ball in the first place it's more likely that you got Box A than Box B in your box choice

So when you select the second, box, you are somehow magically always able to pick the gold ball from it with 100% probability, even though there are 998 silver balls in it?

you pick a box at random and then you pick a ball at random and it's gold, it's more likely that you picked the one with two golds in
should I write a program to show you?

It would be different then. If you got the box with one golden ball and you already got that one then you have 0% of second ball. If you chose golden ball that only has one silver ball then it is 99%. It is not a perfect 50% chance overall due to the 99%. So it would be 49 for golden and 51 for silver.

The first box has no silver balls.

No. The problem postulates that you picked the box at random, so it's 50/50.

It's 50%.

I don't get how anyone got otherwise.

The box has been picked. Check.

The ball taken is gold. Check.

Silver and silver box is out then. Check.

That means either you are in Gold Gold or Gold Silver box. Check.

So Either Gold next or Silver. Check.

Why y'all stressing on the probabilities of the balls altogether and their chance of being picked? That's not the question.

Attached: 1523509782672.jpg (236x231, 11K)

The problem states you pick a ball at random from the box. And yet you state that if you pick the second box, you always pick the gold ball. Do you not see the contradiction?

the problem states that you pick a box at random and then you pick a ball at random
am I being trolled?

If you pick a box at random and then a ball at random from the box and it's gold and someone removes the box with 2 silver balls and you want to pick up another gold ball, should you switch box or pick the other ball in the same box?

You’ve 6 balls: 3 yellow ones and 3 gray ones represented in a fraction of 3/6 for either color (or 1/2 once reduced) Therefore, 50% of the balls are yellow and 50% of them are gray. You’ve a fifty percent of picking either one

Now the question is: after choosing (and subtracting) one yellow ball from the mix, what are your odds of picking another yellow ball?

You’ve now two yellow balls and three gray balls. The remaining yellow balls can be represented in a fraction of 2/5 or
.40 or 40%
You’ve a 40% chance of picking an yellow ball.

2/3rds or 66% is an impossible answer. How do your chances increase when you’ve less yellow balls?

No, the problem states that you picked a box at random _and there was a gold ball_. If I had you choose between a box with a 1000 gold balls and one with 999 silver balls and one gold ball, then you drew a gold ball from the box, which box have you most likely selected?

Technically you already chosen a ball and it removes box 3 all together. To chose box with 2 golden balls is 33% if you started without picking anything. It increases to 50-50 since you chose a golden ball already and there is only 2 outcomes. It would be 66% if you never chosen a ball and you wanted to know the chances of getting only one silver or only one golden ball. But since a golden ball has already been chosen and you are looking for 2 golden balls it would be 50-50 at this point.

Please look up conditional probability, the fact that you drew a gold ball changes the distribution for which box you picked, and hence the final answer.

You pick the box at random so that's 1/3 for each box.
You pick a ball at random. Since you know that ball is gold, you know you have either a gold ball from 1 or from 2. If you picked box 1, your other ball is gold. If you picked box 2, it isn't. You were equally likely to pick either of those boxes, so it's 50/50.

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

>there is only 2 outcomes
there's 3, as people have already pointed out

You're mixing up the steps. You pick a box at random AND draw a gold ball from it. That's the setup.

You pick a box at random. Let's say you pick box 3.
You pick a ball at random. Let's say you pick ball 2.
The ball is silver. Now what?

>You were equally likely to pick either of those boxes
You weren't. Given the fact that your first ball is gold, there's a 66% chance that it's the double gold box

Then you pick again because the first ball isn't gold, numbnut.

>Now the question is: after choosing (and subtracting) one yellow ball from the mix, what are your odds of picking another yellow ball?
That is not the question of the OP.

Someone should make a straw poll to see how many retards or ESL people are on Jow Forums

So obviously the correct answer is to pick the same box, but I wonder what 50%ers make of this, if they think staying give them 50% chance of another gold ball, is switching always correct then? Since switching is now 66% chance of a gold ball.

I was given this problem in an interview for a >$130k job at a fortune 500 tech company. I said 50%. I got hired.
Take that as you will

Do you pick again starting from picking a another box, or starting from picking a ball in the box selected?

Jow Forums struggles with basic calculus, I don't know why I'm surprised so many anons refuse to accept basic probability theory

It's obvious that you chose the boxes with equal likelihood. Now if you chose box 1, you get another gold 100% of the time. If you chose box 2, you get another gold 0% of the time. This gives a net probability of 50%.

This thread is quickly making me lose faith in humanity.

The trick is that the monkey starts a new string every time it mistypes, if the monkey were to write 11 character strings that are then compared then the odds would be identical

Someone set up a gambling site involving this problem, I want to rinse the 50% brainlets for all they're worth

>user literally posts the answer with evidence
>brainlets continue to argue about the answer despite the FACT that it is 2/3

Attached: trip1.png (625x625, 658K)

Boxes[][] = [[1,1],[0,1],[0,0]]

For i = 1 To 100000

pickedBox = Rand(0,2)
pickedBall = Rand(0,1)

//if the picked ball is gold, this is in the subsets of results we're looking for
If Boxes[pickedBox][pickedBall] = 1 Then

//if the other ball is gold, it's a pass, otherwise it's a fail
If Boxes[pickedBox][1-pickedBall] = 1 Then passes += 1 Else fails += 1

End

End

Print("Passes: "+passes)
Print("Fails: "+fails)
Print("Ratio: "+Float(passes)/Float(fails+passes))

Passes: 33565
Fails: 16704
Ratio: 0.667707741

you can just test this shit theory in like 30 seconds, run this as many times as you like or increase the sample size. it is 66%

Attached: Screen Shot 2018-12-25 at 9.32.42 PM.png (1550x724, 150K)

>It's obvious that you chose the boxes with equal likelihood
wrong.