Let's see if you guys are less retarded than Sup Forums. What is the answer to this question?

Let's see if you guys are less retarded than Sup Forums. What is the answer to this question?

Most of Sup Forums got it wrong. No surprise there.

Other urls found in this thread:

en.wikipedia.org/wiki/Bertrand's_box_paradox
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33%. Because that's the chance of you picking the box with 2 balls the first time.

2/5 ?

Nope.

Nigger, get out with your troll. That is correct.

Nope.

The answer is 2/3.

1/2. You're in one of the two boxes with a gold ball, and they're both equally likely

No it's not, you moron. Monty Hall paradox only comes into play if you get to switch boxes.

It's 50/50 for sure

Mathematical proof.

P(2nd Gold | 1st Gold) = P(2nd Gold and 1st Gold) / P(1st Gold)

P(1st Gold) = 0*1/3 + 1/2*1/3 + 1*1/3 = 1/2

P(2nd Gold and 1st Gold) = 0*1/3 + 0*1/3 + 1*1/3 = 1/3

Thus

P(2nd Gold | 1st Gold) = 1/3 / 1/2 = 2/3

Negative. It is 33%. Because that is the chance you have of picking the correct box the first time. 1 in 3. Doesn't matter if the first ball was gold, it's still a 1 in 3 chance.

This.

lolnope

This is a different famous problem. It's the Bertrand's box paradox.

Nope.

So far 0 people have gotten it right on Sup Forums. WTF.

0% cause that box only has a silver ball.

>2 Silver Box no longer matters

>2 Boxes
50/50

inb4 op put some retarded hidden message in this to make this answer wrong.

Wrong.

Prove otherwise then faggot. I know it's right. Refer to: wut

Nope.


Still 0 correct answers.

It's 33% from the beginning, but when you pull the first gold ball you have more information, saying that you are either in the box with one gold ball or the box with two gold balls. You are in one of two possible boxes, so the probability is 0.5

It's literally 1/2 because you're guaranteed a gold ball you dumb nigger

Nope both wrong.


Still 0 correct on Sup Forums.

2/3 god damn it

en.wikipedia.org/wiki/Bertrand's_box_paradox

you dense motherfuckers

You already told me that out of 6 balls, I don't have to worry about 2 of them because they don't have a gold
i took out one gold ball, there are now 3 balls left, a silver and 2 golds = 2/3

>inb4 op put some retarded hidden message in this to make this answer wrong.

/thread

Don't care what your pseudoscience answer is.
The probability of it being the 2 silver box is 0.
Meaning the other 2 have a 1/2 chance each of being the box your hand is in.
If there's 2 boxes you could be in it makes you chance of receiving the ball from 1 of them one out of two.

Lol, you have the right answer, but you are completely wrong in your reasoning. How can you be this retarded?

>3 balls left

That doesn't matter nigger. You have 1 box, and that box either has a Silver or a Gold Ball in it. Its 50/50.

op is just a faggot.

50%

1 correct answer. First!

All the info is there.

You're wrong. It's en.wikipedia.org/wiki/Bertrand's_box_paradox

Wrong.

Wrong.

>next box has the gold = next ball is gold
>thinking this is the same

>calling me retarded when you couldn't even describe the reasoning without doing 2 hours of research on wikipedia

holy shit it's 50/50. you could have picked out of 2 boxes. so there is two boxes you could have picked the second out of. therefore the second ball can only come out of the first two boxes so a 50/50 shot of being a gold ball

>"Don't care what you pseudoscience answer is"
>Links pseudoscience
Just because a bunch of people came to the conclusion that the probability of being in one of two boxes isn't 1/2 doesn't make it remotely true, any laymen or gambling man would assume 1/2.

i'm wrong your wrong there's only 2 balls per box you bitch

It's super easy problem. You misunderstood the problem for starters. You only get 1 box. After you choose your box, you cannot pick from the other boxes. Common, step 1 is understand the problem.

Wrong.

Refer to:

>4 balls remain, 3 golds and a silver
>take one gold out
>2 golds and silver left
>2/3

>"completely wrong in your reasoning"
Are you mentally unsound, moron?

Lol pseudoscience. I will write a simulation in Python to prove it.

In the mean time give me a mathematical proof.

Retard.

looked up Bertrand's box paradox, this shit is why I hate statistics.

Read the problem more carefully. That's not it at all. That's not why it's 2/3.

Yup that's the problem. It's been proven as 2/3.

the probability of the first ball being gold is 100% dumbass. its in the problem, you just told me that one ball is gold. so we know that it can't be the last box

1/2

2/3 if the first ball chosen stays out of the box.
Otherwise, 5/6.

See

Agreed. And the proof bears that, notice the 0*1/3 in both equations. Wrong.

You did not understand the problem at all. That's not why it is 2/3. Please read the problem. Reading comprehension is key.

It's easy to make a mistake when writing a simulation as proof.
I once wrote a simulation of this problem, then realised there were multiple interpretations, and my simulation only covered one interpretation (which was not the one I had in mind)

If you still can't understand how you LITERALLY can only be drawing 2 possible conclusions you're a fucking moron.
You're in 1 of 2 possible outcomes, the third box is irrelevant.

I will code exactly what the game is and then we will run the game 10000 times and see how many of the trials end up with 2 gold balls given that the first ball picked was golden.

You can review the code.

0% because ur all ball grabbing faggots

There are SIX possible ball grabs at the beginning of this problem
Balls 1, 2, 3, 4, 5, 6 (labelled form left to right)
Balls 1 and 2 are gold and in Box 1
Ball 3 is gold, Ball 4 is silver, both are in Box 2
Balls 5 and 6 are silver, both are in Box 3

The odds of any given ball being drawn are 1/6.

After this occurs, we check the ball we've drawn, and see that it is gold. this eliminates the odds that we took ball 4, 5, or 6, so now there is a 1/3 of having drawn ball 1, 2, or 3.

Balls 1 and 2 lead to drawing another gold ball, ball 3 leads to a silver ball. The odds that we drew ball 1 or ball 2 is 1/3 + 1/3 or 2/3rds.

I agree with that. But work it out further now. Knowing you arent in box 3, whats the probability of picking a second golden ball given you have one in your hand.

Retarded question.
If it is asking what the probability is post-gold ball then it's 1/2.
If it's asking what the probability is in total, which is useless. It's 2/3.

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?

Read the problem retard. OMG how can you not even read. Holy shit. How can you solve problems without understanding its premise first. FFS.

It's 2/3, but not at all for the reason you said.

The problem I'm seeing is that you aren't drawing the second ball from three choices. You only have one choice if you are drawing from the same box. In that case you are either drawing from the box with two gold balls or the box with one gold ball, in which case the probability is 1/2. If you were drawing from any of the three remaining balls it would be 2/3.

>Both balls in the first box counted as separate probabilities despite being in the same box.
Retarded considering you pick the box, not the specific ball.

You're wrong. You didn't understand the rules of the game if you think that.

Switch. Always switch. (2/3) chance if you switch as opposed to your 1/3 chance door.

This is a different problem however.

>This is a different problem however.
I know. I thought this was the fuck-people-who-can't-into-statistics thread.

It's 2/3. Read the Wikipedia article.

In short, if you draw a gold ball, you are most likely to have drawn it from the first box. 2 of the 3 ways you could have gotten a gold ball would be by having taken it from the first box.Thus you have a 2/3 chance of having chosen the first box, and thus a 2/3 chance of getting another gold ball.

Ok post more problems.

2 correct answers up until now.

>IF that happens to be a gold coin
OPs gay-ass pic says it IS a gold ball so the correct answer to OPs pic is that OP is a massive faggot

...

Correct, makes 3 right answers

1/2
>Inb4 noo dumbass its "insert retard answer here"
We dont count the box with 2 silver. We are talking about the first and second one because only they can have gold. So from those 2 boxes, it is a split 50/50 chance you will get another gold. Again for retarts who have the attention span of a 2 month old, NOT 1/3, NOT 2/3, we are only counting the first two boxes because the with the third box, the question doesnt even make sense because THERE IS NO GOLF IN THAT ONE

Wrong.

There are 6 possible initial ball withdrawals, which can be represented as scenarios 1, 2, 3, 4, 5, and 6 (balls from left to right).
The question implies that the scenario is 1, 2, or 3.
Scenario 1 and 2 mean that the other ball in the box is gold.
Scenario 3 means that the other ball is silver.

It would be 1/2 if the first gold ball was equally likely to come from box 1 or 2, but the ball is taken at random, so to not weight the boxes by their number of gold balls would be to imply that you can magically make sure that the first ball is gold.

The realistic form of this interpretation is that you pick a ball at random, you put it back if it's silver, and define the scenario as having started if it's gold.

A simulation shows nothing though, because the problem is defining exactly what the game is.
You will get the answer you have in mind, unless you make a mistake with the code.
This isn't like the Monty Hall problem, which is easy to interpret but hard to calculate. This is hard to interpret but easy to calculate.

Wikipedia isnt accurate

You're retarded nigger.
a simulation will give perfect emperical evidence to probability because it's purely mathematical. It's like a numerical solution to an analytical problem.

Well we all know the birthday problem and the three prisoners problem, so I'm fresh outta problems (at least problems that are interesting to the average Sup Forumstard).

its 50/50..
you take out a golden ball, so you can put the 3rd box aside
with 2 boxes left and each with a different ball, there's 50/50 chance youll get another gold one

I like how you're keeping count. You really think you're a special snowflake don't you?

What are you saying. The game is super easy to understand.

You have 3 boxes.
> pick 1 randomly
> pick a ball randomly from that box
If the ball is not gold reject the trial
If it is gold, then count the trial.
> pick a second ball from the same box
If it is gold then that is a success

Total probability = successes / good_trials

Agreed?

Explain what I've done wrong.

That's a better way of explaining it.
Scenarios 1, 2, and 3 are good trials, scenarios 1 and 2 are successes.

Do this one:

If the probability of encountering a car on the highway in 30 mins is 0.95. what is the probability of seeing a car on the highway in 10 mins?

Assume constant uniform probability along the highway.

Had this at an intevriew

OP just say the correct fucking answer so these dumbasses can learn that its 50/50, or 1/2, whatever you prefer

That's just a retarded histogram problem.

Other user here: I like how you obviously were too stupid for this problem and now resort to insults.

▲
▲▲

Wrong.

You start by picking a box, not a ball. Read the problem please.

I think you are misunderstanding the problem entirely. Explain yourself better so I can address it.

You're looking at the two yellow balls as the same ball. They're not. They have the same color but they're still individuals god damn it.

Newfag try force harder.
>try force
>triforce
See what i did there

It's 2/3. I have already given a proof. Read the thread.

No It's not. Retard. Give an answer instead of saying it's easy.

My answer is explained in a different way by

ITT : people with Down"s syndrome that never even understood the Monty Hall problem

>You start by picking a box, not a ball. Read the problem please.

1/3 * 1/2 = 1/6
which is what I said.

Ok simulation results are in it's 66%. Feel free to replicate my result by runing the code. You will need to indent in Python for it to work

Python code:

success = 0
good_trials = 0
total_trials = 0

box1 = [0,0];
box2 = [0,1];
box3 = [1,1];

pick1 = 0;
pick2 = 0;

for i in range(0,1000000):
pick1 = 0;
pick2 = 0;
total_trials = total_trials + 1;

box_pick = np.random.randint(1, high=4, size=1);

if(box_pick==1):
pick1 = box1[np.random.randint(0, high=2, size=1)];
pick2 = 0;

if(box_pick==2):
pick1 = box2[np.random.randint(0, high=2, size=1)];
if(pick1==1):
pick2 = 0;
good_trials = good_trials + 1;
else:
pick2 = 1;

if(box_pick==3):
pick1 = box3[np.random.randint(0, high=2, size=1)];
pick2 = 1;
good_trials = good_trials + 1;

if(pick1 ==1 and pick2 == 1):
success = success + 1

print(success/good_trials)

Wait, doesn't the answer depend on which gold ball you drew though? Making it insufficient information to answer?

Can someone tell me why this is wrong?

I doubt it. You seem to be unable to explain yourself. I think you misread the problem entirely to be honest. Don't lie to yourself, challenge yourself until you truly get it.

Not the same problem retard.

Why *1/2?

lol im guessing you didnt get the job.
should have paid more attention in class.

Theres no number difference between what ball you choose if its from the same box.

No, there's enough info there.

First you pick a box then you are stuck picking from that same box for the second ball.

50%

I did get the job.

Wut??? What's your answer?

>Why *1/2?
Once you've picked a box, there's a 1/2 chance of picking either ball within the box.

If I pick box 1, there's a 1/2 chance of ball 1 and a 1/2 chance of ball 2
If I pick box 2, there's a 1/2 chance of ball 3 and a 1/2 chance of ball 4
If I pick box 3, there's a 1/2 chance of ball 5 and a 1/2 chance of ball 6

This. It's 50/50

Yeah, that way its 1/2

Wrong.