ONLY >120 IQ WILL GET THIS

ONLY >120 IQ WILL GET THIS

Other urls found in this thread:

en.wikipedia.org/wiki/Monty_Hall_problem
en.m.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/SFWRedditGifs

2/5 chance

40%

50/50 faggots

It's asking what the probability is with 2 gold left and 3 silver left are you stupid?

If you pick a gold ball, you know you can only be dealing with Box #1 or Box #2 (from left to right). The chance that you will pick another gold ball is 50%, because Box #1 has two gold balls and Box #2 only has one.

10/10

But what's stopping you from picking a ball from box #3?

In other words, if you already know that you picked a box with a gold ball it's a 50/50 shot that you picked the box with a second gold ball, because you only have two options of boxes with ANY gold balls.

I agree with this because really you know you picked one of two boxes and only one of those two will get you a second gold ball so 50/50

because you already got a gold ball and you should take a ball from the same box
that cant be box 3 then

150% im asian

do u even read?

>The next ball you take from the SAME box

He's right, it's 50/50

One ball is gold, the next you draw will be either the other gold from box one, or the silver from box 2.

This guy gets it. The key phrase here is "the same box".

It says, "What is the probability that the next ball you take FROM THE SAME BOX will also be gold." You've already taken a gold ball out of the box for your first go, so that already eliminates the possibility that you chose Box #3.

2/3 chance

$400,000

Because you've obviously already chosen a ball from either one or two

ITT: only a single Sup Forumstard can critically think

TLDR

What?

1/2
Past outcomes don't affect future outcomes anyone saying anything other than 50% needs to wait to answer until they finish 12th grade and take a stats class.

50%
And that makes my IQ 120+ how?
Oh I forgot it's the internet

3 different gold balls to choose. two ways to choose them so that the one in the same box is gold. 2/3 ezpz

CORRECT.
congratulations.

en.wikipedia.org/wiki/Monty_Hall_problem

Oh man, fail

Desired outcome over potential outcome
2/3

wow this is just as good as the monty hall problem with all these confident and wrong motherfuckers.

it's 66% kids.

There are three situations where you picked a gold ball.

[A, B] [C, D] ABC are gold, D is silver.

If you picked A, you win, if you picked B, you win, if you picked C, the other ball in your box is silver and you lose. 2/3

Dont tell me yet, I'm thinking


>pic related

2/3

The monty hall problem doesn't apply here you fucking retard. You aren't allowed to switch boxes.

Wrong, they don't want the overall chance and still then it would be still wrong.
3 boxxes->i you choose one you have to stay with it->only kne box with two gold balls so it's 1/3 chance

man getting the wrong answer I can understand
but getting the wrong answer and then acting like you're smart is the worst

i estimate ~39%
it gets complicated. theres a 33% chance to have a 75% chance to get have 1 silver ball. so its 50+/- 25%

lol, nice troll.

there's not enough information, guys.

>tries to be smart
>posts something completely unrelated
What this nigga said

Is this a meme? Literally no one is saying 1/3 which to me seems like the only possibility. The general consensus here is 2/3 which is fucking retarded given that there aren't 2 opportunities for the desired outcome possible. Am I taking the bait in an elaborate predetermined troll?

correction i missed a word in this problem. please disregard previous guess,

The Monty hall problem only applies to switching boxes. You are stuck with your choice per the image.

You could be drawing from box A or B with equal chance, this is all the info you gain from the first draw of a gold ball. The remaining ball only has two possible colors wit a 50% chance of it being sliver or gold.

If you could switch boxes the other box would give you a higher probability of winning but since you can't change your choice the odds are 50-50.

If you think the statistical concept is unrelated you're pretty fucking stupid.

it is literally 2/3 m8. This is no troll.

If the box chosen was silver/gold box, then 0%. If was gold/gold then 100% if was silver/silver then op is a lying faggot.

R u retarded

Read more carefully next time. Picking a gold ball is deterministic.

There are two ways to approach this problem: Find Pr(Gold) | Gold or find Pr(gold) | pr(gold)

In one you know that you pick gold, in the other you must find the odds of picking gold and then gold, eliminating cases where you pick silver.

The question is ambiguous. If you asume gold first to be certain then the odds are 50/50.

If you assume gold to be randomly sampled and you are only observing cases where you pick gold then in 2/3 of the cases you pick gold first you are in box #1. Putting the odds at 2/3.

Both 50/50 and 2/3 are valid answers because the question is bad.

2/3 fag here
they are only related in the fact that they involve 3 boxes and dummies can't solve them
you are pretty fucking stupid

No. 2/3 is the only valid answer.

It's 1/3. It's a variation of the Monty Hall problem. Because you picked it when there were 3 choices that's what makes the difference.

HURRR HAS ANYONE EVER HEARD OF POISSON DISTRUBTIONS THEY TOTALLY APPLY HERE

It is because you can't fucking switch after choosing one box.

you were right about the 2/3 part and that's literally it
your explanation for why it's right isn't even right

Since you are picking to start at a 1/3 chance that it will be both gold then you still only have a 1/3 chance that it was both gold even after you pick out one ball.

nope, 50/50. there are only 2 boxes with gold balls in them and you know you picked a box with at least one, so it's 5050 whther or not the 2nd is gold or silver, the silver/silver box is a red herring, it doesnt change the problem at all

3 boxes G1 G2, G3 S1, S2 S3

wath is asking you is in with box is the hand

so if you dont take a a silver ball it cant be SS

for first box g1 g2 p= 2/3 ;
for second is g3 s1 p= 1/3;
for last one is s2 s3 p= 0;

the pro of get an extra gold is 2/3

2/3 standard Bertrand box question

I implore you to get three boxes and 6 coins. You will find you cannot get a 2/3 sample unless if you disregard cases where you pull silver before gold.

There are three possibilities if you picked a gold ball.

One, you picked Gold Ball A from Box 1.
Two, you picked Gold Ball B from Box 1.
Three, you picked Gold Ball C from Box 2.

If one of the first two scenarios is true, then there the next ball from the same box is gold.
If the third scenario is true, then the next ball is not gold.

The possible outcomes for this experiment are
1,G,G
1,G,G
2,S,G
2,G,S
3,S,S
3,S,S

Where the number represents the box chosen, the first letter represents the first ball picked from that box, and the second letter represents the remaining ball that must be picked from the same box.

Since the first ball was gold, we eliminate all possibilities where the first letter is S, leaving us with:

1,G,G
1,G,G
2,G,S

Thus, 2/3 of the time the next ball picked from the same box is gold.

The trick here is that the question is making you think that there is an equal chance that you picked Box 1 or 2, but there is not. If you picked a gold ball, there is already a 2/3 chance that you picked box 1 and your next draw is automatically gold (100%) and a 1/3 chance that your next draw is automatically silver (0% gold). Combine these odds and you get 2/3 chance of gold.

If you read this and don't get then I feel very sorry for you.

>still thinks the boxes and silvers matter

This.
You have 2 outcomes- 1 gold or 1 silver.
It either is or isn't.

50/50

2 to 1 chance that itll be a gold ball

>you put your hand in and take a ball from that box at random. it's a GOLD ball. what's the probability that the NEXT ball you take from the box will also be gold

i will disregard cases where you pull silver before gold, because that's not what the question is fucking asking for

Let an idiot be wrong. Too bad this kind of shell game is illegal or I would make a killing doing this.

"Bruh, you picked the silver ball, but hey there's two possibilities, you either have a gold ball left in there or you don't! Wanna double your bet for another chance??"

Simulating this on matlab gives 2/3

You have a 50/50 chance in yhe context of the question because you start choosing from the second "round".
You get the 2/3 chance if you had actually started choosing from the beginning, because of the odds of succession.
In the question there is no benefit from succession because you never got to make that first choice.

Please, no more of this bait-tier autism.

20%

EZ

There are three boxes shown in the image. Gold balls are represented by G and silver balls are represented by S. Let's call the boxes A, B, and C corresponding to GG, GS, and SS.
Upon retrieving your *first* gold ball you now know definitively you are reaching into box A or B since box C, or SS, has no gold ball.
This means that the box you reached into is GG or GS. You are holding a gold ball so the box has an equal likelihood of containing a G or S.
I can't possibly explain this any more clearly it's 50/50 split.

explained well

50%, read the question

Yep

Where you failed is where you forgot about the OTHER gold ball in box 1.

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


Enough fucking arguing. This should end it

This

Let me simplify this for you retards. The first gold ball most likely comes from the first box therefore its most likely you picked the first box

Thank fuck someone who can read the question.

If your answer is 50/50, you must reply to this post or your mother will die in her sleep tonight.

If your answer is 2/3 you must reply to this post or your mother will die in her sleep tonight.

fk you user

you forget the scenario where you choose the second gold ball from Box A, so you have a 33% chance of gold, 33% chance of gold, and 33% of silver
add it up and you get 66% ~2/3
but this takes a broader view to this problem but the question asks for the specific situation so i might be wrong
heil hitler

You literally have it opposite.

I literally drew the hand holding it for window licking retards like you but you somehow managed to be too fucking dumb to explain this to regardless. I thought to myself do I need to draw the hand or will they be smart enough to know the other gold ball is in the guy's hand and somehow I decided Sup Forums was so fucking retarded they would need to see the hand to know it exists. You saw the hand and still couldn't tell it existed. never go full retard.

Fuck, beat me to it.

It's 2/3 are people arguing otherwise?

Try this one instead

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

...

You're staring your calculartion before he picks the ball. The question is asked while the guy has already determined that he is holding a gold ball and therefore is reaching into a box that definitely has a gold ball. If he had asked before you chose one sure, BUT OP FUCKING DIDN'T

>3 gold balls
>each gold ball can be drawn equally
>still thinks its 50%

You are retarded

you know this thread was the boost i needed to finally kill myself, thank you OP

click on the wiki you autist

This isn't bertrands box paradox.

Bertrands box says that you are exclusively grabbing at random.

This one is implying that YOU ALREADY HAVE a gold ball in your hand, and you are PICKING FROM THE SAME BOX again. Actually fucking read what it says before spouting bullshit.

>You're staring your calculartion before he picks the ball.

how can someone even reach this line of logic?

DONT KILL YOURSELF
ITS OK IF YOU THINK ITS 50
WERE ALL GOOD IN OUR OWN WAYS

1/3. Asking if you get gold, If it asked for silver thats a different question.

the point is that it does not matter the situation at hand because, the guy has a 2/3 chance that the first gold ball he chose came from box (GG), upon which he has 100% chance of getting gold again. 1/3 chance that the first gold ball came from box (GS) with 0% chance of getting gold again. It may seem like 50/50 but statistically it is not. this has been proven. check bertrand's box.
heil hitler

the silvers and boxes don't matter, only the choice of gold that was first chosen.

you have been bamboozled

fucking read what it says before spouting bullshit

>THE SAME BOX
some of you are fucking autistic

>You put your hand in and take a ball from that box at random. Its a gold ball

are you just trolling or are you fucking stupid? the question literally tells you the first ball you pick is a gold ball which eliminates any possibility of first choosing a silver ball.

therefore

>you always have a 2/3 chance of drawing a golden ball

is false

given youve already taken a golden ball the chances of the next one being golden is 50% due to the fact there are only 2 boxes where the possibility of drawing a gold ball first.

50%x2=100%

all adds up you fucking cuck

It's the intuitive answer.
We know that there are only two boxes that contain gold balls. If the box we pick already contains a gold ball, then the only choices we're left with are either a silver ball or a gold ball, so it seems like it's a 50/50 probability because we only have two possibilities.
However, the thing most people overlook (and I did, admittedly) was that if you picked a gold ball in the first place, the odds are better that you picked the box with the two gold balls instead of getting lucky with the box with only one gold ball, so you have higher odds of your next choice being a silver ball.
2/3 to be precise.