This should be easy, but still most of you will get it wrong

>If a man fucks another man,
>Either they will have a baby, or they won't
>Two possible out comes
>Figure it out

OP here, like i predicted, only 1 person has given the right answer, thats 1 out of 10 people, thats 10% of you.
90% of you are idiots.

so 50%. Yes or no

Give it a couple of shots and let us know.

the math checks out

shut up faggot

50%. You have picked either the box with one or two gold balls--the box with two silver balls was ruled out by your first selection.

OP reaches into box, pulls out a pair of black balls and starts sucking on them

27.6%

Did no one read that part were it says you take the next ball from the same box. So if you pulled out a gold then it's 50%

it could be a miscarriage, it could be twins, triplets or more, if they used protection the probability of a child is much lower, which is also the case if 1 of them is infertile.
And these are just the options i came up with, there might be more we dont even know about.

So far still only 1 person has gotten it right.

How fucking stupid are you guys?

You're asking the probability of getting a second gold, given you get the first. So the question is P(G2|G1) = P(G2&G1)/P(G1) = (1/3)/(1/3+1/6) = (
1/3)/(1/2) = 2/3

He gets it. 50percenters don't worry you don't, at least you're with the majority.

Jesus Christ b/tards

50%

Lol no.

You know that you’ve got one of the first 2 boxes.

You can throw out the third box completely.

Now it’s either/or.

50/50

you are wrong, it really is 2/3, i know this, i have evidence.

but before i give it, i want to see if you can figure out why it isnt 50/50.

What if each box had 100 balls, with all gold, one gold, and no gold, respectively?

If you pick a box at random and then the ball you pick at random is gold, then that's probably because you got the all-gold box. It's super unlikely that you picked the one-gold box and got the gold.

So when you pick another ball, it's very likely going to be gold again.

Indeed. And if you put letters on the balls: A B C D E and F. E and F are irrelevant because they're from the two silver box. So you pulled out either gold ball A, B or C. If you pulled out A then the next ball will be B (gold). If you pulled out B then the next ball will be A (gold). If you pulled out C then the next ball will be D (silver). So 2 chances for gold, 1 for silver. This might be another way people might be able to wrap their minds around this.

if you pull out A, the option of pulling B first is just as irrelevant as E or F. the draw is done, the alternatives do not matter. you can either draw the relevant gold or the relevant silver.

Spotted the sped

Back when I taught math it made me cringe so much seeing students waaaay overthink problems instead of simply REREADING THE QUESTION.

Same box son, same box...

100%
I'm taking all the boxes because I'm a nigger

1/3
cant be 50/50 because theres three boxes and only 1 box has two gold, another 1 gold and the last none. so its a 1 outta 3 chance you pic the box with two gold balls

The Ottoman Empire

I can't resist taking the bait
it's 50% you retard

50/50

Okay, say you pick the two-gold box at random. Then the change that you pick a gold *and keep going* is 100%. But with the one-gold box, that's only 50%.

So you're twice as likely to get to the point of even asking whether the other ball is gold if you picked the two-gold versus the one-gold box in the first part.

its neither 1/3 or 50%
and this is not bait, its not a trick question, it has a real answer, im just trying to figure out how few ppl actually get it and im suprised, not many have gotten it.

75 b/tards

dont hear that answer too often, how the the fuck did you arrive at 75%?

That's the second person to give that answer. Probably something like 3 out of 4 of the balls in boxes with gold balls are gold.

This again. It depends on weather you put the ball you took it back.

You can always tell who's stupid because they say %75 because they've seen the problem before, but can't into reading.

If you take the first gold ball away it's 50/50. You know it's not the two silver box and only one ball is left.

20% or 2 out of 5

theres two boxes with 3 gold balls and 1 silver, each balls worth 25%, i pick a gold ball, theres two more gold balls I could pic but only 1 silver

The way it's worded it is 50%. You don't specify you put the gold ball back in the box.

That is the correct answer IF YOU ARE PUTTING THE BALL BACK.

The original problem was about eliminating the box with no good balls from the calculation.

Ironically, the knee jerk no math answer (50%) is the correct one because the problem just says you take a gold ball, nothing about putting it back. There is only one ball in the box.

>not bait
Pretty sure it is
>not a trick question
Pretty sure it's not
>it has a real answer
Yea, it's 50%
If you think you know the "real" answer please post it so we may laugh at you

It's 2/3, which only three posts have gotten:
These anons are trying to explain it:

I think this is a good way to visualize the answer

Attached: IMG_20200322_153858.jpg (4048x3036, 1.41M)

58%

Ok, so the reasoning is that it is more likely to have picked the box with 2 gold balls in the first place?
sure, that checks out

1 of those 3 who got it is me, im OP and i know the answer lol.

Sure, This is an old problem called Betrand's Box paradox. its conditional probability which is really counter intuitive, humans are really bad at it.
youtube.com/watch?v=ytfCdqWhmdg&t=560s
This is the best video about it that i have found, there are more but its a really old problem so its not really talked about outside of math classes that teach conditional probability.

It's 50%. Probability of the next draw is independent of the previous draw. You're trying to make this a Monty Hall but it's not. The first pick getting gold was 50%. Then you have gold, there are only two boxes it could be. You either picked a box with another gold, or a box with silver. 1 gold or 1 silver are the only possibilities. You have not introduced any option for another box. It's 50%.

I have to agree with It's a matter of "logical semantics" if that makes sense. I chose not to include the probability of the first choice in my reasoning

Not simply that you were more likely to pick that box, rather that you were more likely to pick that box AND draw a gold ball

then you are talking about a different problem, in which case you should make your own thread.

Its not exactly Monty Hall, but its close, its conditional probability and we are TERRIBLE at taking the condition into account, watch the damn video i linked.

2/3. The one thing that all the 50/50 people fail to account for (because it's counterintuitive to a degree), is that you take a random ball from a random box. Ask yourselves this: you take a random ball from a random box. It is a gold ball. What is the likelihood there is another gold ball in the same box?

This is the same question, but worded differently. It showcases the conditional probability better than the original wording, although the content remains the same.

It's 2,000
I find it easier to consider running many simulations. So, lets say we're running 6,000 simulations of selecting a box and ball at random.
Each of the 3 boxes gets selected 2,000 times. Box 3 doesn't have any gold balls, so we can be sure this particular instance isn't one of those simulations. That leaves a possible 4,000 simulations.
Box 2 has 1 silver and 1 gold ball, so out of 2,000 simulations, there are 1,000 simulations in which we selected a gold ball from Box 2. While Box 1 has 2 gold balls, so there's 2,000 simulations in which we selected a gold ball from Box 1.
Therefore, out of the possible 3,000 simulations that we are now looking at, 2,000 of them are in Box 1 and 1,000 in Box 2. If we are in Box 1, there is a 100% chance of a second gold ball, and in Box 2 there is a 0% chance. So, in 2,000 out of 3,000 simulations in which we selected a gold ball on the first try, we will also get a second gold ball

whoops, I meant to say its 2/3 not 2,000... The 2,000 was from the simulation thing

yea that's what I implied by saying more likely to have picked
That is also what is fucky, the question is asked like you already have a gold ball, but the answer includes the probability of having picked a certain box.
It makes sense but it's essentially a trick question

Well since you worded it as THE NEXT ball from the same box will be gold, that must mean it will either be the box with 1 gold and 1 silver, or the box with 2 golds, so only 2 choices.

1:2

However if it was a cumulative probability figure from the start, then 1:6

>Be me, 14
>Doing a conditional probability problem for the first time
>Problem is something like, "The chance of thing 1 is X%, ..., what's the chance of thing 1 given other thing?"
>"It's clearly X%, it says so in the problem"
>Smart friend says I'm wrong
>"It's not asking for the chance of thing 1, but the chance of thing 1 *given* other thing"
>My dumb ass just doesn't get it
>Keep saying, "But it says the chance of thing 1 is X% right there in the problem!"

Conditional probability was hard.

that sounds somewhat similar to how i heard about it, my friend was doing conditional probability at school and presented me with the monty hall problem, i argued against him for quite some time, he didnt manage to change my mind, a little later i realised i was wrong all along and i felt stupid, not because i had gotten it wrong (most people do), but because i had been so sure about me being right.

But it's twice as likely you picked the box with the 2 gold balls

I'm not going to watch your video, Kevin
Yes, but the question didn't include "given other thing" I understand why the answer is 2/3, it's not that I'm too dumb, I'm just too autistic not to view this as a trick question

i wish i was Kevin, but im not, but even if i was you should still watch it, not because i tell you to but because you might understand why its not a trick question.