This is high school material in The Netherlands

50%

la creatura...

Very good toothpaste, very good. However...

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it's 2/3, retards

that's an elementary school problem unless you're in some backwards education program for literal retards.

That's not an American highschool problem. That's 6th grade math.

I thought this too, but it said "from the SAME box" so it is 50/50

1/2 (50%)

For a longer explanation

>there are two boxes with gold balls (lol)
so there are 1/2 chance you chose one and 1/2 you choose the other.
>There are no more possible cases (the last box doesn't have a gold ball to pick first)

>if you pick the gold+silver box, then your chance is:
1/2*0 = 0

>if you pick the gold+gold box, then your chance is:
1/2*1/1 = 1/2

>Final result:
0+1/2 = 1/2

both are highschool tier here?

a b and c are all 0 in all cases
Checkmate Muhammad

en.wikipedia.org/wiki/Bertrand's_box_paradox

wrong

christ this thread is going to be painful, but i've seen this thread a million times and the answer is 2/3

fuck didnt see the condition, nevermind then

en.wikipedia.org/wiki/Bertrand's_box_paradox

2/3

Technically you don't know which box you're picking from so you have to consider the entire population. So 3 balls left 2 being gold. Probability is 2/3.

I'm not sure if I'm being baited because I'm too autistic to tell, but I'll write out a solution anyway.
The question is asking what is the probability the second ball you draw from a box is gold, given the first ball you draw from the box is gold.
A = second ball is gold
B = first ball is gold
P(A | B) = P(A and B)/P(B) = (1/3)/(1/2) = 2/3

If you can't follow this, look up conditional probability

I didn't know Fermat's last theorem was high school tier

woah so syrians proved fermat last theorem was false? nice.

0% the game is rigged so you can only get silver balls.

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1. Probability of putting your hand in the double gold is 1/3.
2. Probability of putting your hand in the double gray is 1/3.
3. Probability of putting your hand in the one gold one gray and picking gold first is 1/6.
4. Probability of putting your hand in the one gold one gray and picking gray first is 1/6.

1 and 4 respect the rules so you have 1/3 + 1/6 = 1/2.

Pic related is State of Texas 7th grade curriculum

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66%? no?
maybe i forgot how to do combinatorics

It's 2/3.

t. math major and trader

>adding probabilities
either there's a gold coin or not
if the first pick was a gold coin then the third box is out
that leaves next result as G or S
1/2, 50%

lazy way to solve everything

>1)
a=1, b=1 c=2

>2)
a=b=1 c=2^(1/2)

>3)
a=b=1 c=2^(1/3)

you can do it forever on

hope you're trolling but the length explanation leads me to believe you're not

But the probability that the first gold coin you picked out was from the box with two gold coins is twice as likely as the coin having been from the box with only one gold coin.

>

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Russian high school material is kinda similar, probability theory is a must, but it isn't really looked into. It's just a single question at the final math exam after all.

1 correct question out of 4
it says "25%" two times, but the answers are A, B, C, and D, all different

bleh

a = 4
b = 2
c = 2

These things with combinatorics and euler and stuff are basically last year HS/ first year uni

never hear of the long con?

ITT: Baye's Theory is ignored

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>ywn be a genius french mathematician
how do i get over this feel?

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I wish differential equations was as simple as "100 - x = 40," but why am I thinking that this board has anybody but students with sub-2.0 GPA

Yeah, something like that. Dunno about uni, I study at an art college (yeah I know), so we don't have maths.

>proving it with identities
thats the lazy man's way of doing it, since thats usually assumed anyways

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1*1/3+1/2*1/3 + 0 = 3/6 = 1/2

Probability that you picked the 2 gold box given that you picked a gold ball: 2/3
Probability that you pick a second gold given you picked the 2 gold box: 1

Answer: 1*2/3 = 2/3

> move to paris, drink at the local coffee shop while you do math homework

It would be 2/3 if all four balls were in one box. But its not, therefore 50%.

Even if the other box had zilloins of gold balls its 50%.

that's not what the question asks though

>2. Probability of putting your hand in the double gray is 1/3.
you can't take a golden ball from a box that has two grey ones

2/3. It's not difficult to understand. When you picked a gold ball, you have effectively eliminated the box with only silver balls in it. Therefore, you now have three remaining balls to pick randomly from, two of which are golden. Therefore, the probability of picking another golden ball is 2/3.

yes it is.
it stipulates you pick a box at random. given that you pick a box at random and the ball is gold, it is 2/3 that you picked the box with 2 golds.

I literaly said it was the lazy way, a.k.a. the best way

no

you keep insisting with the box and the total of probabilities
the question asks about probability of getting the golden coin AFTER you picked up a box

You're summing over all boxes anyway. Why the fuck would you use Bayes for this?

p(box_i) = 1/N for i in [1,3]
p(gold|box1)=1
p(gold|box2)=1/2
p(gold|box3)=0

p(gold) = 1/2

sum(boxi|gold)=sum p(gold|boxi) *p(boxi) / p(gold)
= 1/3 + 1/6 = 1/2

it's the same

>the question asks about probability of getting the golden coin AFTER you picked up a box

Read my explanation above. You've gained knowledge in picking a box with a golden ball. You KNOW that there now exist three possible balls that you may pick blindly, two of which are golden and one of which is silver. Therefore, the probability of picking a golden ball is 2/3. Do you see?

doing that wont help you when you truly need another approach

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I prefer the one with the goats

Hush.

the monty hall one?

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generaly speaking, you need to break the energetic barrier to go against any equilibrium
which is to say, there must be a reward in some kind to go against the natural equilibria.

considering there is no incentive/energy/reward/interference, you can't expect to have any other result but the one with the lowest energy

>three possible balls
this is wrong, there are only 2 colors that matter, gold or silver
you either pick gold or you pick silver because the box has 2 balls, 1/2
it's like coin toss bias, if you tossed a coin and got heads, the probability of getting heads in the next toss remains the same because the only that matters is the outcome: either heads or tails, 1/2

Yeah, I prefer goats and cars over balls

>this is wrong

No, it isn't. The proofs are in. I have tried as best I could to explain it to you in layman's terms, but at this point (since others have already started talking about it), I may as well tell you that this is a reformulation of a classic combinatorics problem.

en.wikipedia.org/wiki/Monty_Hall_problem

bleh

blame special functions (or just the error function)

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75% actually.

>high school
I didn't come across partitioned sample spaces until my second semester at university. They dumbed down our nation's maths curriculum a few years ago you see... and people wonder why our national average IQ is 92

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ζ(s), s=n
is the one true solution

3,50

1/2, quite easy desu

>I didn't come across partitioned sample spaces until my second semester at university

You don't usually bother with teaching pupils probability and combinatorics unless they've chosen an academic path that requires you to have an understanding of statistics (physics, engineering, economics, computer science, and so on). In Sweden, we learn some of this stuff in upper secondary school ("high school"), but it's only for people who've chosen an upper secondary program that is preparatory for this kind of academic path.

I'd argue that everyone would be much better off if they understood the basics of this stuff (and of statistics), from a social and political perspective especially, but it's actually quite hard for most people to get their head around problems such as the one posted in the OP. So I do think that we shouldn't waste much time trying to teach 92 IQ individuals these things.

2/4 because two answers are correct and there are 4 in total.

then you'd choose B, which is ONE answer in FOUR (1/4)
paradox

I like you Sweden, you honestly tried to explain, but i guess when the original problem with the goats was put out, very few understood aswell

1/3 1/3 1/3x
1/2 1/2 1/2 1/2x
G G g
66.7% but Monty Hall problems still break my brain

2/3 is my only guess but I am retard with severe alcoholic brain damage.

My thoughts is:
3 Gold balls total (you definitely pick one of them, fuck the other box its a trap).
2 gold balls are in the box with 2 gold balls
1 gold ball is in a box with a silver.

Ergo and therefore 2/3 you pick one of the gold balls in the double gold ball box. 1/3 you picked the gold ball in the silver-gold box.

If there are 1000 doors with 1 car and 999 goats and after you choose one, the moderator discards 998 doors. Would you switch or stay?

Yes, I think this is the best intuitive way of understanding it. Of course you'd want a 1/2 chance to win rather than a 1/1000 chance.

Depends if the moderator knows or not. If the moderator is discarding doors at random, and he doesn't know which one has the car, then it makes no difference if you go or stay.

If the moderator does know then duh you switch but WAIT. What if you really always want a pet goat? You can drink goat milk, make goat cheese, make a goat cart for it to pull around the kids, and even eat the goat.

you should be able to solve this

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which makes perfect sense if you include an "imaginary" unit attached to each 1/4 answer

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The whole point of the Monty Hall problem is that the moderator knows

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Fuck I hate to admit it but the Swede is right
I simulated it

import numpy as np
import pylab as py
from random import randint,shuffle

samples = 50000

result = []
plotX = []
plotY = []
boxes =[[1,1],[1,0],[0,0]]
for s in range(samples):
for b in boxes:
shuffle(b)
box = boxes[randint(0,2)]
if box[0]:
result.append(box[1])
plotX.append(s)
plotY.append(np.mean(result))
py.plot(plotX,plotY)
py.show()

Attached: boxes.png (640x480, 13K)

Yeah but you forgot the crucial detail in your retelling in and in so doing you broke the problem.

This is wrong I misunderstood the problem

P(box1|gold)=p(gold|box1)p(box1)/p(gold)=(1*(1/3))/(1/2)=2/3

Sorry

I feel like this is a Monty Hall problem, and the answer is 1/3

even thought the intuitive answer is 1/2

Naw Syrians suck at maths.

25% because I'm unlucky

Think about it this way. There 2/3 chance that you pick the goat and when he opens the door to the other goat the remaining door have the car if you picked the goat. And since it was 2/3 chance you picked the goat you will get the car with a chance of 2/3.

i was when i was in highscool

The odds that the first one is a gold ball is 1/2 (obviously)
The odds that the second ball is gold is 1/3 (equal to picking box 1)
The odds that the second is gold given that the first one is also gold is (1/3)/(1/2)
The answer is therefore 2/3