Show that you're smarter than /sci/, here is your chance
Probability has applications in computer science.
>>53875326
50?
Gentoo/10
66.6%
50 percent
>>53875368
No I'm pretty sure it's 66%. Because if the first ball is gold, then that means that there are only 3 possible balls left that you could choose. 2 of them are gold, one is silver. So it's 66%.
>>53875385
My thinking was that if you got a Gold ball, you are in bin 1 or 2. No matter which bin it is, it's either gray or gold and those are your only 2 options. Although 50 percent seems to easy so I'm sure there's some bullshit reason why it's not right
>>53875385
No, because you are pulling from the same box as the first draw. So either you got a gold ball from the gold/silver box or the gold/gold box. The next ball must be from that same box so its 1 in 2 odds for gold or silver.
>>53875326
Given the constraints of two boxes with any gold balls you have a fifty percent chance that you picked the box with two gold balls.
>>53875426
This is incorrect
>>53875420
Man this is screwing with my head. I get what you're saying though.
>>53875385
How are 2 gold, only 2 boxes even have gold balls in them. Box 3 is removed entirely due to it having 0 gold balls.
50%
This seems like the monty hall problem but I can't find a reason for it not to be 50 percent.
2/3
>>53875444
Well make sure not to elaborate in any way.
Spelling it out for the retards early:
The initial selection selects box A twice as often as box B.
Box A (picked 2 out of 3 times) has a 100% chance of drawing gold again.
Box B (picked 1 out of 3 times) has a 0% chance.
There's a one in six chance of drawing any particular ball (1/3 for the box * 1/2 for selecting a ball from it).
There are 3/6 outcomes in which we draw a gold ball. Using a bit of conditional probability we find that there's a 1/3 chance of drawing any individual ball.
Given that there's a 2/3 chance that the gold ball you drew was from the first box. Hence a 2/3 chance that the second draw will also be gold.
It's either 0% or 100%. You're not activating a random number generator by reaching into the box again; the second ball's colour has already been determined by the first drae
>>53875463
It doesn't involve us changing boxes, so its just a faux trick question.
>>53875463
Monty Hall functioned on giving the option to switch to a different "box." In this case there is no switching. The results, and odds, are locked at the time the box is chosen.
>>53875467
ARE YOU KIDDING ME I WAS RIGHT? SCREW EACH AND EVERY ONE OF YOU FOR MAKING ME DOUBT MYSELF.
2/3
You pick a box at random, but then you draw a ball at random. The ball is gold. This gives you information as to which box you are in. 2/3 of the time you will be in the box with 2 gold, 1/3 of the time you will be in the box with 1 gold and 1 silver. You will never be in the box with 2 silver.
>>53875467
How do you conclude Box A is picked 2 out of 3 and box B is picked 1 out of 3 times? The problem only has one selection not three.
It is 2/3 because you can't see into either box and thus don't know when you draw a gold if it was the first or second gold. Because of this, there are 3 possible ways that you could draw a gold as your first one: two ways in box 1 and one in box 2. This limits you to two boxes, however you're twice as likely to have picked box 1. Because of that, you're overall chance is 2/3. I'm really tired right now, but you can check this in Bayes Theorem, it's right. Might post later if I can gather my thoughts for it.
>>53875452
FAGGOT.
SEE
>>53875467
AND NEVER SPEAK TO ME AGAIN
>>53875420
It is right. By pulling out a gold ball you rule out the last box, so you're either in the first one or the second one. So the probability of you pulling out another gold ball is 1/2.
>>53875507
Because it's selecting one gold ball at the beginning. 2/3s of the gold balls are in A so 2/3s of the time you'll be in A
>>53875493
Not sure where this magical 3rd box in your chances come from. Only 2 boxes are viable, >>53875507
only 2 selections.
>>53875513
>Because of this, there are 3 possible ways that you could draw a gold as your first one
U wut m8? There are only two boxes which fits the parameters of the question. How do you come up with three possible ways to draw one gold ball from a box when there are only two boxes that contain a gold ball?
If you pick a gold ball, there's a 2/3 chance it initially came from the box with both gold balls. Another way to think about it is that the box with two gold balls offers more chances to pick a gold ball, so the probability is higher
>>53875472
best explanation right here
>>53875541
>If you pick a gold ball, there's a 2/3 chance it initially came from the box with both gold balls.
You forgot the first step. You picked box. The problem doesn't function upon the entirety of gold balls in the equation but the number of boxes which contains a gold ball.
>>53875531
>>53875537
2 boxes, but 3 gold balls. Ignore the boxes for a moment. When you opened the box you found a gold ball. 2/3s of all gold balls belong to A so there's a 66% that you just picked A, and that the other ball is gold
1/2
First gold is locked and you just need to consider 2nd ball.
>>53875326
bout tree fiddy
>>53875559
>Ignore the boxes for a moment.
Let me get this straight. You want to disregard part of the parameters because it doesn't fit your conclusion?
>>53875559
>Ignore the boxes for a moment
that's exactly what you must not do
RETARDS LOOK HERE !! RETARDS LOOK HERE
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
There are three gold balls. Ball 1 and 2 are in a box with another gold ball. Ball 3 is in a box with a silver ball.
All balls have the same chance of being drawn.
If you draw...
Ball 1: Next ball is guaranteed to be gold.
Ball 2: Next ball is guaranteed to be gold.
Ball 3: Next ball is guaranteed to be silver.
^^^^^^^^^^^^^^^^^^^^^^^^^^^
RETARDS LOOK HERE !! RETARDS LOOK HERE
>>53875531
There are 3 boxes. If you read what I said, you will never be in the third box.
Bayes Theorem: P(A|B) = P(B|A)*P(A)/P(B)
So P(2nd gold | 1st gold) = P(1st gold | 2nd gold) * P(2nd gold)/P(1st gold)
P(1st gold | 2nd gold) = probability that you draw gold on your first try given that you drew gold both times, so 1.
P(2nd gold) = probably that your box has both gold, so 1/3 P(1st gold) probability that you draw gold on your first draw=1/2 since half of the balls that you can draw is gold.
This gives you (1/3)/(1/2) = 2/3
>>53875537
Pretty much as >>53875559 said. You draw one ball out of the six. Three of them are gold and three aren't. So your chance of drawing a gold on one draw is 1/2
>>53875537
What if, instead of two balls, you had 100 balls in each box?
One box containing 100 gold balls, and one box containing 99 gray balls and one gold ball.
If you reach into a box and grab a gold ball, which box did it most likely come from?
It's the same concept on a different scale.
>>53875559
It's still a coin flip either way. Gold or silver are the only other outcomes. Just because one box has 2 golds in it changes nothing.
>>53875574
>All balls have the same chance of being drawn.
No, they do not. You are not drawing balls. You are choosing one of three boxes. Once the choice is made the balls within and probabilities are set.
>>53875571
Holy autism guys. Run the experiment yourself in real life, you'll see I'm right.
>>53875474
Frequentists please leave
66.666% percet.vwhat do i winM
>>53875602
You fucking retard, you do not consider that you've ruled out the last box by drawing the first gold ball
Why is everyone here so fucking retarded
>>53875555
But the 2nd box doesn't guarantee a gold ball?
>>53875585
>You draw one ball out of the six.
Wrong. You pick one box out of three.
Why do these idiots keep skipping this critical part of the problem?
>>53875602
I guess I am autist for actually following the entire question to get the correct result rather than disregarding the parts that don't match up with my desired result.
>>53875600
You are retarded or this is bait. You select a box at random and then a ball at random, making the chance of drawing each ball uniformly 1/3 * 1/2 = 1/6.
Once you've done that, in 2/3 of cases you'll have selected a ball from box A.
It's not 66 percent. When part of the problem states that every time you pick a gold ball first, you disregard the 3rd box and one gold ball. After you get the 1st gold ball, which happens every time, you have 2 possible different outcomes for the next ball(the first variable). Either gold or silver, giving you a 1/2 chance that it's gold.
>>53875614
Do it in real life or admit you're an autist. I don't care which.
>>53875614
You are the retard sir, everyone agrees that the last box is ruled out.
>>53875625
read the fucking op image again
>>53875326
You have to pull a gold ball the first time. Which means you either have the box with 2 gold or 1 gold. This means there is a 50/50 chance of you having either of the either of the boxes.
50%
>>53875616
>But the 2nd box doesn't guarantee a gold ball?
Why does it need to guarantee a gold ball? We don't calculate probabilities before we get to the end of the parameters.
That's like saying we should calculate all the aces in a deck even though 1/3rd of the deck is removed from the equation.
>>53875634
>you pick a box at random
>take a ball from that box at random
Wow! It says exactly what it says the first time I read it! And the probability of picking each ball still remains 1/6.
2/3
Anyone who says otherwise is autistic.
>>53875624
Okay. Let's look at every possible way that you can draw two balls from one of three boxes:
G1G2, G2G1, G1S2, S2G1, S1S2, S2S1. Those are all possible ways to draw both balls from any of your three boxes. Now how many elements of that event space involve drawing a gold on the first try? 3 of them do: G1G2, G2G1, and G1S2. So you've now limited your event space to 3 possibilities. In how many of them do you draw 2 gold? 2 of the 3. As such, your probability is 2/3. This is as elementary as probability gets: counting desired outcomes out of your total event space. If you or anyone else still has doubts, than you really need to take a probability/statistcs class.
This should help clear it up for you 1/2 fags
>>53875625
>You select a box at random and then a ball at random
Very good. Someone kept track of the first step. We have three possible results when we choose one box. Do we still have three possible results when we pull one ball from that box and determine the ball is gold?
>>53875354
I agree.
There are 6 balls total but by pulling out a gold ball we know the box with two silver balls is eliminated. Thus tour chance of another gold ball is 2 out of 3 since without knowing what box you have there are a total of 2 gold balls balls in place out of a total of 3 balls including the one silver.
>>53875646
You must pick a box with a gold ball you cunt
I'm genuinely curious, because these threads are always a shitstorm and no-one could possible be this stupid, so the only thing I can think is that people are intentionally spamming the wrong answer to "troll" or piss other people off, so:
How many of you retards posted the wrong answer on purpose? Tell the truth. I wonder if anyone will.
>>53875667
Third box can be eliminated from the equation, leaving 1 in 2 boxes with gold balls.
>>53875667
>B-B-BUT THE'YRE N-NOT EQUAL ANY-ANYMORE
>>53875666
>Now how many elements of that event space involve drawing a gold on the first try? 3 of them do: G1G2, G2G1, and G1S2.
Let me get this straight. You are still calculating for the fixed event of the one box being picked and the one ball being gold?
>>53875667
This
Let the number of balls in each box approach infinity, with all gold in the first, one gold in the second, and no gold in the third.
Do you really still think that you would have picked your gold ball from the middle box?
>>53875669
That's like saying that the probability of winning the lottery is 50/50. After all, there are only two possibilities: either you win, or you don't. Wow! I never knew it was as simple as simply counting the number of different possible outcomes and ignoring every other parameter of the equation!
>>53875676
Gold balls aren't mentioned until after the ball is picked you literal fucking two year old.
Even if you threw the third box in the trash before picking the answer is still 2/3.
>>53875667
Thank you. That's an excellent way to explain the importance of the first step: Pick one box.
>>53875689
Yes. I've shown every possible way to draw two balls from a box. The first two are possible if you picked the first box, the second two if you picked the second, and the last two if you picked the third. With that taken care of, I limited the number of valid events to the 3 in which you've drawn a gold as your first draw. At this point we've drawn from both one of the boxes and it was a golden ball. There are only three event spaces that allow this possibility of which 2 of them happen to be if you chose the first box. These are the only two where you draw both golds, so the probability is 2/3
>>53875707
>That's like saying that the probability of winning the lottery is 50/50.
If we removed all but one of the numbers and there were only two possible results for the last number you might be right but the lottery doesn't work that way. See >>53875667 for an explanation of why it is important to follow all the steps.
50%.
You can reinterpret the question as just asking if you grabbed out of the GG (gold gold) box or the GS (gold silver) box. You got a gold, so you know you didn't grab the SS box. There are only two choices, equally as likely from current information.
>>53875710
You're fucking retarded if you can't comprehend an easy problem like this
YOU PICK A FUCKING GOLD BALL YOU FUCKING RETARD CAN YOU NOT READ, IT IS SPECIFIED
ITT: people who don't understand Bayesian probability
>>53875729
>picking a gold ball is the same thing as picking a ball from a box with a gold ball.
At least be consistent with your retarded dribble.
>>53875715
>Yes. I've shown every possible way to draw two balls from a box.
No, you've shown every possible combination with gold being drawn first without regard to the fixed result once a box is chosen.
See >>53875667
>>53875724
You didn't just choose a box, you also chose a ball. 2/3 and you're an autist.
>>53875730
Probability that they answer the question correct given that they don't understand Baysian probability = 0
>>53875667
Welp, now I feel stupid.
>>53875722
That image agrees with my point though.
If you grab a gold ball, it it much more likely to have come from box 1 than box 2. So, you can safely say that 95% of the time, if you draw a gold ball, it came from box 1. It's not 50/50. Are you trolling or are you really just this bad at grasping basic concepts?
>>53875341
yep
>>53875743
You chose a box _and_ a ball. The 2/3rds results only functions without regards to also choosing a box.
>>53875751
Autist
>>53875730
>ITT: people who don't understand probability
ftfy
Bayesian probability is a nice and for me pretty clear way to understand this problem, however as I showed in >>53875666 you can come to the answer in an even simpler way. Like this is day 1 of a college stat class stuff.
>>53875741
Not sure what your point is, seeing as how both of us agree that it's 2/3. That being said, I am sticking to the same box after picking the first one. If my first pick was from the G1G2 space, then my second pick is the G2. Same if it was from the G1S1 space or the G2G1 space.
>>53875724
Sorry fucko! Looks like you're the one who re-interpreted the question.
>There are only two choices, equally as likely from current
How the fuck do you figure that it's equally likely to have drawn a gold ball from a box with 50% gold balls, and to have drawn a gold ball from a box with 100% gold balls.
>>53875326
until the second ball is removed from the box it it both gold and silver, and has an equal chance of becoming either when removed from the box
>>53875749
>If you grab a gold ball, it it much more likely to have come from box 1 than box 2.
The question is not about how likely you drew a gold ball from box one over box two. The question was how likely is it that the NEXT BALL will also be gold. The results of the first pick is fixed. It is no longer part of the probability equation. You only have one of two boxes that you could have chosen that has more than one gold ball in it.
>>53875754
Wrong
See
>>53875758
>>53875759
>Not sure what your point is
You are still calculating for a static result. The first gold ball is no longer part of the probability equation.
I always find these threads frustrating. Everyone starts posting their "intuition" instead of actually using probability.
Let's rephrase the problem slightly.
There are six boxes each is labelled with a number and contains a ball either gold or silver. The ball has a number of another box.
You're going to open one box at random and then open the box printed on the ball as well.
The boxes are :
1. Gold 2
2. Gold 1
3. Gold 4
4. Silver 3
5. Silver 6
6. Silver 5
Given I've drawn a gold ball after the first draw what's the probability of drawing a gold on the second?
Note this game is functionally identical to the one here >>53875326 because the labels are setup to mimic the boxes from the original problem.
>>53875739
So you're saying you can pick a gold ball from a box with no gold balls
Now I get it, you're retarded
>>53875762
Schroedinger's autist
>>53875770
That's not how it works. The first outcome is very important in determining the second. https://en.wikipedia.org/wiki/Bayes%27_theorem
>>53875780
Not an autist
50/50
>>53875764
Retarded child who can't keep track of more than one piece of information at a time detected.
At no point in time is there a situation where 50% of people would chosen box A and 50% of people have chosen box B. If you assume that you are literally spewing out of your undeveloped brain that, two possibilities are always equally likely, regardless of the process used to select them.
>>53875743
we already have a gold ball. the probability of picking the gold ball over the silver ball in box 2 is thrown out because it's assumed that we first got gold.
Two scenarios. Either we got gold and are in double gold or we got gold and are in gold/silver. The picking of a gold ball is already done.
50/50
you already picked gold, so your box is either GG or GS. So in the box is now either G or S.
what is this my facebook feed?
>>53875681
It's a combination of intuition and pride more than outright trolling.
>>53875463
If you picked the golden ball, it's twice more likely to originate from the first box.
>>53875775
clearly that's not what he said.
>>53875764
If 95% of the gold balls you pick are part of the same box, then if you pick one gold ball, there is a 95% chance that the next ball you pick will also be gold.
There is only a 5% chance that the next ball won't be gold, because the chances that you picked that box in the first place are smaller.
If you had one box with a hundred trillion gold balls, and a middle box with only one gold ball, then every time you picked a gold ball you'd be almost 100% guaranteed the box you picked was the first one and that the next ball would also be gold.
>>53875326
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
The probability is 2/3rd.
>>53875804
He was implying I implied that, I didn't
>>53875775
No, but you can pick a non-gold ball from a box with a gold ball in it.
>>53875816
No you can't. It literally said you picked a gold ball.
>>53875624
>Wrong. You pick one box out of three.
Read OP. You pick the box randomly and you pick the ball randomly.
>>53875791
>At no point in time is there a situation where 50% of people would chosen box A and 50% of people have chosen box B.
Correct. Under the parameters only you chose one box. That is 100% certain. You have only one box. You have one box with only two possible results. GG or GS.
I don't like these paradox questions because they function upon theory and not reality. The results are set once you pick the box in reality so it's 50% but in theory it is 66%.
>>53875816
But your first draw HAS to be gold, so you're either in the first box or the second. The next draw is either silver or gold.
So picking a gold ball in reality is picking box 1 or 2.
>>53875801
>>53875681
true
the correct answer is usually in the first five posts
then the trolls start fighting
>>53875848
Ding ding! Retarded baby found! He just made the drooling preschooler assumption that Box A is picked equally as often as Box B, because his peanut brain failed to even attempt to understand the question.
import random
tries = 1000000
N = tries
gold = 0
silver = 0
while tries > 0:
Boxes = [[1,1], [1,0], [0,0]]
box = random.choice(Boxes)
if box == (0,0):
continue
coin = random.choice(box)
box.remove(coin)
leftover = box.pop()
if leftover == 0:
silver = silver + 1
else:
gold = gold + 1
tries = tries - 1
print("Gold ratio: {}\nSilver ratio: {}".format(gold/float(N), silver/float(N)))
Gold ratio: 0.500467
Silver ratio: 0.499533
well consider me fucking surprised, I thought it was 2/3
2/3
>>53875839
It's 67% in reality.
You can do the experiment yourself.
Get three boxes, put balls into them.
1. Pull out a ball out of a random one.
2. If it's not gold, put it back, go to 1.
3. Pull out the other ball. If it's gold, write G. Else write S.
4. Put it back, go to 1.
After reading step 3 100 times, you'll have about 67 G and 33 S.
>>53875559
I think I understand what youre saying. Except the questions asks for the possibility AFTER you picked one box randomly and of which you got a gld ball, not before in which case I agree 100% with you.
>>53875865
It doesn't have to be picked out equally as often, you rule out that possibility
>>53875871
nope, the box with two silver balls is irrelvant
>>53875866
fuck me I suck
[0,0] is NOT equal to (0,0)import random
tries = 1000000
N = tries
gold = 0
silver = 0
while tries > 0:
Boxes = [[1,1], [1,0], [0,0]]
box = random.choice(Boxes)
if box == [0,0]:
continue
coin = random.choice(box)
box.remove(coin)
leftover = box.pop()
if leftover == 0:
silver = silver + 1
else:
gold = gold + 1
tries = tries - 1
print("Gold ratio: {}\nSilver ratio: {}".format(gold/float(N), silver/float(N)))
Gold ratio: 0.749647
Silver ratio: 0.250353
>>53875866
You aren't even checking if the first coin you remove is gold or not you fucking faggot.
>>53875839
>The results are set once you pick the box in reality so it's 50%
There is NEVER at ANY point a 50/50 decision being made (other than the irrelevant fact there's 2 balls in each box). If you are in Box A there's 100% chance the other ball is gold. If you're in box B is there's a 0% chance the other ball is gold. It's entirely a manufacture of your broken intuition that Box A and Box B are ever selected with equal (50%) probability.
>>53875866
You programmed it incorrectly. Particularly, box == (0,0) part is fishy, though I don't know python.
>>53875891
>You aren't even checking if the first coin you remove is gold or not you fucking faggot.
I am. It is the box == [0,0] part >>53875888
>>53875893
Not him but Box A and Box B do have the same chance to get picked.
>>53875906
There is no hope for you.
>retards actually need to fucking think for more than 5 seconds to realize it's 50%
>total retards/trolls actually try to argue it's 66%
I'm done with this board.
>>53875878
>It's 67% in reality.
No it is not. It is 66% in theory because theory functions under a total set of possibilities outside the physical limitations of the box. Whats in the box you picked is not going to change. Given that you have one gold ball in hand there are only two possible results left. I did the same thing with the coin flip and after a hundred of tests (more than a hundred as results with the first landing tails were discounted as not within the parameters of the required result) it was actually over 50% which is closer to 50% than 66%.
>>53875906
That's checking box. You also have to check coin (and if you don you don't actually have to check the box).
>>53875888
You still fucked up.
You're not counting the times you draw a silver coin from the box first.
>>53875888
you know
I am going to start arguing that it is 75%
>>53875893
The third possibility is no longer a possible results so if you actually run the test you have to discount any results which result in the third box with no gold boxes being chosen. That leaves the results in reality to 50%.
>>53875914
Are you saying that experiment I described will end up with about 50 G and about 50 S written? Or are you saying that experiment I suggested does not reprersent the problem properly? If so, which part? How would you suggest to carry out the experiment?
>>53875906
How does that check whether the ball you picked is gold or not?
>>53875947
>>53875932
>>53875931
>>53875929import random
tries = 10000000
N = tries
gold = 0
silver = 0
while tries > 0:
Boxes = [[1,1], [1,0], [0,0]]
box = random.choice(Boxes)
if box == [0,0]:
continue
coin = random.choice(box)
if coin == 0:
continue
box.remove(coin)
leftover = box.pop()
if leftover == 0:
silver = silver + 1
else:
gold = gold + 1
tries = tries - 1
print("Gold ratio: {}\nSilver ratio: {}".format(gold/float(N), silver/float(N)))
Gold ratio: 0.6667663
Silver ratio: 0.3332337
I quit someone take over
>>53875908
Explain how Box A and Box B have equal chances to be selected, while I explain how it's not:
There are three equally likely box selections: A, B, C.
Once you put your hand in a box there are two more possible selections: Ball 1 or 2.
If you put your hand in Box A (1/3): There is a 100% (2/2) chance the ball picked is gold.
If you put your hand in Box B (1/3): There is a 50% (1/2) chance the ball picked is gold.
If you put your hand in Box C (1/3): There is a 0% (0/2) chance the ball picked is gold.
Now add up the probabilities and you get...
Box A with a gold ball: (1/3)*(2/2) = 2/6
Box B with a gold ball: (1/3)*(1/2) = 1/6
Box C with a gold ball: (1/3)*(0/2) = 0/6
>>53875956
There there. You'll definitely get better with time and effort.
>>53875946
>Are you saying that experiment I described will end up with about 50 G and about 50 S written?
No, I'm saying I did exactly that and ended up with 52 (or so I cannot remember the exact number) of S. Run it yourself if you don't like my result.
>>53875965
but the information given makes one of the boxes irrelephant
>>53875965
Right. I was incorrect. I was looking at the situation before the "you got gold" condition was applied.
This should've been the first reply.
Fuck all you retards.
>>53875956import random
tries = 10000000
N = tries
gold = 0
silver = 0
while tries > 0:
Boxes = [[1,1], [1,0], [0,0]]
box = random.choice(Boxes)
coin = random.choice(box)
if coin == 0:
continue
box.remove(coin)
leftover = box.pop()
if leftover == 0:
silver = silver + 1
else:
gold = gold + 1
tries = tries - 1
print("Gold ratio: {}\nSilver ratio: {}".format(gold/float(N), silver/float(N)))
>>53875989
Thanks for making me realize that 50% is definitely wrong.
See the other coin in the first box.
I am out for lecture, bye.
>>53875965
>If you put your hand in Box C (1/3): There is a 0% (0/2) chance the ball picked is gold.
This is not a valid result once you've pulled the first ball.
>>53875981
That's correct, but your statement is also irrelevant to the fact the boxes are chosen with in-equal probabilities. It just changes the numbers from 2/6 and 1/6, to 2/3 and 1/3, which in both cases is box A being selected twice as often as box B.
>>53875989
lel what if you take the other ball from the first box
>>53875974
There is a program written by anon that does exactly that, in >>53875956, and it gives 0.6667663 gold ratio after 10000000 tries. I also wrote a program myself at some point in past with same results. I also did the experiment with 30 tries, long long ago, and had 19 gold and 11 silver.
You are either horribly mistaken or intentionally lying.
>>53875989
It's incorrect.
>>53875994
Gold ratio: 0.6666013
Silver ratio: 0.3333987
>>53875998
I'm talking about the selection process of the first ball.
I always hated probability in school, mostly because I sucked at it. It seems simple yet I always came up with the wrong answer, made me feel stupid.
>>53876007
Then you're a moron that can't reason or code. The box with 2 silvers is never in the equation because the scenario eliminates it for you.
>>53876016
>I'm talking about the selection process of the first ball.
The question focuses on the selection of the second ball. The selection of the first ball is fixed and invalidates the third box as a viable possibility.
The problem becomes much less ambiguous if you rephrase the problem like this:
> you pick a ball at random out of these 6 and it turns out to be gold. What is the probability that you pulled it out of the box with 2 gold balls?
>>53876029
The second part of the question is predicated on the first part. You don't have a "same box" to pull a ball out of if you're a retard who ignores the first half of the question.
A link to the solution to the problem on Wikipedia was already posted. Why are people still trying to argue that it is 50%?
>>53876028
Misquote?
Anyway, go ahead and post your code that gives 50/50.
We have code that gives 67%. You can also point out the mistake in that as you see it, if that's more preferable for you.
>>53876044
Because this is /g/. Everyone is special snowflake smarty-pants and their intuition could never be wrong.
>>53876029
>selection of the second ball
There's only one random even and it's picking the first ball.
>>53875886
Explain.
I was led to believe that we didn't know the contents of each box. We just picked a box.
For anyone who trusts computers.
https://play.golang.org/p/khHp4_KHH3 gives 66.6%
>>53876060
>There's only one random even and it's picking the first ball.
The one thing that is set is result of the first ball and that is your idea of "random?"
>>53875326
OP dropping this bomb in here to cause some autist rage.
well played, breh.
>>53876077
Yes. There are three gold balls, and you have equal probability to pick any of them. Picking the ball that will end up as gold (as we know because of condition) is a random event with three equally likely outcomes. Anything that happens after that event is determined, with no randomness involved.
>>53876077
If you're a monkey with an IQ of 5 and only read "What is the probability the next ball you take will be gold?", then it would be either 100% or 0% depending on which box was chosen.
You can't just magically make up the fact that two boxes are equally likely to be chosen and ignore the first half of the question.
>>53876071
>boxes := [][]int{[]int{gold,gold}, []int{gold, silver}, []int{silver, silver}}
so.... this.... is....... the power........ of... Go.......
wow..............
>>53876103
>Yes.
So what is my random chance that my first pull will be a silver ball? 0%? That's your idea of random?
>>53876110
There are only two possible results of what the first ball drawn will be but under the requirements set by the question I have a random chance it will be silver, huh?
>>53876116
The probability of obtaining a silver ball during the first and only random event as described by this problem is 0%, yes.
If there was no condition that you pulled out a gold ball, it would be 16.7% instead of 0%.
>>53876044
Bertrand was full of shit.
>>53876067
You know that you just pulled out a gold ball of your previously unknown box, so you can rule out box 3.
Also, anyone who argues without Bayes, is a retard.
>>53876133
>The probability of obtaining a silver ball during the first and only random event as described by this problem is 0%, yes.
0% chance is random!
Wow. I need to go update my dictionary.
>>53876130
>There are only two possible results of what the first ball drawn will be
>only two possible results
>there are three gold balls
EPICCCCCCCCCC
You pick a box, and you pick a ball from the box. The scenario only continues if you picked a gold ball. From which box are you going to pick a gold ball more times? The box with 100% gold balls, the box with 50% gold balls, or the box with 0% gold balls?
>>53876046
You have code that factors silver/silver when it's not even part of the equation, but at least you tried.
>>53876149
>>there are three gold balls
I'm going to pull how many balls on my first pull? Just one? And my chance of pulling a silver ball when the result is already determined it will not be silver is random?
>>53875326
50%
Who cares about the 3rd all silver box, it's useless here
So you either picked the gold ball from the all gold box or gold/silver box
Since you don't change boxes after the initial pick, that means you are left with the other ball being a silver or old
>>53876148
The random event has outcomes. Those outcomes have probability. If you flip a coin, that's a random event. It has 50% that the outcome of toss is heads and 50% that the outcome of toss is tails. It has 0% that the outcome of toss is getting both tails and heads simultaneously. Despite it having 0% probability of something, it's still a random event.
>I need to go update my dictionary.
Absolutely.
>>53876137
I feel very vindicated. 1/2 it is, I can go to sleep happy now.
>>53876153
It's not my code, and his latest iteration is correct.
Here: >>53875956
Again, either provide your code or fix his code. So far you're just empty talk.
>>53876006
The pool is always 1 box, i.e it doesn't matter which gold ball you take
Just what the next ball will be
>>53876172
>The random event has outcomes. Those outcomes have probability.
Seems to me you are trying to fit facts to your conclusion rather than the conclusion to the facts. We have a fact that 100% the first pull is going to be silver. That is set not random.
I'm done wasting time with this idiocy.
ayylmao
>>53876195
Gold, not silver. You have 100% to pull out gold, and 0% to pull out silver. But there are three gold balls, and each has 33% chance to be pulled. Picking the ball is a random event. The event will be not random if, and only if, all involved probabilities are either 100% or 0%.
>I'm done wasting time with this idiocy.
Don't forget to update your dictionary. Those are basics.
>>53875326
50% for the first box, 0% for others, if you want to express all three in one number then probably 33%
>>53876201
I am going to piss on his fucking grave.
>>53875472
the right explanation.
It's more likely to draw a gold ball from the first box than from the second, so the probability of drawing another gold ball is higher than drawing a silver one
>>53876214
fuck, 100% for the first one ofc
>>53875994import random
num_golds = 0
num_grays = 0
tries = N = 10000
while tries > 0:
boxes = [[1, 1], [1, 0], [0, 0]]
box = random.choice(boxes)
i = random.randrange(0, 2)
ball = box.pop(i)
if ball is 1:
num_golds += 1
if box is 0:
num_grays += 1
else:
num_golds += 1
tries -= 1
print str(N / float(num_golds))
>0.661550674782
Confirmed, 50/50fags BTFO
>all the fucking retards ITT who have heard of the paradox before saying 66% instead of actually reading the op image and arguing against anyone saying 50%
>>53876229
Still wrong, first pick has to be a gold ball, you outrule all other possibilities
50% like others have said:
- It's obviously not the box with two silver balls, so it's irrelevant.
- Since one of the balls is a gold ball, the other that's left MUST either be a silver or gold ball
Thus, you have a 50% chance it will be gold.
Remember, you picked the box at random and happened to get one with at least one gold ball.
Also, even though you can't see into the box, it doesn't matter since the question is just about the probability.
I like theses threads a lot.
It's the best kind of bait.
>>53876245
When playing the lottery, you MUST either win or lose.
Thus, you have a 50% chance you will win.
isn't this just the monty hall problem
>>53876245
This is not your first post in the thread. Can I just take that as confirmation that you're simply shitposting?
>>53875463
It's not the Monty Hall problem, it's the coin flip problem.
>>53875722
I can either win or not win the lottery. What other option do you see?
>>53876259
the problem specifies you must pick a gold ball first or are you retarded
the next ball has to be from that same box
what the fuck don't you understand
Let's spice things up:
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?
>>53876284
What the fuck does that have to do with anything?
>>53876201
This. Bertrand was full of shit.
The chance of picking a gold/gold or silver/silver box is 66%, that doesn't change just because he believes in fairies and tosses reality out the window.
>>53876299
you're making a retarded argument
>>53876307
I am using the exact same argument you made.
>>53876284
The condition imposed on you by the problem makes it not equivalent to just picking a box out of 2 at random. Situation where you just pick one box out of 2 at random does not account at all the fact that you are more likely to pick the first box (again, as imposed by the condition that the ball you ended up getting is gold).
>>53875326
It's not 50%. I reread the question. At the first quick read I almost thought that it's a trick, but it's kinda well worded.
The initial selection is from all 3 boxes. The question is if you have picked randomly and when the picked ball is a gold ball what could happen then?
In other words, chose a ball randomly, in the case it's gold what are the future implications of those... aka given that you can only choose silver or gold now, BUT this is only one possible "path". You could have chosen a silver ball first.
What most of you guys think (1/2) would only make sense with the following wording: "You always chose a gold ball first." as in you can't choose any other balls, basically selecting the first ball from one of two boxes. That's not the case though.
>>53876294
yes
>>53876283
What is the probability of winning the lottery without any fixed results?
>>53876294
Suppose you're on a game show, and you're given the choice of three boxes:
a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and a silver coin.
You pick a box, say No. 1, and the host, who knows what's in boxes the doors, opens another box, say No. 3, which has two silver coins. He then says to you, "Do you want to pick box No. 2?" Is it to your advantage to switch your choice?
>>53876294
Yes. But we see half a car behind the door we chose. It's similar but different.
>>53876267
>This is not your first post in the thread.
This is how I know you're incorrect.
This is also how I can safely assume you're the one posting multiple times trying to defend your answer which contradicts mine and several others' which are the correct ones.
Kindly fuck off and stop believing things as fact when they aren't.
>>53876349
>But in these circumstances we see half a car behind the door we chose. It's similar but different.
>>53876357
Yes, the correct answers that directly contradict the Wikipedia article.
>>53876316
When you pick the box with 2 gold balls it doesn't matter which one of the balls you pick. If you pick the box with 1 silver and 1 gold ball, you must pick the gold ball.
So in regard to probability of picking either of the two boxes, they are identical
The problem specifies you must start with a gold ball
>>53876327
That would be valid if the question asked for the probability of that entire scenario, but it doesn't.
The question explicitly only asks for the probability of the NEXT ball.
>>53876335
Suppose you're in a coin box, and you're given the choice of three game shows.
A game show about coins in boxes
A game show called monty hall
A game show presented by Julien Lepers
You pick a box, say No. 1 because you feel related to it and Julien Lepers telles you that he is a square container made of wood for the purpose of storing metallic flat cylinders of denomination, he is, he is?
>>53876294
>>53876370
the wikipedia article contradicts itself
>>53876391
>being this bad at reading comprehension
>>53876370
>using Wikipedia as a factual source
>>53875956
ITT people can't read
The probability must be calculated AFTER you pick a gold ball first, it's the fucking premise. The probability of picking a gold ball as first is 100% because it fucking says so. This rules out the last box and leaves with a 50/50 chance.
>>53876396
>being this bad at physics
>>53876372
>So in regard to probability of picking either of the two boxes, they are identical
In order for me to better understand your reasoning, could you please provide an answer to the following problem:
> You have two boxes. One has 1000 gold coins. The other has 999 silver coins and one gold coin. You pick a box at random and pull out a coin at random. It's a gold coin. What's the probability that the box you chose at random is the one that has 1000 gold coins?
According to your reasoning, the answer to this should be 50% - which is (or, well, should be) quite obviously incorrect. So I eagerly await your answer and explanation.
>>53876244
You're right, I fucked that up.
I still get the same answer thoughimport random
num_golds = 0
num_grays = 0
tries = N = 10000
sample_size = 0
while tries > 0:
boxes = [[1, 1], [1, 0], [0, 0]]
box = random.choice(boxes)
random.shuffle(box)
ball = box.pop()
if ball is 1:
sample_size += 1
if box[0] is 0:
num_grays += 1
else:
num_golds += 1
tries -= 1
print str(num_golds / float(sample_size))
>0.660120240481
>>53876411
>physics
Ok, I'm stopping.
>>53876408
The probability of picking a gold ball is 100% but there are three balls and each has 33% chance to get picked. Two of those three are in a box with two gold balls, for a total of 67%.
>>53876426
>boxes = [[1, 1], [1, 0]
Fixed.
>>53876420
Gambler's fallacy man
>>53876441
Does not really make it any better and also makes it less of a direct simulation of how the problem was presented (although does not change outcome).
>>53876441
Fixed? So you are saying that if we change that, then the program will be correct?
Man, even /a/ is better than you guys.
>>53876444
So, no answer? Or are you not him?
>>53876436
How are there 3 balls if you're supposed to get the other ball from the fucking SAME box
These threads are a great litmus test for how shit this board is.
The problem is easily solved, and anyone who graduated high school can figure it out pretty quickly. Knowing the answer, you can go through the thread and see how it's full of people who legitimately can't figure it out, people baiting with wrong answers, and people taking that bait.
>>53875326
P("second ball golden" | "first ball golden") = P("second ball golden" AND "first ball golden") / P("first ball golden")
P("second ball golden" AND "first ball golden") = 1 / 3
P("first ball golden") = 1/2
=> (1/3) / (1/2) = 2 / 3
The result is 2/3.
>>53876488
Slow down there buddy! The people who think it's 1/2 don't understand conditional probability. You can't explain it that way.
>>53876462
I'm not him and I'm telling you >>53876420
is gambler's fallacy.
Assuming the coin you picked is gold means if you picked the silver box you automatically picked the gold coin. It's still 1/2 to pick either box.
>>53876420
I think I explained my reasoning quite clearly before.
If there was a third box filled with 1000 silver coins, we'd rule it out, we can't pick it because it contains no gold coins.
So we're left with two boxes. If we pick the one with 1000 coins, the next coin also picked from that box is gold.
If we pick the box with 999 silver coins and 1 gold coin, we have to pick that single gold coin to meet the criteria of the original problem. And the next coin picked from that box obviously is silver.
>>53876501
The problem is specified as such
>>53876449
The problem requires that the ball you pick first must always be gold
The third box with 2 silvers is a retard bait
50%
>the next ball is either gold or silver
>>53876488
>P("second ball golden" | "first ball golden") = P("second ball golden" AND "first ball golden") / P("first ball golden")
>P("second ball golden" AND "first ball golden") / P("first ball golden")
> / P("first ball golden")
no, you deliberately ignored part of the problem
>>53875326
2/3?
>>53876524
50%
>you either win or lose the lottery
>>53876524
This
>>53876508
The fact remains that you're still much more likely to have picked the box with 1000 gold coins.
There's a 99.9% chance that, had you picked the box with 999 silver coins, it would have been silver.
So it's not 50/50, the fact that you picked a gold coin at all tells you about which box you're MORE LIKELY to have picked in the beginning.
lmao so many people can't even read the question in the OP image properly
>>53876549
Tell me how I'm wrong.
>all the retards saying 2/3 pretending that the problem doesn't explicitly state in this scenario you've already picked a gold ball and only asks for probability of the next ball
>>53876538
You pick a box first, the problem says so. If you pick a box with 9999999999999 silver coins and 1 gold coin, you must pick the gold coin, because the problem says that you do.
>>53875326
Silver balls?
40%
you are not finding the probability of what box you picked, you are not finding the probability of which gold ball you have, you are not finding the probability of multiple boxes
you are finding out what the probability that the next ball you take *from the same box, after picking a gold ball* is also gold
>>53876436
There aren't three balls. There's only one other ball because you're taking it from the same box you already took a gold ball from. The chance lies in whether you're rooting around in box A or box B. It's a coin flip. 50/50.
73%
