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?
>>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%
>>53876556
It says you take a coin at random, not you take a gold coin. Did you even read the question properly?
>>53876556
No, it says you pick a ball first and in the case it's a gold ball then what happens. It just doesn't care when you pick a silver first.
2/5
>>53876554
Are you seropositive, because I hope you won't procreate.
>>53875326
As the question is stated the answer is 1/2. The probability is calculated _after_ "It is a gold ball."
If the question stated "If the first ball pulled is gold what is the probability that the next ball you take from the same box will also be gold" would be 2/3.
>>53876593
Ad hominem instead of logical arguments. Par for the course I guess.
There's a 0% chance you'll pick a gold ball afterwards because I already ran off with the heaviest box and sold it to Cash4Gold.
>>53876579
>It says you take a coin at random
>Did you even read the question properly?
0/10
>>53876579
Then it is not an analog problem with the OP and can't be used to draw any conclusions
>>53876582
No, not in the case it is gold. IT IS GOLD, you rule out all other cases.
>>53876556
So if you run this problem over and over again, and ONLY count the boxes from which the first coin you picked was gold, it wouldn't change the probability at all
99999999.9% of the time, if you pick a gold coin first, it will be followed by another gold coin.
Every once in a while, you will stick your hand into the other box with mostly gray coins, pull out a gold coin, and then the second coin you pick will be gray.
But that would only happen 0.0000000001% of the time.
>>53876600
Probability does not care about order of events.
>>53876601
There's no discussion needed to make the fact that >>53876533 is the most retarded reply ever any clearer than it already is.
>>53876527
Which part?
This is just the definition of conditional probability:
P(A | B) * P(B) = P(A and B)
Therefore:
P(A | B) = P(A and B) / P(B)
The elementary events were given.
>>53876619
And we don't care about you being an illiterate retard.
>>53876619
You want to factor in an elements that is no longer a function of probability in the equation anyway?
>>53876611
Are you saying that ever time you stick your hand in the second box you are guaranteed to pick a gold coin?
73% guys
>>53876622
That analogy actually perfectly illustrates how stupid you are if you think the answer to OP's problem is 50/50.
>>53876639
This problem doesn't need any complex formulas, just some logic.
>>53875354
>>53875385
>>53875611
>>53875878
>>53875912
>>53875914
Your poor education is showing.
>>53876658
Just like the lottery.
>>53876555
/sci/tards can't read, it is known
fucking 73% listen to me
p=(1/3)/(1/3+1/8)
>>53876664
My fucking god
Y'all morons falling for a conditional probability problem and trying to look at it as if it weren't. Google up the formula.
>>53876611
>No, not in the case it is gold. IT IS GOLD, you rule out all other cases.
>You pick a box at random.
>You pick a ball from that box at random.
>It's a gold ball.
Guess why it's worded that way. It's because it's describes one possible scenario, when you pick a gold ball and not a silver ball.
>>53876635
If your hand was in the second box you would have had to have picked the only gold coin, as the problem is specified in OP
>>53875326
I think many say 50% because the think:
"Oh, we already got one golden ball, so it can't be the right box, so it has to be either the left or the middle box: 50/50"
But this is wrong.
Because the chances to pick the box in the middle and get gold are already smaller than to pick the left box and get gold. So the event that we picked the box in the middle at the first try is more unlikely:
P("pick box left" and "get left gold") = 1/6
P("pick box left" and "get right gold") = 1/6
P("pick middle box" and "get gold") = 1/6
P("pick middle box" and "get silver") = 1/6
P("pick middle box" and "get left silver") = 1/6
P("pick middle box" and "get right silver") = 1/6
So:
P("pick box left" and "get left gold") = 1/6
P("pick box left" and "get right gold") = 1/6
P("pick middle box" and "get gold") = 1/6
And:
P("pick box left" and "get gold") = 2/6
P("pick middle box" and "get gold") = 1/6
That's exactly what this posts says:
>>53876488
These are the threads that make 4chan worth going to.
>>53876701
>Because the chances to pick the box in the middle and get gold are already smaller than to pick the left box and get gold.
That is no longer relevant when the premise fixes the result. "It is gold" not "If it is gold."
>>53876695
Then why does the problem state "take a ball from that box at random"? Is that information irrelevant?
>>53876701
You wrote all that out and you're not even right while other people just had to read the problem and already knew the answer.
>>53876701
Sorry, a typo.. Of course we need to look at the right box too . I meant:
P("pick box left" and "get left gold") = 1/6
P("pick box left" and "get right gold") = 1/6
P("pick middle box" and "get gold") = 1/6
P("pick middle box" and "get silver") = 1/6
P("pick box RIGHT" and "get left silver") = 1/6
P("pick box RIGHT" and "get right silver") = 1/6
>>53876703
What, uneducated scrubs getting trolled? Man, these are the threads that remind me why I hate this place.
>>53876693
It describes the only possible scenario we're concerned about.
You've picked a ball and random and it is GOLD.
You ruled out box 3, there are no gold balls in that box, only 100 silver ones.
Your hand may be placed in box 1 with 100 gold balls or it may be placed in box 2 with 99 silver balls and 1 gold ball, you don't know, but the choice has already been made.
Looking from here on forward, if your hand is in box 1, your next choice has to be gold. If your hand is in box 2, your next choice has to be silver.
Thus making the probability of the next ball being gold 1/2. You don't care about other events, only this one.
This is best I can explaint it.
>>53876701
you are not calculating the probability of what you get as a first ball or on all boxes, just on what is the probability of your next ball being gold from the same box, after already picking gold first
Strip this down to its bare components --
Possibility A: The box you picked at random is the one with two gold balls
- Your next ball is 100% likely to be gold.
Possibility B: The box you picked at random is the one with one gold ball
- Your next ball is 0% likely to be gold.
It's 50/50. The balls themselves have nothing to do with the question.
>>53876681
oh sry p=(1/3)/(1/3+1/6)=2/3
>>53876711
>Then why does the problem state "take a ball from that box at random"?
Fuck if I know. I didn't make the image.
>Is that information irrelevant?
Yes
>>53876247
SHUT THE FUCK UP!
everything you say is wrong. there are bait threads that are 66% better than this.
>>53876639
You drew a gold ball from either box one or two, the questions clearly states your second draw must be from the SAME BOX you drew the gold ball. If you drew the first gold ball from box one your second draw must be a gold ball, if you drew the gold ball from box two your second draw must be a silver ball. The only thing that matters in this question is if you drew a gold ball from box one or two.
>>53876711
it's the same with the third silver box, it's irrelevant
>>53876501
Can you imagine yourself in real life doing that?
You'd have two boxes, one 1000 gold cons, one 999 silver and 1 gold.
You'd be picking a box at random, picking a coin at random, and all in half of cases where the coin is gold it will be from the first box, and in other half of cases, that gold coin will be from the box with silver coins. Do you honestly imagine this is how the experiment will go?
>>53876725
The fact that you know the first coin you took out is gold changes the probability of which box you picked. It's not 50/50 with that added information.
>>53876623
Nah, you're right, dude's clueless.
>>53876571
>There aren't three balls.
There are not three gold balls...? Try reading the problem again.
a 3/6 chance
anyone who tells you otherwise can't into phenomenology
>>53876760
you are stuck with 1 box after picking the box on random
>>53876760
>There are not three gold balls...?
There are three gold balls among the total of six balls in the three boxes. But you drew a gold ball. That is 100% certain. The only factor left uncertain is whether you drew the gold ball from box A with two gold balls or box B with one gold and one silver ball.
I love how it everyone fails to see how 5% of the world population is colorblind, and would confuse a gold ball for a silver one.
#stopcolorblindphobianow
>>53876774
>>53876769
>>53876760
There are no boxes or balls it's just pixels on a screen.
>>53876725
Well, if you look only at that subtree, then it's ok, but my point is exactly that if you look at it in relation to the entire tree it isn't right anymore. I won't argue, because it's pointless (everyone seems to either understand the question this way or the other). A better specified question would make this a whole lot easier.
>>53876769
Of course. But you still pick one ball, out out of three, and all three have the same probability to be picked. Boxes do not have the same probability to be picked.
>>53876774
>But you drew a gold ball. That is 100% certain. The only factor left uncertain is whether you drew the gold ball from box A with two gold balls or box B with one gold and one silver ball.
Yes. That is uncertain. And you falsely assume that both boxes have the same chances ot be picked. Only balls have the same chance to be picked (because of the imposed condition), but the first box is more likely to be picked (again, because of the imposed condition).
>>53876714
>You wrote all that out and you're not even right
Yes I am right.
I already gave the formula (here:
>>53876488 ) and math doesn't lie.
But I thought I shoudl also give an more "intuitive" explanation..
>>53876708
Nah bro, you got mind tricked.
It's similar to the "Monty Hall" problem, just the other way arround:
At "monty hall" you pick a door, then new information gets added so it's a new game with new chances.
Here you THINK you can leave the other states behind when in fact you need to think about the way how you could get into this state in the first place..
>>53875326
Conditional probability mother fuckers.
>>53875559
i feel like ive seen this somewhere before...
>>53876796
>And you falsely assume that both boxes have the same chances ot be picked.
Once it is determined you drew a gold ball Box A and Box B have an equal chance to be picked because it is certain that you did not pick Box C once you drew the gold ball. The premise is clear. You drew a gold ball. With this premise stated there is no possibility you choose Box C with two silver balls.
>>53876750
I bet you're the kind of person who thinks if a number comes two times in a row in roulette the third time is only 1/36^3
>>53876822
Why do you assume that Box A and Box B have an equal chance of being picked?
>>53876798
>It's similar to the "Monty Hall" problem, just the other way arround:
As noted previously, it is similar but different. Under these circumstances we opened the door halfway and saw half of a car. The only question remains if this door has the other half of a car or half of a goat.
>then new information gets added
You DREW A GOLD BALL. What other information gets added after this point?
>>53876779
Maybe if you weren't so transmassphobic you would realize that it's up to the balls to decide if they identify as gold or silver.
>>53876833
There are only three choices and at the point you draw a gold ball one choice was eliminated. Once you draw the gold ball there are only two possibilities of which box you picked. Either Box A or Box B. To believe otherwise would to go against the stated premise. You drew a gold ball. This dictates that Box C is was not drawn and it's probability is 0%. It is no longer an element in to factor into probability.
>>53876796
>but the first box is more likely to be picked
that is false since the question states your first pick is always gold (it can never be silver), but it's also irrelevant since it is just asking for the probability of the next ball from the same box that you randomly picked
>>53876833
because you can never pick a silver ball as first
if it falls to B, the gold ball will always be picked first
>>53876854
If you can eliminate the two silver balls from Box C being picked, why can't you eliminate the silver ball in Box B from being picked?
Riddle me this, 66fags. There are two boxes. One has two gold balls, the other has one gold and one silver. You randomly choose a box. Somebody else looks into the box and informs you that the box contains at least one gold ball. What are the chances that you picked the box with two gold balls?
>>53876878
Because Box B meets the premise "you drew a gold ball" and thus cannot be eliminated like Box C.
>>53876822
This defies reason. See >>53876420. It should be clear even to you that the condition imposed by the problem does not make the chance of picking either of boxes equal. No, this has nothing to do with gambler's paradox. There is only one random even here and probability of its outcome does not change when you do that even multiple times.
>>53876883
It will be 50%. You removed the condition the original problem had that you picked a gold coin, so naturally the answer is going to be different.
>>53876915
>You removed the condition the original problem had that you picked a gold coin
that is this part
>Somebody else looks into the box and informs you that the box contains at least one gold ball
>the ball you picked first is gold
>>53875326
There is a german song that explains something like this:
https://www.youtube.com/watch?v=DWdcupH_p34
>>53876902
>See >>53876420.
See what exactly? His outright rejection of a 50% answer for no explicable reason?
The premise "You drew a gold coin" is not a probable result. It is a certain result. You have a 0% chance of drawing anything but a gold coin when the premise states you drew a gold coin. Something with 0% chance of occurring is no longer a factor of probability. The answer at that point becomes 50%. If the problem stated "IF you drew a gold coin" leaving the possibility of not drawing a gold coin it is no longer 0% possible to draw something other than a gold coin.
>>53876896
Now think about this problem. There is 1 giant box. There are 6 smaller boxes within the giant box. 2 of the smaller boxes are colored gold, and have a sheet of paper with "A" written on them on the inside. Another of the smaller boxes is colored gold, with "B" written on the inside. Another of the smaller boxes is colored silver, with "B" written on the inside. The last two boxes are colored silver, with "C" written on the inside. You reach into the big box, and take a smaller box out. It is colored gold. What is the probability of the other box with a sheet of paper with the same letter written on it as the sheet of paper from the box you picked being gold?
If you answered 66% come to 56 Spruce in Manchester UK I'll fight you. If my dad answers the door say you're looking for Jarett.
Everyone ignore the fact you already drew a gold ball, and they especially ignore that you must draw from the same box you drew your first gold ball.
>>53876923
It's not the same. Considering both the boxes have at least one gold coin, your clarification about someone looking and saying does not add any more information at all. It can be removed, it can be retained and in both the cases it will be the same problem, and will be different from the one with condition that the coin you pulled out is gold.
>>53876933
You've changed the parameters of the question. You don't choose two of six possible choices. You choose one of three.
>>53876959
You choose one of six choices in both variations.
>>53876932
So do you think, that when you set up an experiment like I described, you would be as likely to pull out a gold coin from box A as from box B?
If I was there in person with you, I'd get you to do the experiment. As it is right now, I'm powerless.
>>53876959
>You don't choose *one of six possible choices.
Fixed.
>>53875326
Probability of getting two balls as gold = 1/3
Probability of getting first ball as gold = 1/2
( [1/3] / [1/2] ) = 2/3
>>53876925
Sorry confused the op-image with the Monty-Hall-Problem.
>>53876973
>You choose one of six choices in both variations.
You've just violated the parameters of the OP's question. You choose one of three boxes not one of six balls.
>>53876981
>Probability of getting first ball as gold = 1/2
that's not what happens
you will always get first ball as gold, as stated in the problem
>>53876837
>You DREW A GOLD BALL. What other information gets added after this point?
Yes, but WHICH gold ball?
The probability to get a golden ball from the left box is 2/3.
>>53876883
That's a differnt problem!
Because if you look into the middle box it always contains a golden ball.
But if you picked a ball from the middle box chances are 50% that it's silver..
To explain it one last time:
Assume this:
There are three cities, Paris, Berlin and London.
In Paris and Berlin live 3 million people, but only ONE of them is black. In London everybody is black.
Now you get drugged and when you wake up the first person you see is black..
Now don't you think it's more likely you are in London? Because the chances you met the one black peron in Paris or Berlin is tiny.
So it's NOT 1/3 for every city.
>>53876974
>you would be as likely to pull out a gold coin from box A as from box B?
Irrelevant to the premise in the OP. The likelihood of drawing a gold ball from Box A is just the same as Box B in the premise. You drew a gold ball set it at 100% likelihood. There is on question of not pulling a gold ball.
a lot of morons for a """smart""" board
every single 2/3 fag should be shot
>>53876995
>Yes, but WHICH gold ball?
Why does it matter _which_ gold ball? It is the first pull from whatever box you chose. The fact that you pulled it from Box A or Box B does not change the other ball in Box A or Box B. Thus why the wording "You drew a gold ball" and not "if you drew a gold ball" is critical.
I would say it's 1/5 since the next ball could be gold or silver and you don't know what's in the other boxes.
>>53876992
You choose one of three boxes and then one of two balls inside the box you chose. Resulting in 6 possible, equally likely (before the condition is applied) outcomes. After the condition is applied, there are 3 equally likely outcomes - because you have 3 golden balls to pick from.
>>53876992
You are still choosing one of six balls in the original question. You pick a box at random, then take a ball at random. All of the six balls have an equal chance of being picked, before the fact that a gold ball was picked is conditioned.
>>53877009
Read the rest of my post, maybe you understand it then..
>>53876847
Wow! That is so racist of you.
In fact, when OP has a gold ball in his hand, there's 95% he's not color blind which means the ball is actually gold, and 5% chance that he'd confuse a random ball for a gold ball.
In the 95% non-colorblind scenario there's a 2/3 chance that the gold ball came from the left box, meaning there's 2/3 chance that the next ball will also be and be seen as golden.
In the 5% colorblind scenario, there's an equal chance that any ball was picked, meaning there's an equal chance any second ball is picked, which has 1/2 chance of being gold or silver. However, the silver ball would be confused for a gold ball, making the chance on picking a gold ball, and then another gold ball 100% in the 5% case of OP being colorblind.
This leaves us with 19/20 * 2/3 + 1/20 * 1 = 41/60.
>>53877014
>Resulting in 6 possible,
If you have a 0% chance to draw anything but a gold ball how do you have six possibilities in the OP's problem?
>>53877018
>then take a ball at random
woah there
as stated in the problem, you get a gold ball first, no ifs. It then asks for the probability of the next ball you take (from the same box) being gold
>>53877022
I read the part of your post that responded to my post. Repeating yourself does not make your post any less irrelevant to the OP's problem. You claim there is relevance to which gold ball was pulled but did not explain how it is relevant. Instead you merely presented a repetition.
>>53876994
and thus it is divided away in the last step.
P(A and B) = P (A) × P (B|A)
Thus, to know P(B|A), you can calculate [P(A and B) / P(A)]
>>53877023
Nice bait.
Let's just ignore the "gold silver interchange visual disorder" which makes people see golden objects as silver.
Check your priviledge, man!
>>53877000
It's not irrelevant. It's exactly the same. You picked a box at random. You picked a coin at random. Out of a 1000 in your picked box. If someone asked you, which box do you think you pulled it from, would you answer just randomly because, by your reasoning, both boxes have the same probability? Or would you answer box A, because it has 1000 gold coins in it, while B has just 1 and 999 silver ones?
>>53877031
You get a gold box first, no ifs. It then asks for the possibility that the other box that has the same letter inside the box you got is gold.
>>53877028
>Resulting in 6 possible, equally likely
>(before the condition is applied)
>After the condition is applied, there are 3 equally likely outcomes
Please use your reading skill.
>>53877059
>You picked a coin at random.
And it stays a random pick after you determine the result? You still have a 33% chance to pick a silver ball after you 100% drew a gold ball?
>>53877045
same box
>>53877076
If the event is random, it stays random after it's finished. There is an outcome, and a probability of that outcome. Toss a coin. It comes heads up. That's an outcome. Tossing a coin is a random event. It has, had and will always have 50% probability of falling heads up.
>>53877072
>>After the condition is applied, there are 3 equally likely outcomes
How are there three likely outcomes when one of the three is 100% determined? You drew a gold ball. There is a 0% chance you drew Box C.
Why do you continue to treat the gold ball you drew as conditional when it is determinate?
>>53877095
>If the event is random, it stays random after it's finished.
[citation needed]
I'm sorry. If I shoot a cat in a box and observe the cat is shot there is no longer a random chance I didn't shoot the cat.
>>53877042
>You claim there is relevance to which gold ball was pulled but did not explain how it is relevant. Instead you merely presented a repetition.
Yes I did.
OK, let's explain it like this. You should think about this thoroughly:
You take all balls from the original problem out of their boxes, under every ball you make a small note from which box you took it.
Now you have two golden balls with a "1" underneath and one golden ball with a "2" underneath.
You pick one golden ball. What are the chances to get a "1" underneath?
Yes, it's 2/3.
>>53877095
>If the event is random, it stays random after it's finished.
You're a retard. You have a one in 52 chance to pull the ace of spades from a deck of cards. When you pull a card and it is the ace of spades you do not still have a one in 51 chance to pull the ace of spades from a deck of cards.
>>53877101
>How are there three likely outcomes when one of the three is 100% determined?
There are three outcomes and each has a 33% probability. That stays true even after you carry out the random even and actually pick one golden ball. You have it, but the probability of it for the random event is still 33%.
>You drew a gold ball. There is a 0% chance you drew Box C.
Not arguing otherwise (although if you think A and B both have 50% of being picked, you are incorrect about that).
>Why do you continue to treat the gold ball you drew as conditional when it is determinate?
Which out of three balls you draw is not determined. It's an outcome of a random event with three equally likely outcomes.
>>53877113
>[citation needed]
lol
Here: https://en.wikipedia.org/wiki/Event_%28probability_theory%29
So much autism ITT. Thread only proved /g/ can't read for shit. Props to everyone who said 1/2.
>>53877133
>You pick one golden ball. What are the chances to get a "1" underneath?
You are trying to input likelihood again in a problem that removed it.
>>53877101
Because that is the only way that the question makes sense. The question states that you choose a box at random. Box C has no silver balls. The question cannot be consistent if you assume no "if", because it is saying you are guaranteed a gold ball, but also you first pick a box at random, which could be Box C, which only has silver balls.
>>53877136
It's a different random event.
In the first one you have 52 cards in the deck.
In the second one there is 51 cards.
>>53877139
>There are three outcomes and each has a 33% probability. That stays true even after you carry out the random even and actually pick one golden ball
Except the question is not about the likelihood of your first pull being a gold ball. That is 100% set. You are still focusing on something that is 100% determined and no longer a factor in the result of the second pull. You do not put the first ball pulled that is determined to be gold back into the same box. You only pull the second ball from the same box. If there are three gold balls in total and you remove one how many other gold balls are there? Three? No. There are two. The likelihood of pulling a gold ball on your first pull is not a probability when the premise states "you pulled a gold ball." Stop trying to put the gold ball back into the pool.
>>53877156
the third box is completely irrelevant to the question
ignore it completely
whats the probability of a box that has one gold having another gold?
A: 50%
whats the probability of first picking one gold then picking another gold out of the same box?
A:face palm
B:60%
if your asking:
whats the probability of picking the box w/ two gold?
A: your an idiot
B: 33%
whats the probability of picking two gold when the box with only silver is removed?
A:your fucking retarded
B:75%
>>53877182
How is it irrelevant? The question STATES that there are 3 boxes, and you choose one RANDOMLY. How can this be consistent with the notion that you are GAUARANTEED to pick a gold ball?
>>53877146
No, you didn't "remove" it..
It's exactly the same as before:
[BEFORE YOU DRAW YOU FIRST BALL]
Chances are 2/3 to get a golden ball from from box 1.
Chances are 1/3 to get a golden ball from from box 1.
[BEFORE YOU DRAW YOU FIRST BALL]
Chances are 2/3 you just got a golden ball from from box 1.
Chances are 1/3 you just got a golden ball from from box 1.
>>53877156
>Because that is the only way that the question makes sense.
There is a 0% chance that the first ball you pulled will be the second ball you pulled but let's put it back in anyways to make it possible that the second ball pulled will be one of three gold balls?
Once you pull it you can only pull the second ball. The likelihood of pulling a gold ball on your first pull is 100% that leaves how many gold balls? Three still?
>>53877188
>60% oops 66%
>>53877197
Sorry, "1/3 from box 2"..
>>53877162
>It's a different random event.
If we put the gold ball back into the box you might be right. But we don't. The first pull is a gold ball. It will always be a gold ball. It will not randomly change to not a gold ball. It is not a random event in the first place. It is a set and fixed event because the premise specifically states "you draw a gold ball."
>>53877197
Sorry, a fucking mess..
Once again:
>[BEFORE you draw your ball]
Chances are 2/3 to get a golden ball from from box 1.
Chances are 1/3 to get a golden ball from from box 2.
>[AFTER you draw your ball]
Chances are 2/3 you just got a golden ball from from box 1.
Chances are 1/3 you just got a golden ball from from box 2.
>>53877199
Yes. There are 3 gold balls. Two of them are from Box A. One of them is from Box B.
>>53877177
>. The likelihood of pulling a gold ball on your first pull is not a probability when the premise states "you pulled a gold ball."
Yes. But which out of three golden balls you pulled fully determines if there is another one in that box or not.
>>53877209
Does not specify which golden ball. Which out of three golden balls you pulled fully determines if there is another one in that box or not.
>>53877177
The question not really about balls, it's about boxes.
It's about the likelihood a random gold ball being in box A.
>>53877223
Yes, that's it.
I already explained it here:
>>53877133
I guess it's just trolling now.
>>53877218
Let's try this another way. Let's say we have the same problem but this time you are given a glove that has enough attraction to pull a single gold ball. Even from outside the box. You choose a box at random and you feel your glove get heavy meaning you picked a box with a gold ball. You pull that box from the other two and as you pull your hand a way the gold ball attracted by the glove is pulled. Thus you pulled a gold ball. What is the chance you choose the box with two gold balls?
>>53877218
Two gold balls and one silver ball remaining between both boxes A and B is irrelevant to the binary choice between if you picked from box A or B.
>>53877219
>But which out of three golden balls you pulled fully determines if there is another one in that box or not
how? You will always pull a gold ball first in either box A or B
>>53877219
>which out of three golden balls you pulled fully determines if there is another one in that box or not.
How so? If you have a 0% chance to pull a silver ball, as the premise states, what relevance is there in that you didn't pull a silver ball from Box A or Box B? See the problem in >>53877236 for example.
>>53877223
>It's about the likelihood a random gold ball being in box A.
Look at the problem from another way. You have a 0% chance to pull a sliver ball on your first pull. What is the likelihood that you pulled a gold ball from Box A or Box B?
>>53877236
This is a differnt problem..
It would be the same if you explained it like this:
>"Let's say we have the same problem but this time you are given a glove that has enough attraction to pull a single gold ball. Even from outside the box."
Now you put one ball in every corner.
>"You choose a box at random .."
but are only allowed to touch one corner.
>"..and you feel your glove get heavy meaning you picked a box with a gold ball"
..at the right corner.
Now we know the box with the golden balls has TWO corners with golden balls in it.
The other box has only ONE corner with a golden ball in it.
>>53877244
>>53877264
You pulled out gold ball 1 (from box A), gold ball 2(from box A), or gold ball 3(from box B)
case 1: the second ball is gold
case 2: the second ball is gold
case 3: the second ball is silver
Really simple.
>>53877055
Thank you for pointing this out. You're right, as compensation I've slit my wrist 14 times.
So let's put the facts together:
18/20: not colorblind
1/20: perceive-gold-as-silver syndrome
1/20: perceive-silver-as-gold syndrome
We have been deceived by OP, which sneakily, inherently excludes people with perceive-gold-as-silver syndrome by stating "it's a gold ball".
Disgusting! There is no way someone with perceive-gold-as-silver syndrome identifies a picked ball as gold. This is exactly what is wrong with America today.
>>53875326
Fuck off racist!
>>53877278
Best post in this whole thread.
I'm out..
>>53877244
>>53877252
As we gave names to boxes (A B C), let's give names to balls - 1, 2, 3, 4, 5, 6 (1 2 are in A, 3 4 are in B). Pulling out gold ball 1 means the other one is gold also. Pulling out gold ball 2 means the other ball is gold also. Pulling out gold ball 3 means the other ball is silver. Hence, which golden ball you picked (out of three - 1 2 3) fully determines if there is another one in that box or not.
>>53877272
>This is a differnt problem..
No. It is the same problem. Only we remove the confusion caused by "random pull" as we have a 100% chance to get a gold ball or a 0% chance to get a silver ball depending on how you look at it. Either way the premise is determinate. You draw a gold ball and failed to draw a silver ball. If your equation is correct it should work both ways but it does not.
>>53877277
>You pulled out gold ball 1 (from box A), gold ball 2(from box A),
You can only draw one ball. The fact that there are two balls does not mean you can draw two gold balls from Box A. It is irrelevant which of the two gold balls in Box A is drawn just like it is irrelevant which of the one gold ball in Box A is pulled.
>>53877293
>let's give names to balls - 1, 2, 3, 4, 5, 6
That would involve an order which is not provided for in the problem. You do not get to arbitrarily add a new premise.
>>53877277
If you drew a gold ball from box one at the start of the question there can only be one gold ball remaining in box A. There can only be two cases, you drew the other ball from box A or B.
>>53877295
>just as it is irrelevant which of the one gold ball in *Box B is pulled.
The question is not about the likelihood of the first drawn gold ball.
>>53877305
There is no requirement for order. 1 and 2 are in box A in any order. 3 and 4 are in box B in any order. 5 and 6 are in box C in any order. 123 are gold, 456 are silver.
>>53877306
>If you drew a gold ball from box one at the start
Then the second one you pull is gold. 100%. It doesn't matter how many are left.
>>53877236
It would be the same answer. Instead of pointless bickering, just carry out the experiment yourself and discover that it is, indeed, 66%.
>>53877337
>There is no requirement for order.
Then why are you trying to number the balls unless it is to impose an unrequired for basis of order?
>>53876408
>ITT people can't read
>>53876408
>it's the fucking premise
that is literally what the two continue statements does you inbreeding asstractor
>>53877355
So that I can refer to them by name, just as you named the boxes so you can refer to them by names.
>>53877355
To make it simpler for you to understand conditional probability.
I could post a wikipedia link but there's no way you'd understand anything.
>>53877354
>It would be the same answer.
If we had a glove that automatically eliminated one box from being randomly chosen, one without any gold balls in it. There is still a one in three chance that we will pick a box with two gold balls?
>>53877295
You aren't understanding the point of the OP question. You are ultimately picking balls, not boxes. As I stated before, the OP is probably misworded in the way that it should be "if it's a gold ball", or else it cannot have a consistent narrative. In that case, you should do the experiment, as specified by the OP, ignoring the trials where you get a silver ball, and see the results for yourself. I can't be bothered to argue anymore.
>>53877369
>So that I can refer to them by name
Why would you want to refer to them by name? Is not a name used to distinguish one similar object from another, or to put it as stated previously, impose an unrequired for basis of order?
>>53877373
>To make it simpler for you to understand conditional probability.
You continue to claim it is conditional. You do not put the first ball pulled back into the box. How is that conditional?
>>53877295
Look, you just aren't right.
Many guys have tried to explain it in many differnt ways, there's a mathematical proof as well as some others. Why don't you try to listen?
In statistics peopel often fall for it, when the excercise sound as if you where looking for A, when you are actually looking for B..
>What do we know?
-We just pulled a golden ball.
-There are three golden balls in this game.
>What do we want to know?
Is the next ball in this box golden or silver?
--> The next ball is golden ONLY IF we pulled a ball from box one.
-> The next ball is silver ONLY IF we pulled the single golden ball from box two.
We have no further clue which box we took.
We have no further clue which ball we took.
So we must assume it's 1/3 for each ball.
This means:
If we took ball 1 or ball 2, we are in the first box. This means the next ball will also be golden.
If we took ball 3, we are in teh middle box. This means the next ball will be silver.
Since we don't know about which ball we grabbed at first, we can only assue it's 1/3 for each ball, meaning with a probability of 2/3 we grabbed ball 1 or 2 (and the next bal in the box is also golden).
>>53877346
Well obviously, but there is no way there are 66% odds. If you draw a gold ball from box A there can only be a single gold ball left in box A and it is your only option. If you drew the gold ball from Box B all that will remain in box B is a silver ball, and that is your only option.
The question boils down to did you remove the ball from Box A or Box B.
>>53877402
there are two golden balls in box A
>>53877391
>You aren't understanding the point of the OP question
I say the exact same thing about you.
>You are ultimately picking balls, not boxes.
Wrong. The contents of the box are fixed. You pick a box. Which ball you choose from the box is already determined. There is no relevance to picking a ball. The result is determined the only question left is what is the other ball in the box you picked.
>>53877402
>The question boils down to did you remove the ball from Box A or Box B.
No, the question boils down to did you remove the ball 1 from Box A or the ball 2 from Box A or the ball from Box B.
>>53877381
Right. I reread your problem. The answer for your problem is 50%. I will now explain the difference between those two problems for you.
The glove does not eliminate the box from being chosen. When carrying out the experiment, in reality, you will sometimes pick box C. In that case, you will discard that experiment and try again. This will allow you to calculate the probability which will end up being. In your problem, you only discard experiments where you picked box C.
In original problem, you will discard all experiments where you picked box C and ALSO somew of experiments where you picked box B. This is what causes the answer to be different for these two different problems.
>>53877402
And CONDITIONING on the fact that you drew a gold ball, the chances of you drawing it from box A is 2/3 and Box B 1/3. How is this hard to understand?
>>53877398
>Many guys have tried to explain it in many differnt ways, there's a mathematical proof as well as some others. Why don't you try to listen?
I've tried to explain it in various ways. Why don't you try to listen?
>So we must assume it's 1/3 for each ball.
What?
>--> The next ball is golden ONLY IF we pulled a ball from box one.
>-> The next ball is silver ONLY IF we pulled the single golden ball from box two.
Where did this phantom third choice come from to assume 1/3rd when you only list two possible outcomes?
>>53877393
The order does exist, we just don't know it.
>Why would you want to refer to them by name?
To prove the statement that you asked me to prove few posts above (obviously).
>>53877434
>you can either win or lose the lottery
>there are only two options, so it is 50/50
>>53877408
And if you selected box A you removed one of these balls. There are only two boxes to draw from, your odds are still 50% regardless of the balls originally in the box at the start of the problem.
>>53877434
You are assuming there is equal likelihood to pick either box with the condition imposed by the problem, and that assumption is incorrect. there have been many explanation posted why the likelihood is not the same, and you are ignoring all of them.
>>53876488
>>53876981
This is true, yet P("second ball golden" | "first ball golden") is not the probability the ill-worded OP's image is about.
Written as is, P("second ball golden" | "first ball golden") is the probability when you're excluding the box with silver balls from the second experiment, because of the infamous "it's a gold ball" statement. Your probability is actually P("second ball golden in an experiment excluding the silver box" | "first ball golden in an experiment including the silver box").
Yet, if you keep the silver box in the second run, you can apply the law of total probabilities to P("second ball golden" | "first ball golden").
Let's say:
- there are boxes A, B and C in OP's image order
- G1A is "I pick up a golden ball at first in box A", G1B the same in box B, G1C the same in box C
G1A, G1B and G1C form a partition of G1, with P(G1A|G1) = 1/2, P(G1B|G1) = 1/2 and P(G1C|G1) = 0. Thus, it follows from the law of total probability that:
P(G2|G1) = P(G2|G1A)*P(G1A|G1)+P(G2|G1B)*P(G1B|G1)+P(G2|G1C)*P(G1C|G1)
Now, given OP's image, P(G2|G1A) = 1 (box A is only gold balls), P(G2|G1B) = 0 (box B only contains a silver ball if G1B) and P(G2|G1C) = 0 (duh).
It follows that P(G2|G1) = 1/2.
>>53877445
in one universe you removed the first ball from box A
in another you removed the second ball from box A
in a third you removed the ball from box B
in two of those universes, the remaining ball is the other ball from box A
in one of those universes, the remaining ball is the silver ball from box B
>>53877421
>The glove does not eliminate the box from being chosen
If we want to know the answer the question, yes it does. We will never choose box C if we want to know the chance of picking a second gold ball from the box.
>You are assuming there is equal likelihood to pick either box with the condition imposed by the problem
You explicitly stated there were only two possible results yet you want to calculate for an impossible choice of the third box?
Are you daft?
>>53877452
Forgot to define G2 as "I pick up a golden ball at second turn).
>>53877434
>I've tried to explain it in various ways. Why don't you try to listen?
Because studied this shit and you don't.
> when you only list two possible outcomes?
There are two outcomes, but three possibilities.
Or more precisely:
There are two "events":
1) You took a golden ball from the left box
2) You took a golden ball from the middle box
But there are three "elementary event":
1) You took the first golden ball from the left box
2) You took the second golden ball from the left box
3) You took the golden ball from the left box
The probabilities result from the "elemenatry events", not from the events..
>>53877452
There are two gold balls in box A, and one gold ball in box B. It's not 1/2.
>>53877451
Whoops! See second part of >>53877466 because I forgot the link.
>>53877469
You'll excuse me if I scoff at your claim at superiority and elitism and disregard the rest of your post as completely baseless shit.
>An impossible result is still a possibility.
I don't even have to pretend it isn't baseless shit.
>>53877469
>1) You took the first golden ball from the left box
>2) You took the second golden ball from the left box
You're still separating these two as a result for the first pull which is irrelevant. It does not matter which of the two gold balls you pulled from Box A just like which of the one gold ball you pulled from Box B when you 100% pulled a gold ball from whatever box you chose.
>>53877466
>We will never choose box C if we want to know the chance of picking a second gold ball from the box.
It is impossible in a random event. The method to account for conditional probability when carrying out experiments is to discard experiments whose outcome are forbidden by the condition.
>>53877466
>You explicitly stated there were only two possible results yet you want to calculate for an impossible choice of the third box?
The likelihood of picking first box is not the same as the likelihood of picking second box, with the condition imposed. The likelihood of picking third box is 0 with the condition imposed.
>>53877393
That's what conditional probability is. The probability of A(second ball is gold) -assuming- B(first ball is gold)
>>53877490
>It does not matter which of the two gold balls you pulled from Box A
why
>>53877495
>That's what conditional probability is.
It's a condition because I say it is a condition.
The only condition is which box you chose. It does not matter which ball you pulled from the box. It is a gold ball. You have the same 100% chance to pull a gold ball from Box A as you do the 100% chance to pull a gold ball from box B. The only question left is did you pull the 100% gold ball from Box A or Box B. There is no third possibility as you cannot pull a 100% gold ball from Box C.
There is no if, and, or but. You pulled a gold ball. That is 100% certain. You can call it a condition just like I can call you daft.
>>53877510
Why are you even in this discussion when you don't know what conditional probability is?
>>53877510
alright
https://en.wikipedia.org/wiki/Conditional_probability
>>53877510
If you follow the procedure as described in the OP post, there is no 100% chance of getting a gold ball. That is why we have to condition on that event.
>>53877500
>why
Because the premise states you draw a gold ball. There is no probability of which ball is drawn. Whatever ball you pulled first is gold regardless of which of the three gold balls it is or which of the two boxes with gold balls in them you pulled it from. That result determines the other result based upon which box you chose. The content of the box does not change because you believe claim it can.
>>53877528
The boxes do not have an equal probability of being chosen under the conditional that a gold ball was drawn.
>>53877520
>In probability theory, conditional probability is a measure of the probability of an event given that another event has occurred.
What probability is there to condition the second ball upon? It's value is set relative to the box you chose. The box you chose is not relative to the ball you pulled because regardless of if you chose Box A or Box B you pulled a gold ball from it. That is not a probable event but a static event. Stop trying to present the premise "you drew a gold ball" as something other than a static event.
Let 'GG' denote the outcome of two gold balls in a row.
Let 'G' denote the outcome that the first ball drawn is gold.
Let 'P(x|G)' denote the probability that the ball was provided from box 'x' provided that it is gold
With boxes numbered 1,2,3 in order of the image:P(GG,G) = P(GG|1,G)P(1|G) + P(GG|2,G)P(2|G) + P(GG|3,G)P(3|G)
P(x|G) = P(G|x)P(x)/P(G)
P(G) = (1/3)*1 + (1/3)*(1/2) + (1/3)*0 = 0.5
P(x) = (1/3)
x P(G|x) P(x|G)
1 1 1*(1/3)/(0.5) = (2/3)
2 1/2 1*(1/3)/(0.5) = (1/2)
3 0 1*(0)/(0.5) = (0)
P(GG,G) = 1*(2/3) + 0*(1/2) + 0*(0) = (2/3)
How did I do /g/?
>>53877528
>That result determines the other result based upon which box you chose. The content of the box does not change because you believe claim it can.
This only lets you enumerate outcomes. Yes, there are two of them - picking box A and picking box B. This does not let you know the probabilities of those outcomes, and in particular you assuming that those two outcomes are equally likely because there are two of them is incorrect. You have no basis to claim that under imposed condition in the problem in OP A is as likely to be picked as B - to claim in, you have to prove it.
>>53877526
>If you follow the procedure as described in the OP post, there is no 100% chance of getting a gold ball.
What? The premise "You drew a gold ball" means you may not have drawn a gold ball?
>>53877542
>The boxes do not have an equal probability of being chosen under the conditional that a gold ball was drawn.
Really now? There are three boxes. One box has no gold balls in it. That leaves two boxes. But they do not have an equal chance of being chosen if you pull one gold ball from those two boxes?
>>53877548
Oops, should beP(2|G) = (1/2)*(1/3)/0.5
>>53877543
>In probability theory, conditional probability is a measure of the probability of an event given that another event has occurred.
>an event given
This event is you finding the second ball in the box you picked.
> given that another event has occurred.
This event is you pulling a gold ball when picking a random box and a random ball in it.
>>53877490
>You're still separating these two as a result
God fucking damnit, are you even reading my posts befre answering?!
That's why I wrote:
"The probabilities result from the 'elemenatry events', not from the events"
In the OP we learn about an EVENT (golden ball drawn). An EVENT consists of ELEMENTARY EVENTS.
As an example:
"You have a dice. You just rolled an even number. How big are the chances that there's a higher number?"
Again we have the EVENT of an even number. It soncists of three ELEMENTARY EVENTS: 2,4,6.
If it's 2 or 4, we have a higher number.
If it's 6 we have no higher number.
That's why the statement about the EVENT didn't give us enough information to rule out 2 or 4. So we must assume each of them is possible.
It's exactly the same here.
The EVENT of drawing a golden ball has happend.
Yet we don't know which ELEMENTARY EVENT has happend, therefore we can't rule out anything.
A difference would be:
First box has gold and iron ball.
Second box has gold and aluminium ball.
Now the probablity for iron is 1/2.
But since both balls are golden, EACH OF THE GOLDEN BALLS in the first box can be the golden ball in question..
>>53877568
Yes.
If there was no condition, they would be equal.
But the condition is there, so box A is more likely to be picked.
>>53877557
"You drew a gold ball" is a premise, or a condition, in the language of probability. The procedure in the experiment are separate. The procedure states what you DO. The condition specifies part of the results. The procedure is that you pick any box at random, Box A, Box B, or Box C. You then pick any ball in the box. Naturally, this procedure does not guarantee that you get a gold ball. Now the problem asks what happens IF you get a gold ball.
>>53877580
>This event is you finding the second ball gold in the box you picked.*
fixed
>>53877580
>This event is you finding the second ball in the box you picked.
The probability of the box you chosen is not conditional to the first gold ball pulled. Nor is the second ball in the box you chose conditional to the first gold ball you pulled.
There is no probability event conditional to the first gold ball being pulled.
>>53877584
Is 0% a probable event?
import random
class Coin(object):
def __init__(self, type):
self.type = type
class Box(object):
def __init__(self, coinA, coinB):
self.box = [coinA, coinB]
def pick_coin_from_box(self):
choice = random.choice(self.box)
leftovers = [c for c in self.box if not (c is choice)]
return choice, leftovers
def Experiment():
#wait for choice of golden coin to satisfy premise
while True:
boxA = Box(Coin('gold'), Coin('gold'))
boxB = Box(Coin('gold'), Coin('silver'))
boxC = Box(Coin('silver'), Coin('silver'))
box_choice = random.choice([boxA, boxB, boxC])
coin_choice,leftovers = box_choice.pick_coin_from_box()
if coin_choice.type == 'gold':
gold_leftovers = len([c for c in leftovers if c.type == 'gold'])
silver_leftovers = len([c for c in leftovers if c.type == 'silver'])
return gold_leftovers, silver_leftovers
else:
#no golden coin, experiment is not valid
continue
tries = 10000
N = tries
gold = 0
silver = 0
while tries > 0:
g,s = Experiment()
gold = gold + g
silver = silver + s
tries = tries - 1
print("Gold ratio: {}\nSilver ratio: {}".format(gold/float(N), silver/float(N)))
Gold ratio: 0.6589
Silver ratio: 0.3411
>>53877596
I ask you as well. Is 0% chance of a result a probable event?
>>53877617
There is no such thing as "is conditional". You can apply a condition on anything. I think what you are looking for is "independent of", which is also false. It is clear that the fact that seeing a gold ball means you did not pick box C. Why do you think it is a stretch to think that it affects the chance of picking box B?
>>53877648
>You can apply a condition on anything.
We are not discussing a condition. We a discussing a conditional probability. If we have no probability then we have no conditional probability as well?
>>53877637
You are confusing terms. An event can have 0% probability, yes. You have 0% chance to toss a coin and end up with it being both tails and heads up. And experiment can have outcomes with 0, 100, or anything in between probabilities.
>>53877663
There is a probability for everything. 0 and 1 are also probabilities. The probability that 1 + 1 is equal to 2 is 100%. You don't seem to have a background in probability judging from your misuse of vocabulary, so I ask that you make yourself more clear when describing the terms you use.
>>53877648
>I think what you are looking for is "independent of", which is also false.
And I should have gotten to this as well.
The result of pulling a gold ball from a box is indepdent of the number of gold balls in a box if it is greater than zero. The only thing having a 0% chance of pulling a silver ball determines is that we have no chance of failing to pull a second silver ball from the box.
>>53877672
>An event can have 0% probability, yes.
Then we are discussing a certainty not a probability. Thus there is no conditional probability involving the certain event of drawing a gold ball.
>>53877682
>There is a probability for everything for certain events.
[citation needed]
There seems to be a clear distinction between a probable event and a certain event.
>>53877689
It is not. You can run the experiment yourself. Multiple people have posted programs that simulate the experience. You can look on Wikipedia for more proofs that the solution is 2/3. At this point, you are simply showing off your lack of knowledge on the foundations probability.
>>53877617
>Is 0% a probable event?
No it's not. Every event has to have a probability.
Thats why you can't just tal intuitively and say "there's a 0% of getting a silver ball in the first draw".
This effectively means "there are no silver balls".
You can't just remove them in the first draw and then add them again, that isn't correct mathematically..
You need to understand the actual question:
"Something has just happend (golden ball) BUT we don't know what exactly (which box from)."
The andwer of the question is realted to the state in the beginning (including the silver balls), we can NOT say "there's a 0% chance in the first round", it's just not how statistics work..
>>53877631
Beat me to it, I was doing in it Java.
But the guy won't accept it anyway..
>>53877631
No Your Wrong !!
>>53877689
>Thus there is no conditional probability involving the certain event of drawing a gold ball.
My answer has no relation to what we were discussion before I asked an isolated question here: >>53877637
If you want to apply this "knowledge" to the problewm in OP, you'll have to fully elaborate on it.
>Then we are discussing a certainty not a probability.
Certainty and probability are not mutually exclusive. Probabilities of 0 or 1 are certainty - and you can still reason about them as probabilities.
>>53877711
>No it's not. Every event has to have a probability.
Thus the first pull of a gold ball is certain and cannot be an event for a conditional probability.
>Thats why you can't just tal intuitively and say "there's a 0% of getting a silver ball in the first draw".
It is not intuitive. It is observed and determinative based upon the premise. There is a 0% chance that first pull with be a silver ball. What is the chance that the next pull from the same box will not be a silver ball?
>>53877722
>Certainty and probability are not mutually exclusive
I didn't claim that the were. I claimed that they were distinct. That you cannot base conditional probability upon a certain event thus the repeated claim that the first gold ball pulled is a conditional probability is null and void.
>>53877723
The first pull of a gold ball is not certain, it is a condition, how many times must we explain this to you?
Run this procedure in the OP 300 times. Around 100 times, you will pick Box A. Around 100 times, you will pick Box C. Around 100 times, you will pick Box C. Every time you pick Box C, and around half the times you pick Box B, you will end up with a silver ball.
>>53877697
I will give you this link again.
https://en.wikipedia.org/wiki/Event_%28probability_theory%29
Probable event is not even a thing.
>>53877737
You can calculate conditional probability of any combination of events, including any number of possible, impossible or certain events. The conditional probability is calculated using a simple formula.
>>53877750
>The first pull of a gold ball is not certain
Go read the OP's problem again. Specifically:
>It's a gold ball.
Are you claiming that the problem is wrong? There is anything other than a 0% chance of what you pulled was a silver ball?
>>53877723
>Thus the first pull of a gold ball is certain and cannot be an event for a conditional probability.
The first pull is not "certain", it's what just happend. You need to find out WHAT EXACTLY happend.
>Certainty and probability are not mutually exclusive. Probabilities of 0 or 1 are certainty
A probabiliy of 1 is called certainty.
A probabiliy of 0 is called "impossible event".
The problem is (I'm repeating myself) you can't just "shift" probabilites (like you are doing) and say "now it's certain, now it's uncertain"..
>https://en.wikipedia.org/wiki/Conditional_probability
>>53877753
>A single outcome may be an element of many different events
May be? May be it was an independent event? May there be no elements to the event contrary to what was previously claimed?
>An event defines a complementary event, namely the complementary set (the event not occurring), and together these define a Bernoulli trial: did the event occur or not?
There are independent events and the gold ball being pulled is one of them.
>>53877766
You are not familiar with probability theory. "It's a gold ball" is a way to specify a condition, exactly the same as "If it's a gold ball". Go read the OP's problem again. Tell me, how, if you follow the procedure as described in the OP for picking a box, then picking a ball, how you can GUARANTEE that you will pull a gold ball.
>>53877766
>Are you claiming that the problem is wrong?
I'll spell it out for you:
"In probability theory, conditional probability is a measure of the probability of an event given that (by assumption, presumption, assertion or evidence) another event has occurred."
>https://en.wikipedia.org/wiki/Conditional_probability
>>53877780
>The first pull is not "certain", it's what just happened.
What is the difference in a single event. You picked one box. You pulled one ball. It was gold. That is not a certain event?
>>53877784
You have been given solid reasoning. You asked for references, there they are. Address my arguments directly or don't post.
>>53877785
>You are not familiar with probability theory.
I think I'm familiar enough to realize when you are trying to go outside the problem as it was it was set. No matter how many times you run the same problem you cannot ignore part of it because you don't like it. Every time you run the equation that fits within the premises as set by the problem you will find that every time you pull a gold ball on the first pull. That is static. The only way you will "happen upon" a different result is if you go outside of the problem as it was presented.
>>53877791
> That is not a certain event?
No, it's not a certain event.
It's a reduction of possible events, about the space where you are in.
It's like a map where you zoom in.
>>53877802
>You have been given solid reasoning. You asked for references, there they are
Your own provided reference undermines your claim of some _necessary_ related events. There may be elementary events but you present it as there are definitely elementary events. That the gold ball being pulled cannot be independent to whether it was pulled from Box A or Box B. Just because you present it does not make it true. You admit that you do not know what those elementary events are but state they must be there when your own reference states they do not have to be there.
>>53877807
Your lack of understanding of terms regarding portability is extremely evident. But let's assume your made up definitions then. Then the problem in the OP is inconsistent. It makes no sense that you can follow the procedure described in the OP and be certain that you pull a gold ball in the first pull.
>>53877824
None of what you wrote is related to my reasoning or my claims. Try again.
>>53877821
>It's a reduction of possible events
A reduction to one possible event? A reduction to one possible event fits the definition of a certain event to me.
>>53877829
>Then the problem in the OP is inconsistent.
The problem doesn't fit my result so the problem must be wrong.
>>53877848
Answer this question, or I assume you are giving up. How is it that you can follow the procedure as described in the OP and be certain that you will pull a gold ball?
>>53877848
>>The problem doesn't fit my result so the problem must be wrong.
Maybe you should read the entirety of his post. The problem is inconsistent according to your understanding of it.
>>53877845
Perhaps I have been responding to a different anon. Would you like to clarify what your position is or should I assume you are merely presenting a false rejection?
>>53877858
Just answer >>53877854. I share his sentiment.
>>53877857
>Maybe you should read the entirety of his post.
Excuse me. I thought you were still trying to be constructive in your posts. My mistake. I'll just assume you are being an ass because you cannot handle being wrong and trying to hide the fact that you are wrong in some baseless sense of superiority.
You keep trying to put the gold ball you pulled back into an equation to come up with a phantom third probability greater than 0%.
>>53877854
>How is it that you can follow the procedure as described in the OP and be certain that you will pull a gold ball?
The problem states explicitly you pull a gold ball. What else am I supposed to do but be certain that event happens exactly as the problem stipulates? Can I free assume that I changed boxes even though the problem states the second ball comes from the same box?
>>53877888
Ok. That's it. I'm done.
>>53877848
> A reduction to one possible event fits the definition of a certain event to me.
Yes, that's correct!
But in the original problem it's not a reduction to one possible event..
We have 6 balls in the game, three of them golden. We pulled 1 ball, it was golden.
So what we can rule out:
We did NOT pull the silver ball from box 2.
We did NOT pull a ball from box 3.
The rest of the events are still possible events. We just ruled out that we get a silver ball at first, we make the possibility space smaller.
In other words, we could describe OP's question as:
"Let's ignore all cases when someone draws a silver ball at first."
Or also:
"You draw balls and put them back in until you get a golden one. NOW: which box is this ball from?"
>>53877897
>But in the original problem it's not a reduction to one possible event
Pardon? We have reduced the event of pulling the first ball to pulling a gold ball. It seems you want to unreduce the event from its certain result.
>>53877897
>"Let's ignore all cases when someone draws a silver ball at first."
If we are going to reword what was actually written. Which is a more accurate reflection of what was written in the problem: "Ignore all cases when someone draw a silver ball at first."
Or
"There is a 0% chance the first pull from your chose box is silver."
>>53877895
Should I now freely assume I can answer a different question than was presented in the OP as well?
>>53877937
I'm not going to bother trying to explain probability terms to someone at the level of a 7th grader anymore. You can do what you want.
>>53877927
the former, of course
>>53877904
>We have reduced the event of pulling the first ball to pulling a gold ball.
Yes.
The EVENT of pulling a golden ball occurs, when one of the three following ELEMENTARY EVENTS happens:
Pull ball 1, ball 2 or ball 3.
>It seems you want to unreduce the event from its certain result.
It's not a certain result.
It's an ASSUMPTION.
That's a differnce.
Example:
-Every year there's only 1 day where it snows.
-When it snows, the street is frozen with 50% probability
-When it soesn't snow, the street is frozen with 1% probability
-It snows. Is the street frozen?
Now we don't CARE how often it snows.
And we don't say "it's 100% certain that it snows".
We just ASSUME it snows and look at how the odds are now.
>>53877948
>You can do what you want.
Thanks. I'm so happy I have gotten your permission to repost what was already stated long ago.
>After a box has been chosen, but before a box is opened to let you observe a coin, the probability is 2/3 that the box has two of the same kind of coin. If the probability of "observing a gold coin" in combination with "the box has two of the same kind of coin" is 1/2, then the probability of "observing a silver coin" in combination with "the box has two of the same kind of coin" must also be 1/2.
We have observed that one of the two coins is gold.
>>53877956
I'd agree with you if the question asked "If the first ball drawn is gold" but instead it states explicitly "It is gold." That seems more accurate to reword it as "0% silver." The wording could have but did not include any possibility but the first result being gold.
>>53877962
>The EVENT of pulling a golden ball occurs, when one of the three following ELEMENTARY EVENTS happens:
>Pull ball 1, ball 2 or ball 3.
Except we don't have a ball 1, ball 2, or ball 3. We have three gold balls. The only difference is which box they are in. You are incorporating something that is not provided for in the problem. Exactly as I pointed out hours ago. You are imposing order on a problem that does not involve order.
>>53877978
>it states explicitly "It is gold."
See:
>>53877786
> the probability of an event GIVEN that (by assumption, presumption, assertion or evidence) another event has occurred
>>53877978
you'll never understand probability so I'll give you life advice instead
once you hit high school, don't argue with your math teacher. it won't help you.
Nice highschool questions....
But intuition is a tricky thing, especially in statistics it is misleading you.
>>53877997
>conditional probability is a measure of the probability of an event
You are saying that the event of the pulled gold ball is given by another event?
>>53878005
Are you resorting to trolling when you've got nothing of substance to add to the discussion?
>>53877990
> You are incorporating something that is not provided for in the problem.
No, I'm not.
We coudl draw tiny numbers on the balls, it would still be the same problem.
We couls also call the balls "ted, ned and fled" and it would still be the same problem.
Of course we don't need to know which ball we pull out. Also there's no "order".
It's enough to know there are two golden balls in box one, which makes 2/3 probability that a golden ball originates form box 1.
>>53878021
>You are saying that the *measure of the event of the pulled gold ball is given by another event?
>>53878024
>No, I'm not.
>We coudl draw tiny numbers on the balls, it would still be the same problem.
It would be similar but a different problem as we would have an additional premise.
That's like saying I could draw tiny cocks on the wheel of my car and I will still be considered just as straight and heterosexual.
>>53878021
>You are saying that the event of the pulled gold ball is given by another event?
No, the event that we pulled a golden ball is given..
If I was Obama my dick would be huge.
Question:
"You wake up and are Obama. Is your dick huge?"
That's how to read the question in the OP.
>>53878021
No, I just gave up on you. End yourself moron.
>>53878024
>which makes 2/3 probability that a golden ball originates form box 1.
The question is not which box did you pull the first gold ball from.
We have three boxes. One has no gold balls. What is the chance of pulling one gold ball from any of the three boxes? 50%. There is no way you can pull a gold ball from the third option. What are the chances that the other ball from the same box is also a gold ball? There are only two boxes you could have pulled the first gold ball from. There are only two solution sets that would be the same box you pulled the first gold ball from. Where do you get this third possibility to pull another gold ball from?
>>53878043
>No, the event that we pulled a golden ball is given..
Then there is no conditional probability for the first pulled gold ball?
>>53878053
I like to extend the benefit of the doubt. Given your trolling I have little choice but to believe you are resorting to trolling because you have nothing of substance to add. Otherwise you're just an ass who cannot handle disagreement in a mature manner.
>>53878107
There is no disagreement. Only a moron who doesn't understand probability, and clearly does not want to understand it.
I'm talking to a child to says 2+2=5 despite the four apples in front of his face. I'm glad I'm not a teacher.
>>53878145
who says*
>>53878089
> There is no way you can pull a gold ball from the third option.
Ignore the third box. It's not relevant for the question.
>Then there is no conditional probability for the first pulled gold ball?
"Conditional probability" means:
How are the odds that something will happen IF WE ASSUME something differnt has happend before?
We are interested in the odds that we have one of these boxes and the second ball in the box is golden IF WE ASSUME the the first ball was already golden.
> What is the chance of pulling one gold ball from any of the three boxes? 50%
That's correct.
But it's not interesting, since we already assume this has happend. Even if the chances for the first ball would have been only 1%, we could still assume it has just happend.
>There are only two boxes you could have pulled the first gold ball from.
Correct.
>There are only two solution sets that would be the same box you pulled the first gold ball from.
That's correct, but incomplete.
The question is about the "other element of the set" (the remaining ball). So if we got the first OR second element of the first set, the other ball would be golden.
>Where do you get this third possibility to pull another gold ball from?
It's just the second ball from the first box.
It's two differnt objects, so we could have pulled either of them.
Let's say you have the sets {A, A} and {A, B}.
You get the element "A".
How many possibilities you see?
A is present three times, so it's three possibilities.
>>53878190
>How are the odds that something will happen IF WE ASSUME something differnt has happend before?
What something different has happened before in regards to the first ball being pulled?
It seems intuitive to assume we can attribute a specific box to the event of pulling a gold ball.
If we only pulled one ball from a chosen box and it always, every time, no difference in any event, there was a 0% chance of pulling anything but, results in pulling a gold ball does it matter that there is only one ball in Box B? Would we not invariable pull the one gold ball from Box B?
>>53878250
Look, the questions says:
>"You pick box at random. You put your hand in and take a ball from that box at random. It's a golden ball."
That's the assumption. This is where the story starts. We imagine a world where this has just happend.
The phrase "take a ball from that box at random" is not relevant here. It's just to show that we have no knowledge about the boxes, just that we just got a golden ball under the given circumstances.
>It seems intuitive to assume we can attribute a specific box to the event of pulling a gold ball.
We have two boxes:
Box 1: 999 silver balls, 1 golden ball
Box 2: 999 silver balls, 1 golden ball
Now we pull balls (alternating from each box) and put them back in if it's silver until we finally get a golden one: which box is it? The odds are 50/50.
Now we change the first box, it's now:
We have two boxes:
Box 1: 1000 golden ball
Box 2: 999 silver balls, 1 golden ball
Again we pull balls (alternating from each box) and put them back in if it's silver until we finally get a golden one: which box is it?
Of course we would assue it's the box with more golden balls.
Yes, the question says nothing about "alternating turns", but since we don'T have any further information it's absolutely unclear, where we get the ball from.
>results in pulling a gold ball does it matter that there is only one ball in Box B?
Yes it does matter.
There's a differnece between:
1) "Someone opens boxes until there's at least one golden ball in it. Then he give the ball to you."
2) "Someone grabs one ball form the boxes until he has a golden one. Then he give the ball to you."
The former would result in 50/50 (as you said).
The latter is the question of the OP and result in 2/3.
>>53878250
>Would we not invariable pull the one gold ball from Box B?
This is why I made a specific reference to the language. The wording chosen by the OP indicates an independent event. If the intent was for the first ball pulled being something but an independent event then the question would be something like, "If the first ball pulled is gold what are the chances the other ball in the box is also gold?"
I may not know probability theory as well as some anons but I do know the English language and the conscious choice of words. Given the language chosen in the problem, we can draw no inferences from first pulled gold ball other than it cannot come from the box with two silver balls. It does not leave room for any other result but that one of the two balls picked from the box was a gold ball.
Oh well, I'm tired now..
KTNXBYE
>>53877452
Damn, I finally fucking got it right.
The error is there:
>P(G1A|G1) = 1/2
>P(G1B|G1) = 1/2
>P(G1C|G1) = 0
There is actually a 1/3 probability of picking G1A and a 1/6 probability of picking G1B, which actually makes P(G1A|G1) = 2/3 and P(G1B|G1) = 1/3. And I finally get P(G2|G1) = 2/3.
Kill me.
>>53876813
Yes hi I'm >>53875559 and that was one of the most amazing things I've read all day. What board was that on?
>>53875754
No you've already chosen a BALL, write out the sample space.
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
