>>8347660 >crypto intellectuals

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Anonymous Thu Mar 15 01:13:49 2018 No.9591739 [Reply] [Original]

>>8347660

>crypto intellectuals

>>	Anonymous Thu Mar 15 01:15:13 2018 No.9591741 >>>/biz/8347660 fucked up the link lmao

>>	Anonymous Thu Mar 15 01:18:13 2018 No.9591747 >>9591739 1/3

>>	Anonymous Thu Mar 15 01:21:05 2018 No.9591750 >>9591747 its 2/3 anon

>>	Anonymous Thu Mar 15 01:21:37 2018 No.9591753 File: 16 KB, 220x285, 220px-Einstein-formal_portrait-35.jpg [View same] [iqdb] [saucenao] [google] Not this shit again Last time was over 300 posts of people trying to explain conditional probability to morons who thought the answer was 1/3rd

>>	Anonymous Thu Mar 15 01:24:41 2018 No.9591757 >>9591750 yea i fat fingered it, sorry

>>	Anonymous Thu Mar 15 01:25:02 2018 No.9591759 am I dumb for thinking its 50%?

>>	Anonymous Thu Mar 15 01:27:47 2018 No.9591763 >>9591759 just a little but you're better off than anyone who said 1/3

>>	Anonymous Thu Mar 15 01:29:50 2018 No.9591767 >>9591759 No. You know that there's a gold ball in the box, so it can't be box 3. It's a 50/50 shot you're in the goldibocks, from there. It's only at this point that the question is posed

Anonymous Thu Mar 15 01:32:23 2018 No.9591770
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>>9591739
>grab a box
>pull a gold ball out of box
>I now know that I either have the box with two gold balls or the one with both kinds.
Sorry how could it be anything other than 50%

>>	Anonymous Thu Mar 15 01:41:30 2018 No.9591792 >>9591770 >>9591767 embarrassing

>>	Anonymous Thu Mar 15 01:42:38 2018 No.9591793 >>9591770 >>9591767 jsut take the total number of gold balls and divide it by the ones that are left 2/3. it's not 50% that's not ho w it works

>>	Anonymous Thu Mar 15 01:58:45 2018 No.9591814 File: 264 KB, 701x576, 1521090829157.png [View same] [iqdb] [saucenao] [google] here's your math homework

>>	Anonymous Thu Mar 15 02:06:08 2018 No.9591819 >>9591814 Ah. I feel stupid for thinking 50/50 now. I understand why I had the thought process so I am not that mad though.

>>	Anonymous Thu Mar 15 02:16:32 2018 No.9591839 >>9591770 Because sometimes there can be two choices with unequal odds. This is one of those tomes.

>>	Anonymous Thu Mar 15 02:19:14 2018 No.9591844 File: 14 KB, 728x441, bertrandBox.png [View same] [iqdb] [saucenao] [google] >>9591739 2/3

Anonymous Thu Mar 15 02:24:34 2018 No.9591853

No, this is wrong. It's absolutely 50%. Once you've pulled the first gold ball out, you are in a state where the box has just one other ball.

That other ball is either a gold ball, or a silver ball. It's 50%.

The 2/3 answer assumes you would be able to reach into the other box as well, but that's not how the problem was phrased: "What is the probability that the next ball you take FROM THE SAME BOX will also be gold?"

>>	Anonymous Thu Mar 15 02:25:20 2018 No.9591856 >>9591814 What the fuck. You removed a critical aspect of the original question. >>9591819 You should be. He removed the part that said "from the same box" ffs. The answer is 1 in 2.

>>	Anonymous Thu Mar 15 02:25:41 2018 No.9591858 >>9591770 as in any good con you're fooled in the very beginning "grab a box" isn't as simple as you think

>>	Anonymous Thu Mar 15 02:26:18 2018 No.9591859 >>9591770 You're right. The other answers are just trying to confuse you with their convoluted bullshit

>>	Anonymous Thu Mar 15 02:26:45 2018 No.9591862 File: 3 KB, 125x118, bertrandBox_diagram.jpg [View same] [iqdb] [saucenao] [google] >>9591739 2/3

Anonymous Thu Mar 15 02:27:49 2018 No.9591864

>>9591856
>>9591853
Check this out. You have 1 box with 999 silver balls and 1 gold ball. And another box with 1000 gold balls. You close your eyes and pull out a gold ball from one of the boxes, you don't know which one. What's the chance of pulling another gold ball.

>>	Anonymous Thu Mar 15 02:28:20 2018 No.9591868 >>9591853 Exactly. You have to reach into the SAME box, and the question is posed at the point of only 2 possible outcomes. Anything that happened previously is irrelevant (think gamblers fallacy)

Anonymous Thu Mar 15 02:28:47 2018 No.9591871

>>9591856
Except you don't know what box you picked.

If you picked from the first box you'll have a 100% chance

If you picked from the second box you'll realistically have a 0% chance

But there's a 2/3 chance to pick the first box, so the answer is 2/3.

>>	Anonymous Thu Mar 15 02:29:30 2018 No.9591873 >>9591864 Not enough information. Do I have to reach into the same box again?

>>	Anonymous Thu Mar 15 02:29:54 2018 No.9591874 >>9591873 Yeah.

>>	Anonymous Thu Mar 15 02:30:31 2018 No.9591876 >>9591871 >but there's a 2/3 chance to pick the first box how, exactly each box has a 1/3 chance of being picked at random lol

>>	Anonymous Thu Mar 15 02:30:57 2018 No.9591878 >>9591864 50% because there's only two boxes next

>>	Anonymous Thu Mar 15 02:31:56 2018 No.9591879 >>9591873 what part of "next ball you take from the same box" was hard to understand?

>>	Anonymous Thu Mar 15 02:32:10 2018 No.9591880 >>9591874 One in two, then. I've either already depleted the box of gold (zero chance of picking another) or will definitely pick another, assuming the second box has 1000 gold balls and nothing else in it.

Anonymous Thu Mar 15 02:32:15 2018 No.9591881

>>9591864
If you have to reach into the same box again, it's 50%.

Either you grabbed the 1 gold ball, so the remaining 999 balls are going to be silver, or:

You reached into the box with gold balls, so the remaining 999 balls are going to be silver.

50% chance.

>>	Anonymous Thu Mar 15 02:32:20 2018 No.9591882 >>9591876 Because ANY time you pick a silver ball it doesn't count. You HAVE to pick a gold ball. And the only way to do that is either the double gold or split box. And you have a higher chance of picking gold from the double gold box. See>>9591864

>>	Anonymous Thu Mar 15 02:33:11 2018 No.9591885 >>9591879 reread his post, he didn't specify like OP did. I was just being specific.

Anonymous Thu Mar 15 02:33:25 2018 No.9591886

A person first picks a box from the 3 and then draws both balls from it one at a time. Let E be the event that the second ball he draws is gold and G be the event that the first ball he draws is gold. We are looking for the probability

P(E|G)=P(E,G)/P(G)

P(E,G) = 1/3 because it would mean that the person chose the first box out of 3.
P(G) = 1/3*1+1/3*1/2 = 1/2. If the person chose the first box then it is 100% chance that 1st ball is gold. If the person chose the 2nd box then its 50% chance the first ball is gold.

so the final answer is (1/3)/(1/2)=2/3

Anonymous Thu Mar 15 02:33:37 2018 No.9591887

>>9591881
FAIL, sorry, I meant gold. Let me try that again:

Either you grabbed the 1 gold ball, so the remaining 999 balls are going to be silver, or:

You reached into the box with gold balls, so the remaining 999 balls are going to be gold.

50% chance.

>>	Anonymous Thu Mar 15 02:34:16 2018 No.9591891 >>9591882 There's no "have to" anywhere. The problem specifies the following given information: 1.) you have already picked out a gold ball 2.) you must pick from the same box There is no rule that says you HAVE to pick a gold ball.

Anonymous Thu Mar 15 02:35:31 2018 No.9591892

Frequentist approach:

ok, say we do this 1500 times

500 times you pick from SS
500 times you pick from GS
500 times you pick from GG

SS: 0 favourable
GS: 250 favourable
GG: 500 favourable
--------------------------------------------
250+500= 750 favourable ("It's a gold ball")

500 of those 750 times you have locked into the GG box.
500/750 = 2/3

>>	Anonymous Thu Mar 15 02:36:14 2018 No.9591896 >>9591886 Except that's fucking wrong lol. You (the person solving the problem) haven't chosen the first ball. The ball has already been chosen for you.

>>	Anonymous Thu Mar 15 02:36:58 2018 No.9591899 >>9591739 There are 3 gold balls I could have picked and 2 gold balls remaining to pick. The odds should therefore be 3/2 I think.

>>	Anonymous Thu Mar 15 02:37:14 2018 No.9591900 >>9591892 No, read the OP again. Your math is assuming you have a chance to pick from the box with two silver balls in it.

>>	Anonymous Thu Mar 15 02:38:19 2018 No.9591907 >>9591899 winner winner chicken dinner

>>	Anonymous Thu Mar 15 02:38:42 2018 No.9591908 >>9591896 You...take a ball...It's a gold ball. lrn2read

Anonymous Thu Mar 15 02:38:49 2018 No.9591909
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>>9591880
>>9591891
Right. But the only way to pick a gold ball is to choose from one of the boxes. What has a higher chance: picking a gold ball from the all gold ball box or the 1:999 chance box. If you pick a silver then you try to again until you get a gold ball. If you pick the 1:999 box you have a 1 in 1000 chance of getting a gold ball. If you pick the other box it's 100%. So we can see if you DO manage to find a gold ball, it's really likely that it came from the all gold box. And if you pick another ball from the SAME box you got the first one from, it's really likely to be gold since it's likely you got the first gold from the all gold box.

>>	Anonymous Thu Mar 15 02:39:01 2018 No.9591910 >>9591896 This guy gets it!

Anonymous Thu Mar 15 02:39:24 2018 No.9591911

>>9591876
I don't think you understand. The question clearly states that you already picked a box that has a gold ball, But you don't know which box. So we have to find the probability of picking the box with the gold ball.

The probability of picking the first box with the gold ball is 2/3

The probability of picking the second box with the gold ball is 1/3

The probability of picking the third box with a gold ball is 0/3

The probability that the second ball in the first box is gold is 100%

The probability that the second ball in the second box is gold is 0%

The probability that the second ball in the third box is gold is 0%

Therefore:

2/3 * 100% + (1/3 * 0%) + (0/3 * 0%) = 2/3

If we knew which box we first picked, it'd either be 0% or 100%.

Anonymous Thu Mar 15 02:40:50 2018 No.9591913

>>9591896
>i'm right but only if i make up a totally different problem than the one in the op
Are ALL 50/50 fags just stupid to begin with? I always want to give benefit of the doubt since even the smartest people can be fooled by unintuitive problems, but for some reason they inevitably back themselves into a corner with increasingly ridiculous reasoning to get their wrong answers.

>>	Anonymous Thu Mar 15 02:40:50 2018 No.9591914 Y'all are hopeless idiots. It's 2/3. True, you either have Box A in which case the other ball is gold, or you have Box B in which case the other ball is silver. But, the probability that you have Box A is higher, and that is why the answer is not 50%.

Anonymous Thu Mar 15 02:42:43 2018 No.9591917

Here I was thinking /sci/ was supposed to intelligent. I got taunted on other boards saying I'd never be smart enough to be allowed to post on /sci/ or /lit/. Low IQ they said. Way too low they said. And now I see you "high IQ" individuals can't even figure out this simple math problem? Sad!

>>	Anonymous Thu Mar 15 02:42:57 2018 No.9591919 Wait a fucking second. After we pick the first ball, do we put it back in the box or is it assumed to have been removed?

>>	Anonymous Thu Mar 15 02:43:33 2018 No.9591920 >>9591900 Yes, when you look at the whole situation. Once you have that under control, you can move on and inspect the subset of when the first ball taken was gold. That's why the favorables are 250+500=750 (and not 500+500)

>>	Anonymous Thu Mar 15 02:43:36 2018 No.9591921 >>9591919 Typically questions like these it's assumed that you don't put it back in, it's removed from the equation.

Anonymous Thu Mar 15 02:44:43 2018 No.9591924

>>9591911
Except that is irrelevant. Your first probability of picking a box with a gold ball in it is 100%: The question is only asking for the cases where a gold ball was picked on the first pick.

Since a gold ball was picked (as stated in the problem), that probability doesn't matter.

From there it's simple to see it's 50%.

>>	Anonymous Thu Mar 15 02:45:30 2018 No.9591925 >>9591739 https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>>	Anonymous Thu Mar 15 02:46:31 2018 No.9591927 >the probability that you have Box A is higher No it isn't. Gambler's fallacy

>>	Anonymous Thu Mar 15 02:48:01 2018 No.9591929 >>9591924 >>9591927 See>>9591864 and >>9591909

Anonymous Thu Mar 15 02:49:39 2018 No.9591933

>>9591911
>the probability of picking the first box with the gold ball is 2/3
No, it isn't. The probability of picking either the first or second box with either one or two gold balls in it is 100%, because it has ALREADY BEEN DONE BEFORE THE PROBLEM HAS STARTED.

Anonymous Thu Mar 15 02:50:02 2018 No.9591935

>>9591914
>>It's 2/3. True, you either have Box A in which case the other ball is gold, or you have Box B in which case the other ball is silver. But, the probability that you have Box A is higher, and that is why the answer is not 50%.

I think this is the best explanation I've heard yet.

>>	Anonymous Thu Mar 15 02:51:05 2018 No.9591936 >>9591927 >>9591924 >>9591933 based retards

Anonymous Thu Mar 15 02:57:27 2018 No.9591953

It's easy to do this experiment yourself if you don't believe the math:

Get 4 marbles or coins or something of identical shape but different color.

Then mix them up, take two into each hand, and set one down. If the one you set down was a gold one, check the color of the remaining one in that hand.

Mark down if it's silver or gold. Repeat 30 times. You'll see you get more gold answers than silver. Science!

Anonymous Thu Mar 15 03:16:51 2018 No.9591978

the problem OP posted stated the first ball will always be gold (It's a gold ball). As such the possibility of picking a silver ball first is always 0%, no matter which box he picks from at first. If you always pick a gold ball you either picked one of two golds from Box A or the only gold ball from Box B.
Much like how you can't pick Box C because it is has only silver balls and therefore no gold ball to pick you can't pick the silver ball from Box B first as it is not gold.
50%.

>>	Anonymous Thu Mar 15 03:20:27 2018 No.9591982 >>9591739 75%, this is 4th grade shit.

>>	Anonymous Thu Mar 15 03:21:03 2018 No.9591983 >>9591978 You had a 100% chance to choose a golden ball, you didn't choose a box that has a golden ball, therefore implying probability of getting which box with the golden ball. 2/3.

Anonymous Thu Mar 15 03:29:45 2018 No.9591995

>>9591983
The problem states the first ball will always be gold. Box B could have 1 gold and 999999 silver balls but the probability of picking a gold ball first from said box would be 100%.
The use of random selection of balls at first refers to the random selection of gold balls within a box.

>>	Anonymous Thu Mar 15 03:34:42 2018 No.9592006 >>9591953 tried this and got 22 gold, 8 silver. 2/3

>>	Anonymous Thu Mar 15 03:38:06 2018 No.9592014 >>9592006 Because you don't have probability defying abilities like the OP is assumed to and is capable of choosing a silver ball first. Remove all the silver balls and boxes without gold balls (what OP has a chance of picking) and try again.

>>	Anonymous Thu Mar 15 03:40:38 2018 No.9592018 >>9592014 >>Remove all the silver balls and boxes without gold balls Probability of choosing a gold ball: 100% 10/10 would chance again

Anonymous Thu Mar 15 03:42:43 2018 No.9592021
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New problem:

There are two boxes.
Each box has 10 cupcakes in it.
In one box, 9 of the cupcakes contain small bombs that will explode exactly 10 minutes after consumption, spilling out your guts in front of you.
In another box, only one cupcake has a small bomb.
It is impossible to tell the difference between a normal and loaded cupcake until eaten.
The boxes are unlabeled and it is impossible to tell which is which.

You pick a box at random and open it. The cupcakes all look the same to you, so you pick a random one and eat it. The cake is fresh and moist, and the cream cheese frosting is perfect, but the sprinkles are stale and the thought enters your mind that you might have preferred it without them.

5 minutes pass. So far so good, but you won't know for sure what will happen until another 5 go by. As you anxiously await your fate, you notice that the door to the chapel, where you're enjoying your dessert, is unlocked! All of your family members should be fast asleep in the guesthouse, but you suspect your chubby nephew likes to take some late night strolls so you take action, just in case. The door locks with a satisfying *click* and you briefly consider that it would be quite a mystery if you ended up dead here in this locked room. Amused by your little fantasy, you notice that 10 minutes have finally passed and you are somehow still whole.

You sit back down. One cupcake is not enough for you; you knew this from the start. But you also don't want to eat any small bombs. You contemplate for a good minute before hunger forces you to act. You open the OTHER box and eat a cupcake from it. Your reasoning is as follows:

>I just eliminated a safe cupcake from the open box. This reduces the chance of another safe cupcake coming from that same box. It may even be that I ate the ONLY safe cupcake in the box! At least if I eat from the other box, I must have some chance of picking a safe cupcake.

What are the odds that your family wakes up to a locked room mystery?

Anonymous Thu Mar 15 03:44:32 2018 No.9592025

>>9592018
Yes the probability of choosing a gold ball first is 100% as it clearly says here.
>>9591739
>It's a gold ball

It is
>I flip a coin. What are the odds of me flipping heads twice in a row
vs.
>I flip a coin. It is heads. What are the odds of me flipping heads twice in a row.

>>	Anonymous Thu Mar 15 03:47:02 2018 No.9592029 >>9592021 >What are the odds Pretty good odds... You should have eaten from the same box again. (since you didn't asplode the first time, you probably picked from the box with only 1 bomb. so eat more from that box!)

Anonymous Thu Mar 15 03:48:50 2018 No.9592030

>>9592025
I just mean that you said to remove all silver balls, and boxes with no gold balls. That leaves one box with two (gold) balls, and 1 box with 1 (gold) ball.

It's impossible for the second pick to be a silver ball, since you specified there aren't any.

Anonymous Thu Mar 15 03:56:03 2018 No.9592039

>>9592030
Only remove them for the first pick. If you want to put the silver one back in the box with the single gold to represent the second pick you can but it should be obvious at that point what the next ball you will pick given the number of gold balls left.

>>9592021
If you picked the dangerous box first you have a 90% chance of living. If you picked the safe box first you have a 10% chance of living.
Given that you lived long enough to know you ate a normal cupcake both scenarios are equally likely so you have a (10+90)/2 = 50% chance of living.

>>	Anonymous Thu Mar 15 03:59:05 2018 No.9592044 >>9592039 >Only remove them for the first pick. In that case that's exactly what was described in >>9591953

>>	Anonymous Thu Mar 15 03:59:12 2018 No.9592045 >>9591844 >>9591862 stage #1: 3 ways to have picked the first gold stage #2: Next ball, 1 silver, 2 golds

>>	Anonymous Thu Mar 15 04:00:40 2018 No.9592048 >>9592044 That includes the possibility of choosing a silver ball first, therefore it's a different problem.

>>	Anonymous Thu Mar 15 04:04:21 2018 No.9592053 >>9592048 No it doesn't. Read again: >>9591953 >If the one you set down was a gold one, check the color of the remaining one in that hand. You don't mark anything down if you pick silver first.

Anonymous Thu Mar 15 04:05:04 2018 No.9592056

>>9592048
No it doesn't, if you read what's written:

>>If the one you set down was a gold one, check the color of the remaining one in that hand. Mark down if it's silver or gold.

The check is only done if the one you picked was gold. In other words, it does not include the possibility of choosing a silver ball first.

>>	Anonymous Thu Mar 15 04:08:09 2018 No.9592059 >>9592053 >>9592056 >If the one you set down was a gold one >If So there are times where he selects a Gold-Silver pair but picks silver first. Which is impossible given that he always picks gold first.

Anonymous Thu Mar 15 04:18:52 2018 No.9592075

Here's the problem explained for retards

>You saw the three boxes and noted the balls and their colors in the three boxes
>You pick out a ball from a random box that you don't know about
>It's a gold ball
>You remember that only two boxes have gold balls
>The first box has two out of three gold balls, meaning you had a 2/3 chance of picking this box
>The second box only has one gold ball, meaning you had a 1/3 chance of picking this box
>You are asked to guess the probability picking ANOTHER gold ball
>Well if you picked the first box, you have a 1/1 chance of picking a gold ball
>If you picked the second box, you'd have a 0/1 chance of picking another gold ball
>So a 2/3 * 1 = 2/3 chance of picking a gold ball from the first box and a gold ball again
>And 1/3 * 0/1 = 0/3 chance of picking a gold ball from the second box and a gold ball again
>Add the probabilities up: (2/3 + 0/3) = 2/3 chance of picking a gold ball on the second try, given that you chose a gold ball on the first try

Was that so difficult?

Anonymous Thu Mar 15 04:19:01 2018 No.9592076

>>9592059
>So there are times where he selects a Gold-Silver pair but picks silver first.
Yes, but fortunately they are not included in the results of the experiment.

Your obsession with probability-defying powers is getting tiresome. You're lucky you didn't have to eat that cupcake because it would have been extremely likely that you picked an exploding one the second time around

Read the OP more carefully next time. It's not telling you that you have magic gold picking powers. It even specifically says that you cannot look inside the boxes. It's just telling you what happened. You picked a box at random. You picked a ball at random. The ball happened to be gold.

OP is not predicting what will happen, it's telling you what already happened and asking you to use this information to figure out the probability of what will happen next. The fact that you RANDOMLY picked a gold ball tells you something about the box you picked (and the boxes you didn't pick), much in the same way that Monty opening a door that he knows contains a goat gives you information about the doors he chose not to open.

>>	Anonymous Thu Mar 15 04:34:11 2018 No.9592094 >>9591770 It's more likely that you picked it out of the first box than the second. Since it's more likely that you picked it from the first box it's not 50%.

>>	Anonymous Thu Mar 15 04:35:46 2018 No.9592098 >>9591739 https://onlinegdb.com/SJ7-knDFz >Events: 3334921 >Out of: 5002330 >Raito: 0.666674

Anonymous Thu Mar 15 04:38:45 2018 No.9592102

I understand what you 2/3-ers are saying, but as the problem states, you pick the box first, so it's irrelevant what the number of balls in the box are.

If for example we have 2 boxes where one has 1000 golden balls and the other has 1 golden and 999 silver balls, you first pick one of those two boxes at random, and then the invisible hand of fate gives you a golden ball from either of them. So now you either have a box of 999 golden balls or 999 silver balls.
We're not counting the number of balls at all, we're counting the likelyhood that the next ball in the box which you picked randomly is a gold one. And the likelyhood can only be 50/50 because there are only 2 boxes. If you picked the first box, you're 100% going to get the golden ball, if you picked the second box it's 100% going to be a silver ball.

The "you put your hand in and take a ball from that box at random" is a red herring, because it's already decided that you get a golden ball no matter which box you picked.

I hope this clears up my thinking.

Anonymous Thu Mar 15 04:49:21 2018 No.9592125

Alright let me propose a similar but clearer problem.
You have 2 chickens, one of them lays golden eggs, the other one lays regular eggs, but it has a golden egg stuck up its butthole. You pick a random chicken and it lays a golden egg, what's the probability that the next egg is going to be golden?

Anonymous Thu Mar 15 04:54:01 2018 No.9592137

>>9592102
After you've picked the gold ball, one of two things have happened and you don't know which one. When you've picked a gold ball, it's more likely that the outcome of the first choice was that you picked the first one.
2/3rds of the time when you pick up a gold ball it will be because you picked the first box.

Anonymous Thu Mar 15 04:54:14 2018 No.9592138

>>9592125
How about three chickens, one lays only golden eggs, one lays golden eggs 50% of the time, and the other lays no golden eggs.
You pick up a golden egg from one of the nests but you don't know which chicken lays in that nest. What is the probability that the next egg the chicken lays is going to be golden?

>>	Anonymous Thu Mar 15 05:01:50 2018 No.9592146 >>9591927 If you made the first choice many times you'd have 50% times when you got a gold ball and 2/3 of those times would be after choosing the first box.

>>	Anonymous Thu Mar 15 05:04:46 2018 No.9592150 >>9592102 The code says you're wrong. https://onlinegdb.com/r1kDI2vKf >Events: 50000222 >Out of: 50050267 >Raito: 0.999000105234

Anonymous Thu Mar 15 05:04:53 2018 No.9592151

>>9592125
Same guy here.
I think I understand what the issue is.
The 2/3 argument would be that you have a golden egg laying chicken, and a chicken who lays golden eggs 50% of the time and regular eggs 50% of the time.
If you picked a chicken at random and it lays a golden egg, then of course there's a 2/3 possibility that you have a 100% golden egg laying chicken.

But the thing is that golden egg laying is predetermined for both chickens as they are presented to you. The word "randomly" threw people off.

Imagine that it's the same chickens, but the regular egg laying one lays golden and regular eggs cyclically.
You pick a chicken, but it has already laid its golden egg.
The night before, one of the chickens laid a regular egg, but the chicken peddler closed shop that day because he wants both chickens to have laid golden eggs, because then people might buy the cyclical chicken, which is valued less.
You simply have a choice between 2 chickens with a golden egg beside them, you don't get meaningful info from this.

Anonymous Thu Mar 15 05:08:43 2018 No.9592154

>>9592138
You assume that the precedent doesn't have external agency involved, as the problem states, we know that you were going to pick up a golden egg from the nest anyway, all regular eggs are discarded, the crooked chicken peddler exchanges all regular eggs in all the nests with golden ones.

>>	Anonymous Thu Mar 15 05:13:09 2018 No.9592159 >>9592137 You don't actually get to pick a ball though, it's an illusion. You pick the box.

Anonymous Thu Mar 15 05:15:55 2018 No.9592163 [DELETED]

>>9592102
let's tweak >>9591892
to match your example:

Frequentist approach:

ok, say we do this 3000 times

1000 times you pick from SS
1000 times you pick from G(1000xS)
1000 times you pick from 1000xG

SS: 0 favourable
G(1000xS): 1 favourable
1000xG: 1000 favourable
--------------------------------------------
1+1000= 1001 favourable ("It's a gold ball")

1000 of those 1001 times you have locked into the 1000xG box.
1000/1001 = [math]0. \overline{999000}[/math]

Anonymous Thu Mar 15 05:17:40 2018 No.9592165

>>9592151
>The word "randomly" threw people off.
It would be more accurate to say that people who came up with the incorrect 1/2 answer failed to comprehend the premise of the question and would rather make up a completely new question that fits their solution.

The answer to the OP's question as written is undeniably 2/3.

If you assume you have a magic gold tractor beam that always picks a gold ball first in any box that has at least one, then the answer is undeniably 1/2. Feel free to turn that corrupted version of the question into a fun image and see how many people you can bait with it in a new thread. It would actually be pretty interesting: I bet some will skim it and try to answer 2/3 because they think it's something they already solved.

Anonymous Thu Mar 15 05:22:04 2018 No.9592171

>>9592163
You don't pick the ball!
You pick a box, and a golden ball gets magnetically pulled out of it.
What the premise is that SS gets completely discarded because there are no golden balls to take out of them first.
If you pick G(1000xS) then the G gets deducted, you have 0 chance to pull a G from it now.
If you pick 1000xG, a G gets deducted, you have 999xG left, nothing else, you have 100% chance to pull a G.
It entirely depends on which box you picked.

>>	Anonymous Thu Mar 15 05:23:51 2018 No.9592174 >>9591896 Yeah but there's a 2/3 chance the first gold ball came from the box with two gold balls

Anonymous Thu Mar 15 05:26:35 2018 No.9592180

>>9592102
>The "you put your hand in and take a ball from that box at random" is a red herring, because it's already decided that you get a golden ball no matter which box you picked.
Rephrasing into an equivalent problem:
I'm going to hand you a box.
0.1% of the time, it's going to have a silver ball in it.
99.9% of the time, it's going to have a gold ball in it.

Now I hand you the box and you claim the chance of the ball being silver is 50% because you either have it or you don't?
lol

Anonymous Thu Mar 15 05:26:40 2018 No.9592181

>>9592165
Well that's exactly how my mind comprehends the situation, your own agency in choosing a ball gets taken away when it's directed that you get a golden ball regardless. Thus for me the "you randomly get a golden ball" is contradictory any element of randomness is taken away at that level when there are conditions of result for the entire set up of the problem.

Anonymous Thu Mar 15 05:27:57 2018 No.9592183

>>9592102
let's tweak >>9591892
to almost(***) match your example:

Frequentist approach:

ok, say we do this 3000 times

1000 times you pick from SS
1000 times you pick from G(999xS) (***)999 instead of 1000 silvers makes math easier
1000 times you pick from 1000xG

SS: 0 favourable
G(999xS): 1 favourable
1000xG: 1000 favourable
--------------------------------------------
1+1000 = 1001 favourable ("It's a gold ball")

1000 of those 1001 times you have locked into the 1000xG box.
1000/1001 = [math]0. \overline{999000}[/math]

>>	Anonymous Thu Mar 15 05:29:29 2018 No.9592186 >>9591739 1/2 m8

>>	Anonymous Thu Mar 15 05:31:45 2018 No.9592191 >>9592180 That's not equivalent at all. First of all, the boxes don't generate balls, each one has a distinct number of balls, second, there are two boxes.

Anonymous Thu Mar 15 05:35:57 2018 No.9592196

>>9592181
Honestly I think it's just a simple case of misreading the OP. It's not saying "you will get a gold ball if you pick one randomly." It's just saying "you picked one randomly and it was a gold ball."

You don't need to add extra shit to the question, there's no crooked ball fondlers involved switching up the boxes. No color ball was preferentially selected.

Hell, if you're so concerned about agency let's go all the way. Forget about you. Jimmy picked a random ball. Jimmy used the random.org app on his smartphone to pick a random box and then again to pick a random ball. He holds up the ball he picked.

If the ball he holds up happens to be gold, what is the probability the remaining ball in the box is gold?

If the ball he holds up happens to be silver, what is the probability the remaining ball in the box is silver?

The answer is 2/3 in both cases. All OP does is put you in the position of Jimmy after he lifted up that gold ball. If you're saying there's no information to be gained about the box Jimmy picked by looking at the color of the ball he picked, then I don't know what else to say. You might just have trouble with hypothetical scenarios in the first place.

>>	Anonymous Thu Mar 15 05:37:59 2018 No.9592200 >>9592165 But that's what it is. You have magical fingers that find you the golden ball first every time.

Anonymous Thu Mar 15 05:39:36 2018 No.9592203

>>9592200
Like I said, put that fundamental modification to the premise in the image and post it in a new thread. It'll make for good bait. As long as you keep trying to discuss a completely different problem than the one at hand, there's nothing left to be said.

>>	Anonymous Thu Mar 15 05:39:36 2018 No.9592204 >>9592200 the prompt says nothing about magic

>>	Anonymous Thu Mar 15 05:44:16 2018 No.9592209 >>9592203 the fucking question states that the gold ball is magically found every time

Anonymous Thu Mar 15 05:45:09 2018 No.9592211

>>9592159
Yeah you pick a box and don't know which one you've picked. 2/3rds of the time when you get a gold ball first it'll be after picking the first box.
Some see it like we already know the results of the first choice but we've really just narrowed it down to some of the possibilities.
We know that either something with probability 1/3*1 or something with 1/3*0,5 happened. It's more likely that the first thing happened.

>>	Anonymous Thu Mar 15 05:46:30 2018 No.9592216 >>9592209 Read the fudging question again. Carefully.

Anonymous Thu Mar 15 05:49:36 2018 No.9592219

>>9591739
"Random" is a red herring. There is no random in this question other than whether you selected the 2 gold or 1 gold 1 silver box

The question simplifies to "out of two boxes, what is the probability of selecting one at random"

The randomness starts after you already have a gold ball in your hand. What happens before that is pre determined by the question

1/2

>>	Anonymous Thu Mar 15 05:50:52 2018 No.9592222 >>9592216 see >>9592219

>>	Anonymous Thu Mar 15 05:51:45 2018 No.9592223 >>9592219 Random implies that you chose a random box that had a gold ball. 2/3

>>	Anonymous Thu Mar 15 05:53:14 2018 No.9592226 >>9592203 >>9592204 >>9592211 Read the chicken peddler analogy. The 2/3 chance would be valid if there was actually chance to pick the silver ball first, but there isn't. OP is a jew trying to sell you 2 golden balls in a rigged game.

>>	Anonymous Thu Mar 15 05:54:53 2018 No.9592229 >>9592223 There is 0% chance of selecting the 2 silver box There is 50% chance that you selected the 2 gold box and 50% chance that you selected the 1 gold 1 silver box 1/2

Anonymous Thu Mar 15 05:55:02 2018 No.9592230

>>9591933
>The probability of picking either the first or second box with either one or two gold balls in it is 100%

>The probability of picking the first box with the gold ball is 2/3
>The probability of picking the second box with the gold ball is 1/3

2/3 + 1/3 = 1. The total probability of the things that might have happened is of course one but there's different probabilities for those different things.
The choice has been done but you still don't know exactly what happened. If you just do the first choice many times you'll have gold balls 1/2 of the times and out of those times, 2/3 will be because you picked the first box and 1/3 from the second.
You just know you're in those 1/2 from the first choice so it's a 2/3 chance that you have picked the first box.

Anonymous Thu Mar 15 05:56:09 2018 No.9592232

>>9592219
>>9592226
So the "trouble with hypothetical scenarios" it is then. Fair enough, it's a big leap when you're used to just thinking about practical everyday situations, but a very important one to take if you ever want to get involved with science or philosophy.

Anonymous Thu Mar 15 05:58:51 2018 No.9592236

>>9592191
I have two boxes. One has 999 silver balls, one has 999 gold balls. I know which one is which. You don't.
I'm going to hand you one of them.
I'm stacking the game though: I'll give you the 999 silver balls 0.1% of the time, and the 999 gold balls 99.9% of the time.
After you have the box in hand, what is the chance you draw a gold ball?

Is it 50%, since at this point, you either have the gold box or you don't, and the prior probabilities don't matter, even if you know them?

>>	Anonymous Thu Mar 15 06:00:09 2018 No.9592239 There are a total of 3 Gold balls and 2 out of the 3 of them will lead to choosing a gold one and 1 out of the three will lead to a silver. People are getting the probability confused with the number of different results.

>>	Anonymous Thu Mar 15 06:03:54 2018 No.9592247 >>9592229. The pretense is that you selected a ball at random and it happened to be gold. There's 3 gold balls and 2/3 happen to be in one box. Therefore we can conclude that there's a 2/3 chance of selecting the two gold box. 2/3

>>	Anonymous Thu Mar 15 06:07:30 2018 No.9592251 box 1 100% box 2 50% box 3 0% you picked box 1 or 2. box 3 is completly out. 2/3 chance you get another gold ball 1/3 chance you get silver ball 0.00.......01% chance you grap your professors testicals

Anonymous Thu Mar 15 06:08:40 2018 No.9592252

>>9592226
There was a chance to get a silver ball but it didn't happen. A random event has happened and you don't know which one but have some information which narrows it down to some of the possibilities. There was one outcome with a 1/3 chance and another with a 1/6 chance that both would have given this signal. You know one of them happened but it's more likely that the first has happened.

Anonymous Thu Mar 15 06:10:35 2018 No.9592257 [DELETED]

>>9591739
50%
explanation:
The given information of "You have a gold ball" means we don't have to worry about the probability of that happening, it is assumed to be true.

(ALL THIS OCCURS AFTER DRAWING YOUR FIRST GOLD BALL)
so your box could either be the one with originally 2 golds (which now has 1 gold) or the one that originally had 1 gold and 1 silver (which now has 1 silver)

Two options for which one is yours, each just as likely, because we started off with that original condition of having a gold ball.
So your chance of having the double gold box is 50%

>>	Anonymous Thu Mar 15 06:12:17 2018 No.9592259 >>9592247 Visualise the situation 1/2

>>	Anonymous Thu Mar 15 06:14:15 2018 No.9592261 >>9592259 Sure thing, refer to >>9592075 since no one was able to disprove this yet.

Anonymous Thu Mar 15 06:15:16 2018 No.9592262

>>9592257
>means we don't have to worry about the probability of that happening, it is assumed to be true.
You do have to worry about the probability of it happening since it could have happened in different ways. There are different things that can have happened and they had different probabilities.

>>	Anonymous Thu Mar 15 06:17:46 2018 No.9592263 >>9592261 0% chance of selecting 2 silver box, not 33.3% 1/2

>>	Anonymous Thu Mar 15 06:25:01 2018 No.9592269 >>9592263 It was never implied that there was a chance of selecting the 2 silver box. It is a established that you selected a gold ball. Knowing what box you selected it from (aka only box 1 and 2) is finding the probability. Are you just trolling?

>>	Anonymous Thu Mar 15 06:26:30 2018 No.9592273 you have 1 box with 1ball in it it can be gold or silver 50% chance

>>	Anonymous Thu Mar 15 06:26:55 2018 No.9592274 P(picking up gold ball \| 1st ball was gold) This means only box {GG} and {GS} are possible. Assume both have 1 gold ball picked up. This happens because you are picking from the same box. Therefore, they become {G} and {S}. The result is 50% chance.

Anonymous Thu Mar 15 06:28:05 2018 No.9592277

A random event has the following possibilities.
A:1/3
B1/6
C:1/6
D:1/3
If A or B happens you will know it was one of them but not which one it was. If C or D happens you'll know it was one of them but not which one.
Then you make an artificial event where you've grouped different outcomes together with the possibilities
E:1/2 and F:1/2 where E is the event that you get the signal that A and B produces.

If the event happens and you get E, you don't know if A or B happened but 2/3 of the times that E happens it's actually A and 1/3 of the times it's B.
Now the event is has happened and the result was E, what's the probability that it was A that happened?

>>	Anonymous Thu Mar 15 06:36:07 2018 No.9592288 you all are wrong lmao >choose a gold ball from a random box (3/6) >put the ball back >choose a ball again >2/2 chance on box 1, >1/2 chance on box 2, > 1/2 + 2/2 + 3/6 = 4/2 chance to select a gold ball your welcome

>>	Anonymous Thu Mar 15 06:44:38 2018 No.9592296 If you’re legitimately having trouble with this then search Bayes Theorem. It makes these types of problems completely trivial.

>>	Anonymous Thu Mar 15 06:55:27 2018 No.9592312 >>9592229 the GG box can be selected two ways so it's: There is 66.6% chance that you selected the 2 gold box and 33.3% chance that you selected the 1 gold 1 silver box

>>	Anonymous Thu Mar 15 07:13:40 2018 No.9592339 >>9592226 the prompt says nothing about chickens

>>	Anonymous Thu Mar 15 07:20:34 2018 No.9592351 File: 49 KB, 800x600, tree.jpg [View same] [iqdb] [saucenao] [google]

>>	Anonymous Thu Mar 15 07:24:05 2018 No.9592354 >>9592351 >>9591862

>>	Anonymous Thu Mar 15 07:27:46 2018 No.9592356 >>9592296 I knew that a few years ago but hadn't thought about it so I was trying to do kind of the same thing but reinventing the wheel for it. ((1/3)*1)/0,5=2/3

Anonymous Thu Mar 15 08:43:12 2018 No.9592457

>>9591868
>Anything that happened previously is irrelevant
What happened previously is relevant. Because we don't know what happened previously we have to count the probabilities of the things that might have happened.
It's what happened in the first draw that decides what ball you get now. One of two things happened and the sum of their probabilities was 1/2, one had 1/3 and the other was 1/6. The probability that the first thing happened is then 0,33/0,5=0,66.

>>	Anonymous Thu Mar 15 08:44:30 2018 No.9592459 >>9591876 2/3 of the times you pick a gold ball first will be from picking the first box.

Anonymous Thu Mar 15 09:02:24 2018 No.9592480
File: 130 KB, 2598x625, received_10211126193012045.png [View same] [iqdb] [saucenao] [google]

The thing you have to find is the probability that you have chosen the first box, here I took an example and replaced the numbers and you see that the probability is 2/3.

>>	Anonymous Thu Mar 15 09:16:41 2018 No.9592494 >>9591739 The color of the balls exists only as a cloud of probabilities until you look into the box. Since OP's image explicitly states that looking into the boxes is not possible, then there is no way to ever know.

>>	Anonymous Thu Mar 15 09:48:11 2018 No.9592541 >>9592494 >looking into the boxes picking out of boxes is permitted try again

Anonymous Thu Mar 15 10:28:14 2018 No.9592589

the chances that you picked the silver/gold box and grapped the golden ball is
50/50 for the box and 50/50 for the ball
1/4 total

the chances that you picked the golden/golden box and grapped the golden ball is
50/50 for the box and 1/1 for the ball
1/2 or 2/4 total

1/4 for the first example and 2/4 for the second

3/4 overall that you picked the golden ball

OPs question started here

you have to take out the other ball from your box
how likely is it that it's a golden ball?

there is 1 box and 3 possible balls
2 golden and 1 silver

the chances that you get another golden ball is
2/3

>>	Anonymous Thu Mar 15 10:32:29 2018 No.9592597 >>9591739 Just ran 5 million simulations of this, got 0.666921 as the answer, hopefully this convinces all the idiots in the thread.

Anonymous Thu Mar 15 11:02:27 2018 No.9592655

>>9591892
Frequentist approach, general case:

ok, say we do this N times

box #1 SS
box #2 A Gold balls, B Silver balls
xox #3 C Gold balls

N/3 times you pick from box #1
N/3 times you pick from box #2
N/3 times you pick from box #3

box #1: N/3 * 0 = 0 favourable
box #2: N/3 * A/(A+B) favourable
box #3: N/3 * 1 = N/3 favourable
------------------------------------------------------------------
N/3*A/(A+B) + N/3 = N/3(A/(A+B)+1) favourable ("It's a gold ball")

N/3 of those N/3(A/(A+B)+1) times you have locked into box #3.

Anonymous Thu Mar 15 11:09:10 2018 No.9592675
File: 126 KB, 499x356, Schroeders Cat.jpg [View same] [iqdb] [saucenao] [google]

>>9592541
Brainlet.

The science clearly says that the probability field does not collapse until you open the box to look. It says nothing about "reaching in and pulling the cat out."

>>	Anonymous Thu Mar 15 11:24:29 2018 No.9592728 Holy fuck I hope this is a troll thread the answer is 50%.

>>	Anonymous Thu Mar 15 11:28:28 2018 No.9592735 >>9592675 prompt > shitpostnut

>>	Anonymous Thu Mar 15 11:30:12 2018 No.9592738 File: 101 KB, 700x729, catSchrodinger.jpg [View same] [iqdb] [saucenao] [google] >>9592675

>>	Anonymous Thu Mar 15 12:21:28 2018 No.9592825 You roll a 6 sided die while keeping your eyes closed, with the intention of guessing whether you rolled from {1,2,3} or {4,5,6}. Your friend, whom you trust, tells you you rolled an even number. What are the odds you rolled from {4,5,6}?

>>	Anonymous Thu Mar 15 12:38:01 2018 No.9592857 >>9592728 Run the code >>9592098

>>	Anonymous Thu Mar 15 12:41:24 2018 No.9592863 File: 1.09 MB, 1037x1037, 1510362378498.png [View same] [iqdb] [saucenao] [google] >>9592857 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA IT TOOK ALL MY LINKS DO NOT OPEN!

>>	Anonymous Thu Mar 15 13:48:59 2018 No.9593001 The problem is the same without the third box. You pick a box at random. 50% chance you get two gold balls, and 50% chance you get one gold ball. You will always pick a gold ball first. That leaves a 50% chance to get a second gold ball.

>>	Anonymous Thu Mar 15 15:17:55 2018 No.9593241 File: 91 KB, 645x729, IMG_1176.jpg [View same] [iqdb] [saucenao] [google] >>9591739 50%

>>	Anonymous Thu Mar 15 15:22:26 2018 No.9593248 File: 52 KB, 768x1024, image.jpg [View same] [iqdb] [saucenao] [google] >>9592094 You're retarded

>>	Anonymous Thu Mar 15 16:01:20 2018 No.9593334 hard mode 1: what if there were 5 golds in the first box hard mode 2: what if there were 4 silvers one gold in the second box dmd: what if both of the above

>>	Anonymous Thu Mar 15 16:06:48 2018 No.9593356 >>9591917 >Sad! who are you, fucking Trump?

>>	Anonymous Thu Mar 15 16:13:57 2018 No.9593383 >>9593334 Just follow this example >>9591862, it's very straight forward.

>>	Anonymous Thu Mar 15 16:20:48 2018 No.9593415 >>9593334 >>9592655

>>	Anonymous Thu Mar 15 16:23:28 2018 No.9593425 It's a straightforward application of Bayes' rule. If you have trouble with this, I really don't understand what are you doing here.

Anonymous Thu Mar 15 16:53:56 2018 No.9593494

>>9593383
>>9593415
I reasoned it out myself and coded it and came to the same conclusions

5 golds first box = 2/3
4 silvers 1 gold second box = 5/6
both = 5/6 again of course

interestingly, if you have say 4 gold 1 silver in the second box the chance is 8/9

>>	Anonymous Thu Mar 15 17:19:19 2018 No.9593548 >>9591862 Are we treating the balls as distinct or indistinct?

>>	Anonymous Thu Mar 15 18:28:30 2018 No.9593764 File: 57 KB, 701x368, 1521090829158.jpg [View same] [iqdb] [saucenao] [google] Eliminated redundant information.

Anonymous Thu Mar 15 18:46:19 2018 No.9593809

>>9593001
As long as you're making ridiculous assumptions about what ball you get to pick first, you're lacking in consistency by preserving the "random" choice of the box. Why not go all the way and say that you always pick the two-gold box which makes it 100%? It follows the same logic you do but to its actual conclusion rather than stopping halfway.

>>	Anonymous Thu Mar 15 19:24:30 2018 No.9593905 >>9593248 It's already explained here so writing an explanation would be a waste of time >>9592480

Anonymous Thu Mar 15 19:50:32 2018 No.9593966

>>9591750
I think so too. You have 6 balls, box 1- 2 gold, box 2-half and half, box 3- 2 silver. Once you pull out a gold one you only have to deal with 4 balls as the two silver balls in box 3. So out of 4 balls you have 1, the remaining 3 are 2 gold and 1 silver so the odds of it being gold is 2/3. If the question was what are the odds the next ball is silver is 1/3.

Anonymous Thu Mar 15 19:55:12 2018 No.9593974

>>9591864
Total balls, 2000. You pull out a gold one, now you have 1999. 1000 of those are gold, therefore the odds are 1000/1999. Approximately 50% but not exactly.
Now here's the catch. The way the question is phrased, "you don't know which one" makes me think. Can the boxes be switched? Will they be switched every time? If that's the case then we'd need to pull out a few more to confirm what are the contents of each box.
Assuming you're pulling from the same box, if you pull out 2 gold then it's all gold box. If you pull out 1 gold 1 silver then it's the mixed one.

>>	Anonymous Thu Mar 15 20:05:01 2018 No.9594000 Fuck Bayes and fuck all of you (((statisticians))).

Anonymous Thu Mar 15 21:12:25 2018 No.9594126

>>9593974
The questions simply asks what are the chances of pulling another gold ball. If you have the all gold box then it's 100%. If you have the other one then it's 0%.
>Inb4 Hur dur average is 50%.
No. Because what's the chance of you pulling the gold ball from a box in the first place. You have a MUCH higher chance of picking a random box, reaching your hand inside and pulling a a gold ball from the all gold box than the nearly all silver box.

Anonymous Thu Mar 15 21:14:58 2018 No.9594130

>>9594126
To clarify, I'm talking about the first gold ball you pick. Read the question again. It says you pick a random box then a random ball. You just so happen to pick a gold ball. The chance that you picked a random gold ball from the 1/1000 box is a lot smaller than the one that's entirely gold.

>>	Anonymous Thu Mar 15 21:15:33 2018 No.9594134 >>9591739 Let me guess, 2 thirds? I don't like to think more than 5 seconds about these things

>>	Anonymous Thu Mar 15 21:37:09 2018 No.9594175 you're way overthinking it my dude you pick out a gold, there's a 1000/1001 chance it came from the all gold box, and a 1/1001 chance it came from the mostly silver box

>>	Anonymous Thu Mar 15 21:44:28 2018 No.9594190 >>9591739 50%

>>	Anonymous Thu Mar 15 21:47:09 2018 No.9594195 It's 50% you retards. It cannot be the s/s box, therefore it's either the g/g box or g/s box, which gives 50% chance of getting a double gold.

Anonymous Thu Mar 15 21:53:07 2018 No.9594202
File: 161 KB, 1152x648, 1000 hours.png [View same] [iqdb] [saucenao] [google]

There is only one sentence that begins with an interrogative word (e.g. Who, What, When, Where, Why, How) and ends with a question mark as its punctuation. This would be the last sentence, which is predicated upon the information from the previous sentences.

There is only one question in OP's picture and it is dependent upon the previous statements made prior to the question.

>What is the probability that the next ball you take from the same box will be gold?
Is dependent upon:
>It's a gold ball
Is dependent upon:
>You put your hand in and take A BALL from THAT BOX at random (emphasis is mine)
Is dependent upon:
>You pick a box at random.
Is dependent upon:
>One box contains 2 gold balls, another box contains 2 silver balls, and the final box contains one gold and one silver ball.

The above sentence dictates that there are three boxes so the above sentence is not dependent upon the statement "There are 3 boxes"

All of the information, except for the very first statement, is relevant to the only question asked in OP's picture.

>But you're picking boxes!
No, the question is picking boxes for you. If you pick the double silver box at random the question is never asked. Being that the question is picking boxes for you we only look at the boxes on the left.

>>	Anonymous Thu Mar 15 21:56:06 2018 No.9594207 >>9592863 wut?

>>	Anonymous Fri Mar 16 01:06:28 2018 No.9594653 File: 43 KB, 859x515, 1493165760905.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Fri Mar 16 05:38:00 2018 No.9594928 >>9591739 kys /sci/, leave my biz alone >>/biz/thread/S8347660#p8355276

Anonymous Fri Mar 16 06:33:10 2018 No.9594976

>>9593764

But you don't pick a GOLD ball at random.

You pick a ball at random.

Having picked a ball - any ball - what is the chance that the other ball in the box is made of the same metal?

2/3rds.

Scenario doesn't change if you specify that the ball you picked at random was gold.

>>	Anonymous Fri Mar 16 06:56:07 2018 No.9595002 >>9591767 it's not about the gold balls it's about the boxes wtf you have 2 boxes 50-50

>>	Anonymous Fri Mar 16 07:01:18 2018 No.9595003 >>9591862 >2 boxes >therefore 2 choices >"2/3" how is this even real life? you don't treat the 2gold ball box as 2 different choices. it's 1 box.

>>	Anonymous Fri Mar 16 07:50:45 2018 No.9595043 >>9595003 ts kiddo, that's how it goes. write and run a basic program and let it run 100000 times

>>	Anonymous Fri Mar 16 08:06:15 2018 No.9595073 >>9595043 it's just a coinflip my man. 1. pick a box 2. pick a ball 2a. if the box had a gold ball, you now picked the gold ball 2b. if the box didn't have a gold ball, you pick again 3. pick the second ball 50-50

>>	Anonymous Fri Mar 16 08:08:09 2018 No.9595074 >>9595073 sure thing buddy, happy that your happy

>>	Anonymous Fri Mar 16 08:10:15 2018 No.9595077 >>9595074 glad you came to your senses

>>	Anonymous Fri Mar 16 08:11:00 2018 No.9595080 >>9595073 >3. pick the second ball >50-50 Only if there's an equal chance between having picked either of those boxes (Hint:

>>	Anonymous Fri Mar 16 08:12:21 2018 No.9595081 >>9595077 sure thing buddy, happy that your happy

>>	Anonymous Fri Mar 16 08:12:32 2018 No.9595083 >>9595080 you cannot pick the silver ball. If you do, you go again from the start. What is so hard to understand here

>>	Anonymous Fri Mar 16 08:13:55 2018 No.9595087 >>9594653 look at this picture and tell me there's a 2/3 chance to pick the gold ball. You can ignore the 2silver box, it is irrelevant. >>9595080 >>9595081 Please

>>	Anonymous Fri Mar 16 08:18:21 2018 No.9595098 >>9595087 >>9591892

>>	Anonymous Fri Mar 16 08:20:53 2018 No.9595104 >>9595098 >>9591892 you don't pick from the ss, it's irrelevant. if you pick from the gs, it's always g. 50-50

>>	Anonymous Fri Mar 16 08:24:32 2018 No.9595111 >>9595104 it's in the prompt, and >>9591892 deals with it correctly you are using a mental short cut that gives a wrong answer

>>	Anonymous Fri Mar 16 08:26:56 2018 No.9595116 >>9595111 >you pick a ball at random. it's a gold ball. = if you did pick a silver ball at random, then the case is irrelevant and you need a re-do. there is no mental shortcut here. Reading comprehension, more like. 50-50 is the correct answer here.

>>	Anonymous Fri Mar 16 08:29:38 2018 No.9595123 >>9595116 sure thing buddy, happy that your happy

>>	Anonymous Fri Mar 16 08:32:06 2018 No.9595125 >>9594653 If a gold ball is taken then there should either be two in the first box with none in the second or one in the first and second box.

>>	Anonymous Fri Mar 16 08:34:51 2018 No.9595128 >>9594653 Only five balls in the picture. This is a BIG red flag. Sloppy thinking.

>>	Anonymous Fri Mar 16 08:37:14 2018 No.9595136 >>9595125 >>9595128 brainlets

>>	Anonymous Fri Mar 16 08:43:00 2018 No.9595149 >>9595136 >i have no argument

Anonymous Fri Mar 16 09:02:02 2018 No.9595172

>>9591739
As the question in the image is formulated is 50/50 because it asks you what's the probability AFTER you pulled out the golden ball.
The Bertrand paradox is "if that happens to be a gold coin", making the choosing if the box count and making it be 2/3s.

Anonymous Fri Mar 16 09:02:31 2018 No.9595173

>>9595083
>you cannot pick the silver ball. If you do, you go again from the start. What is so hard to understand here

And if your algorithm is "start over until you pick a gold ball," there's a 2/3rds chance that the gold ball you picked was from the 2-golden-balls box and a 1/3rd chance it was from the 1-gold-ball box.

>>	Anonymous Fri Mar 16 09:03:11 2018 No.9595175 >>9595173 >>9595172

>>	Anonymous Fri Mar 16 10:24:42 2018 No.9595291 >>9591770 combinations. there are two gold balls in the first box. this means there are two ways to pick a gold ball first and second. the other option is that you have the middle box. -> 2/3

>>	Anonymous Fri Mar 16 10:53:24 2018 No.9595340 >>9595116 >= if you did pick a silver ball at random, then the case is irrelevant and you need a re-do. And if your algorithm is "keep re-doing until you find a gold ball," you're gonna end up with the box with more gold in it. Duh.

Anonymous Fri Mar 16 11:27:35 2018 No.9595401

>>9592174
This is the key concept actually. The second time you either have a 2/3 chance of a 100% chance of getting a gold ball (box 1), or a 1/3 chance of a 0% chance of getting a gold ball (box 2), so at the end is still 2/3.

This is a remake of the Monty Hall problem actually.

>>	Anonymous Fri Mar 16 13:45:01 2018 No.9595637 >>9594130 Okay that's fair.

Anonymous Fri Mar 16 13:48:47 2018 No.9595645
File: 274 KB, 1097x1646, 1519695443299.jpg [View same] [iqdb] [saucenao] [google]

>>9591900
His math is perfect. You are wrong. Once you have 1 gold the odds of the box being SS is 0 so you discount it. Then there are 3 balls, where 2 are gold. Hence, 2/3.

pic unrelated

>>	Anonymous Fri Mar 16 19:31:12 2018 No.9596569 i know its 2/3 beacuse when i ctrl f it says 2/3 the most

Anonymous Fri Mar 16 21:14:21 2018 No.9596771

>>9591770
Now turn the balls over and realise they have letters on the back. Box 1 has gold balls A and B, box 2 has gold ball C now you notice that 2/3 of the time you pulled either A or B so from box 1 and not box 1 50% of the time like you mistakenly assumed before you noticed the letters

>>	Anonymous Sat Mar 17 02:29:38 2018 No.9597286 >>9595401 >This is a remake of the Monty Hall problem actually. Other way around

Anonymous Sat Mar 17 03:28:07 2018 No.9597363
File: 32 KB, 640x640, 1517916981928.jpg [View same] [iqdb] [saucenao] [google]

The thought process that made sense to me is that once it says you have a gold ball, you know you have either the left or middle box. The rest of the question is just a convoluted way of saying what are the chances you picked the left box by chance. There are 3 gold balls, 2 of which are in the left box, ergo 2/3 chance.

Anonymous Sat Mar 17 04:18:20 2018 No.9597415

>>9595116
>if you did pick a silver ball at random, then the case is irrelevant and you need a re-do.
Correct, which is exactly why it's actually 2/3. You'll be forced to redo more when you pick the GS box than when you pick the GG box, therefore you will pick the GG box in all of your counted results more often than the GS box.

I'm amazed that you actually have the process down perfectly but still come up with the wrong result in your head. Actually try what you're saying either by writing a simulation or just doing it in real life with coins or some shit and you'll see.

Anonymous Sat Mar 17 04:24:36 2018 No.9597424

>>9595172
Not sure if you're failing at reading comprehension or you are actually retarded enough to think there's a meaningful difference in probabilities by adding the "if".

After pulling out the golden ball, you can use the fact that you pulled out a golden ball to reason that you are twice as likely to picked the double-gold box than the single-gold box. It's not saying that you magically get a gold ball every time, it's just saying that the ball you got happened to be gold. Any other way of reading the question is not just an alternate interpretation; it's simply wrong.

>>	Anonymous Sat Mar 17 15:16:08 2018 No.9598489 >>9591739 1/2 i've already picked a gold ball, there are two boxes with gold balls in them. you can remove the 2 silver box because it will never be a choice.

>>	Anonymous Sat Mar 17 15:18:47 2018 No.9598491 >>9595645 you have two boxes you possibly could have picked from.

Anonymous Sat Mar 17 16:40:16 2018 No.9598686

>>9595401

With the Monty Hall problem, you have a choice to switch, which is why it eventually comes out to 2/3. In this case, the first choice has been made for you, at least in my interpretation of the instructions.

So you either have another gold ball in the box, or a silver ball since you can't switch, which gives you a 50-50 chance.

Honestly the reason why there is this huge debate IMO is caused by how you interpret the question. If you count the first draw, then it will be 2/3, if you don't, then it's 1/2.

>>	Anonymous Sat Mar 17 17:35:27 2018 No.9598812 >>9598686 someone gets it

>>	Anonymous Sat Mar 17 17:44:35 2018 No.9598837 File: 4 KB, 221x250, 1521008347366s.jpg [View same] [iqdb] [saucenao] [google] >>9591739 does he put the ball back or not, not enough info brian letterman

Anonymous Sat Mar 17 17:46:00 2018 No.9598839

>>9592480
>The thing you have to find is the probability that you have chosen the first box
no you fucking don't, retard
this is a conditional probability, conditional on HAVING PICKED A GOLD BALL

the final probability for OP's question is 1/2 and anyone saying otherwise hasn't passed undergraduate probabilities

>>	Anonymous Sat Mar 17 17:47:01 2018 No.9598841 File: 13 KB, 400x400, 1521237121653.png [View same] [iqdb] [saucenao] [google] >>9592589 >there is 1 box and 3 possible balls retard

>>	Anonymous Sat Mar 17 17:47:18 2018 No.9598842 >>9591739 It's 50/50 you fucking plebs.

>>	Anonymous Sat Mar 17 17:49:04 2018 No.9598848 >>9597424 >you can use the fact that you pulled out a golden ball to reason that you are twice as likely to picked the double-gold box than the single-gold box holy shit the absolute state of this fucking board

>>	Anonymous Sat Mar 17 17:50:06 2018 No.9598849 >>9595645 >Then there are 3 balls jesus christ you are such an arrogant retard learn how to read

>>	Anonymous Sat Mar 17 17:51:51 2018 No.9598850 File: 2.62 MB, 1000x1341, 1521245110637.png [View same] [iqdb] [saucenao] [google] >>9592102 thank god someone in this thread has a functioning brain inside their skull

Anonymous Sat Mar 17 17:53:20 2018 No.9598854
File: 43 KB, 585x644, 1521230227759.jpg [View same] [iqdb] [saucenao] [google]

>>9592252
>There was a chance to get a silver ball but it didn't happen
THAT MEANS THERE WAS NO CHANCE OF PICKING A SILVER BALL, BRAINLET

LEARN ABOUT BASIC CONDITIONAL PROBABILITIES BEFORE YOU EVER DARE POST IN THIS THREAD AGAIN

Anonymous Sat Mar 17 17:57:06 2018 No.9598862

>>9591739
Scenario 1 : I pick Gold Ball I in Box 1 and then Gold Ball 2 in Box 1

Scenario 2 : I pick Gold Ball 2 in Box 1 and then Gold Ball 1 in Box 1

Scenario 3 : I pick Gold Ball 1 in Box 2 and then Silver Ball 1 in Box 2

How many scenarios result in a gold ball being chosen after a gold ball = 2

How many total scenarios = 3

P = 2/3

Anonymous Sat Mar 17 18:49:38 2018 No.9598983

>>9598848
I know it's not easy for a brainlet but I'm confident you can get it if you think REALLY hard.

Let me make a clearer example: let's say there were only two boxes with one ball each. One box has a gold ball and one box has a silver ball. Pick a random box and then a random ball from said box. Grats, you picked a gold ball. You can now reason with 100% accuracy that you must have picked the box with the gold ball in it. Simple enough, right?

The same applies here, except we are dealing in probabilities instead of absolutes because there is more than one box that contains one or more gold balls. If the random ball you pick up happens to be gold, then you can correctly reason that the box with the highest likelihood of being picked was the one that is most likely to result in gold balls from random selections, i.e. the one with the highest ratio of gold to non-gold balls.

The really interesting thing about this is that it applies regardless of HOW MANY boxes there are or how many balls of what color are in each box. If you added a silver ball to all 3 boxes for example, making it GGS/GSS/SSS, the exact same logic applies: you are twice as likely to randomly select a gold ball in the box with 66% gold balls than you are in the box with 33% gold balls, and thus there is a 2/3 chance that you picked the former.

TL;DR the color of the ball you find helps you determine what box you are most likely to have picked. Even if you have trouble believing it yourself, you can easily prove it to yourself by coding a simulation or even performing the experiment in real life.

Anonymous Sat Mar 17 18:58:13 2018 No.9599005

>>9598854
>not knowing what conditional probabilities are yet asking others to read up on them
Jesus fucking christ this board gets more retarded every week. Sorry friend, but "conditional probability" does not mean "anything that didn't happen had a 0% chance to ever happen." That statement is simply determinism, which may even be a true fact of the universe depending on whether quantum mechanics can ever be fully explained without fundamental randomness, but is a meaningless statement when it comes to probabilities in terms of our actual predictive power; flipping a coin and getting heads doesn't mean tails was fucking impossible.

>>	Anonymous Sat Mar 17 19:32:10 2018 No.9599083 >>9599005 >this entire post holy Dunning Kruger, batman!

>>	Anonymous Sat Mar 17 19:35:16 2018 No.9599099 >>9599083 meant for >>9598854

>>	Anonymous Sat Mar 17 19:45:04 2018 No.9599131 >>9599099 retard

Anonymous Sat Mar 17 19:47:29 2018 No.9599137

>>9598854
>THAT MEANS THERE WAS NO CHANCE OF PICKING A SILVER BALL, BRAINLET
No it doesn't.

>LEARN ABOUT BASIC CONDITIONAL PROBABILITIES
If you did yourself then you'd realize what a simple problem this is. Maybe you misread it as saying "if you pick a random ball that ball will always be gold" but that's hardly an excuse considering you should have immediately recognized that it's a self-contradicting statement; it can't be random if it's always gold. You would have then read it correctly and realized that it's a simple narration: "all this stuff happened so far, now use what happened to figure out the probability of this related thing."

When you read it like that there is no fucking excuse for misunderstanding. A random ball was picked and it was gold. Do you need the image to be a fucking choose your own adventure game or some shit just so you can accept that the result was random? Are you that fucking incapable of entertaining hypothetical situations?

>>	Anonymous Sat Mar 17 19:49:01 2018 No.9599143 >>9599131 Feel free to explain your reasoning, then. I love seeing all the variety of ways that people can rationalize getting the wrong answer to this classic conditional probability puzzle.

>>	Anonymous Sat Mar 17 19:49:48 2018 No.9599144 >>9599137 >No it doesn't. explain how the probability is non-zero i'll wait

>>	Anonymous Sat Mar 17 19:51:12 2018 No.9599148 >first grabbing gold ball #1 and then gold ball #2 treating differently than first grabbing gold ball #2 and then gold ball #1 And that's why statistics are fucking dog shit.

Anonymous Sat Mar 17 19:57:41 2018 No.9599168

>>9599144
You picked a random ball from a random box. Some boxes contained one or more silver balls, therefore it was possible to randomly pick a silver ball. Furthermore, one of the boxes that could possibly contain the ball you picked also contained a silver ball. Therefore, with the limited information you have, you know that it was possible that you may have been able to pick a silver ball from the box you picked.

Hope you didn't have to wait long.

>>9599148
You may not like it but the cool thing is that it's actually legitimately true. This isn't just some math trick that only deals with abstractions. If you literally found yourself in the situation the OP describes exactly as it's described, you would indeed find a second gold ball twice as often as a silver ball after a gold ball.

Anonymous Sat Mar 17 19:59:59 2018 No.9599170

>>9599168
>You picked a random ball from a random box. Some boxes contained one or more silver balls, therefore it was possible to randomly pick a silver ball.
But this can't be the case since the ball chosen was gold.
At the point where OP asks us to compute the probability, the ball has been chosen and already determined to be gold.
There is no possibility of this ball having been silver, since it's already gold.

The conditional probability problem is:
Find the probability that the next ball chosen from this box is gold GIVEN THAT one ball already chosen from this box is gold.

>>	Anonymous Sat Mar 17 20:01:03 2018 No.9599176 >>9599168 still waiting, btw

>>	Anonymous Sat Mar 17 20:06:18 2018 No.9599190 >>9599168 IT MAKES NO SENSE I HAVE EXACTLY TWO SCENARIOS LEFT EITHER I PICK A SECOND GOLD BALL OR I DON'T

>>	Anonymous Sat Mar 17 20:08:42 2018 No.9599195 >>9599190 don't expect a Dunning Kruger brianlet to be able to give reasonable explanations for anything

Anonymous Sat Mar 17 20:10:02 2018 No.9599200

>>9599170
>But this can't be the case since the ball chosen was gold.
Yes, obviously you DID NOT pick a silver ball. The fact that a silver ball was among the options that could have been picked is significant, however, specifically BECAUSE it did not happen. The fact that the ball you picked was gold instead of silver is exactly the condition in question.

>Find the probability that the next ball chosen from this box is gold GIVEN THAT one ball already chosen from this box is gold.
Mostly correct, just one small thing that you critically left out: the gold ball was randomly selected. This is the crux of it, really. Youzre choosing to willfully ignore that part of the OP because you know that only by changing the question completely can you turn your wrong answer into a right one.

You're mistaking OP's description of past events as a prescription of the only possible events that could have happened. You're arguing semantics rather than probabilities and the worst part is that your semantic argument itself is wrong.

Anonymous Sat Mar 17 20:15:09 2018 No.9599207

>>9599200
>Youzre choosing to willfully ignore that part of the OP because you know that only by changing the question completely can you turn your wrong answer into a right one.
>You're arguing semantics rather than probabilities
you're the one who seems to think that past events are somehow non-deterministic
get back to me when you've taken a remedial undergraduate course in probabilities and statistics

Anonymous Sat Mar 17 20:22:44 2018 No.9599217

>>9599207
Come on, I really don't know how much more simply I can explain this. Part of me thinks you're deliberately trolling at this point, but just to entertain the argument, I would like to ask you a straightforward question:

Why do you believe that every simulation of the OP's post finds a second gold ball 2/3 of the time? Can you point out the critical bug in them that is causing this supposedly wrong answer? Let me outline the algorithm in plain English to make it easier to discuss:
Step 1. Set up three boxes: GG, GS, and SS
Step 2. Pick a random box.
Step 3. Pick a random ball.
Step 4a. If the ball you picked is NOT gold, revert and record NO DATA WHATSOEVER because this is a completely irrelevant situation to the OP.
Step 4b. If the ball you picked IS gold, record the color of the remaining ball in the picked box.

I guarantee you that every time you run this algorithm, it will converge on "gold" 2/3 of the time.

Let me change that to two direct yes or no questions:
1: Do you believe this algorithm accurately captures the OP's question? Why or why not?
2: Do you believe this algorithm, if properly coded, would NOT result in "gold" 2/3 of the time?

>>	Anonymous Sat Mar 17 20:24:34 2018 No.9599222 >>9591739 2/3

Anonymous Sat Mar 17 20:26:39 2018 No.9599225

>>9599217
Just one small clarification: what I mean by getting "gold" 2/3 of the time is that there will be 2 "golds" recorded for every "silver" recorded, so we're not actually counting those no-data results as anything in case you were thinking that was why. As far as anyone is concerned those no-data results never happened in the first place.

Anonymous Sat Mar 17 20:31:51 2018 No.9599232

>>9599217
to be frank i don't give a shit about simulations since i'm a mathematician and i'm looking at this as a probabilities question
can you point out where the flaw in my critique of your argument is without being an autist?
because you have yet to do so

Anonymous Sat Mar 17 20:47:03 2018 No.9599278

>>9599232
I cannot point out anything more that I haven't already pointed out. I'll give it one more try though since I'm obviously not very good at explaining things, so I'll try wording it in different ways to see if we can reach an understanding:

Yes, the chance of having picked a silver ball is 0%. That is established by the fact that you picked a gold ball. Indeed "you could have picked a silver ball" is really more of an intuitive shortcut to understanding why the answer is as it is, rather than an actual statement of truth. Of course you can't change the past. But the information of what color ball you DID pick allows you to calculate the probabilities of which box that ball came from as long as you accept that the ball itself was the result of a random selection. And when you calculate those probabilities, you'll find that there are two possible scenarios that result in you finding another gold ball, but only one possible scenario that results in you finding a silver ball.

And no, you don't get to say that the ball wasn't the result of a random selection, because the OP literally states that the selection was random. This is because it's a hypothetical problem: the text in the jpg won't randomly change every time you open it. It's asking you to accept that a random selection took place and the result of that selection was as described. If you aren't willing to accept that then you aren't answering the same question at all, and so your misreading of the question will result in a miscalculation of the answer.

Regarding the simulation question, I hope the algorithm was clear enough to discuss because I think it could be much more fruitful at this point. After all, it basically means we can cheat by running it and know that the answer is in fact 66.(6)%. Once we agree on that maybe you can use your math skills to better explain why that is in fact the case. Sometimes it's easier to figure out the explanation when you already know the result.

Anonymous Sat Mar 17 20:56:14 2018 No.9599297

>>9599232
>>9599278
Actually as I was rereading my post I think I might have found a "correct" way to word that. It's not that you "could have picked" a silver ball, it's that silver balls were "available to pick". In other words, you're using what you know the probability of ANY GIVEN random selection to be as information along with the ACTUAL choice to build a more complete picture of your current situation.

To put it another way with a similar example: there are two fields nearby, one full of yellow grass with a few patches of green and one field of nothing but green grass. You find someone's shoes on the sidewalk and they are covered in green grass. You can't find any yellow grass on them, but you can't rule out the possibility that they walked on the yellow/green field and only happened to pick up green grass on their shoes. If you know that yellow and green grass are equally likely to stick to a shoe, then you can reason that it is more likely that these shoes tread in the green field, i.e. the chance is higher than 50% even though there are only two possible fields they could have tread on.

Anonymous Sat Mar 17 22:12:17 2018 No.9599444

>>9599207
>get back to me when you've taken a remedial undergraduate course in probabilities and statistics
kek please tell me you're the same guy crying dunning-kruger because this post is as dunning-kruger as it can get

this is conditional probability 101
>i picked a box from a set of boxes
>one of those boxes is all gold, the other is partially gold
>i got gold on my first blind pick from the box
>its more likely i picked an all gold box because it had a higher chance of giving me gold on my first pick
or fuck just manually count the possibilities if you want to do it the long way:
>there are two ways i could have ended up with a gold ball from all gold box
>there is only one way i could have ended up with a gold ball from the partially gold box

2
over
3

Anonymous Sat Mar 17 22:17:22 2018 No.9599455

>>9599444
Not sure if you read the whole conversation but I think it's been established that his problem was with the flawed argument of "you could have picked a silver ball." Not necessarily the answer of 2/3 itself.

He was absolutely right to call it out because there is no way you could have picked a silver ball -- you plainly did not pick one. This is also an application of conditional probability. We know a silver ball wasn't picked, therefore there is no chance that a silver ball could have been picked. The real key point that I think the original person positing that argument was trying to make was that the gold ball was selected from a pool of balls that contained non-gold balls, and therefore the fact that it was gold instead of silver is useful in determining the probabilities. See my attempts are correcting/clarifying that original flawed argument: >>9599278 >>9599297

>>	Anonymous Sat Mar 17 22:19:45 2018 No.9599460 >>9599455 >gold ball was selected was randomly* selected. Definitely not helping that I keep leaving out essential parts of the scenario, since without saying "randomly" it could be validly interpreted in a completely different way.

Anonymous Sat Mar 17 22:30:20 2018 No.9599486

>>9599455
thats hardly flawed, considering its pretty damn obvious what was meant by it. any random choice of ball might be silver or gold. your ball happened to be gold. if you went back in time and tried it again it is equally likely that it would have been silver. if you get gold then you probably got it from the box that always gives gold instead of the box that only sometimes give gold, only way around that is to start assuming shit that is unrelated at best or contradictory at worst

>>	Anonymous Sat Mar 17 22:31:06 2018 No.9599489 >>9594976 That's not the question. Given the first ball is gold, you must have picked from one of the boxes with old in it. Given you piked from the box with a gold ball, whats the probability the other ball is gold? The answer is 50%

Anonymous Sat Mar 17 22:34:42 2018 No.9599501

>>9599486
Yes, if you actually read the rest of my post you'd know that I agree with your reasoning and the overall 2/3 answer. I'm just saying that, since this is a conditional probability problem, it's important to note (as the guy you quoted correctly did) that conditional probability also forbids the "it could have been silver" argument, because the main condition is that you picked a gold ball. It couldn't have been silver. You can argue that it's just semantics and that's exactly what I did at first, but it's far more important to get the right answer with the right arguments. Getting the right answer with the wrong arguments is just as bad as being wrong. Or maybe even worse since people will believe your fucking argument and start using it elsewhere because "it must be right."

So even if it's obvious to you what was "meant" by that argument, it's still an essentially invalid argument.

Anonymous Sat Mar 17 22:48:49 2018 No.9599533

>>9599489
You're missing an important detail when you reduce the information. You're changing "you randomly picked a gold ball" to "you picked a box that contains at least one gold ball". The latter is one conclusion of the former, but it is not an equivalent statement. The former actually gives you even more information, even if it doesn't seem obvious at first.

When you consider everything up to the point where the question is posed, you actually know TWO things about the box you picked:
1. It contained at least one gold ball. (This eliminates SS and leaves you with a 50% probability if considered alone)
2. The first ball you randomly picked from the box is gold. (This allows you to consider the probability of a box giving a gold ball on ANY random first selection. We know that you DID get gold, now we're just considering how likely it is that you got gold from the first or second box. It might make it easier if you make it more extreme by saying the first box had 1000 gold balls and the second box had 999 silver balls and 1 gold. Picking the one gold ball from the second box is a less common event than picking any ball at all from the first box. Since the first box more commonly satisfies the conditions that lead to your current situation, you correctly reason that the first box is more likely to be the box you've picked.)

So to come to the correct conclusion, you don't just want to consider "what boxes could have possibly satisfied this condition?" you want to consider how likely it is that each box would satisfy that condition on the first random selection of a ball from that box, because that's the REAL situation you're in.

Anonymous Sat Mar 17 23:50:58 2018 No.9599654

It depends on whether you remove the ball or not. I interpreted the problem as meaning you remove the ball and don't put it back.
In that case, the only boxes relevant are ones containing gold balls to begin with: the gold-gold box, and the silver-gold box

If you remove a gold ball from the gold-gold box and draw again, your odds are 1/1, or 100% of drawing a gold ball again.
If you remove a ball from the silver-gold box, your odds are 0/1, or 0%.

The average of 100% and 0% is 50%. If you disagree, you are an idiot with a very poor grasp of logical analysis, or you fundamentally misinterpreted the problem.

Anonymous Sat Mar 17 23:55:12 2018 No.9599663

>>9599654
Edit: My argument for why the ball is being removed and not put back in is because the problem states, "what is the probability that the ball is also gold"
Imagine replacing and drawing the exact same gold ball. How can be "also gold"? How can something also be the same thing which is already is?
Similarity necessitates two distinct objects to be compared between

Anonymous Sun Mar 18 00:29:36 2018 No.9599728

>>9599654
>>9599663
I'm pretty sure everyone agrees with that interpretation of the premise. The question says nothing about putting the ball back so it's a pretty safe bet that it stays out of the box.

And also literally everything you said in your post is factually correct. However one of the statements seemed to have come out of nowhere:
>The average of 100% and 0% is 50%
Yes, that is indeed the average of 100% and 0%. But that's not what the question asked. The question asked what the probability of the next ball you pick from the same box being gold is.

You can rephrase the question as such:
>What is the probability that you picked the gold-gold box?
The answer for that is, unintuitively, 2/3. This is because the first ball that you randomly picked from your selected box was gold. That means it's twice as likely that you selected the gold-gold box than the gold-silver box, because randomly picking a gold ball from the gold-silver box is twice as rare. Another way that might help you conceptualize it is to not get hung up on the initial choice of box. There's no difference whatsoever in probabilities between just picking a random ball from the set of all balls compared to picking a box first, then a ball from it. All balls are equally likely to be picked from the start. You can then eliminate the three silver balls that you know were not picked. Two of those three remaining gold balls are in the same box, so two out of three times the next ball you take is gold.

Basically, you stopped one step short. You correctly deduced what drawing the gold ball told you about the boxes you could not have possibly picked (silver-silver because it contains no gold balls), but you didn't realize what drawing the gold ball told you about the boxes you could have picked (that one was more likely to have resulted in your first choice's color than the other).

For a bit of detail and history of the riddle, see: https://en.wikipedia.org/wiki/Bertrand's_box_paradox

>>	Anonymous Sun Mar 18 01:08:48 2018 No.9599836 File: 50 KB, 631x796, 1518308601614.jpg [View same] [iqdb] [saucenao] [google] 2/3 you stupid brainlets lmfao.

>>	Anonymous Sun Mar 18 01:16:01 2018 No.9599854 >>9599836 Pretty sure you're the brainlet.

Anonymous Sun Mar 18 01:33:06 2018 No.9599893

>>9599728
Very interesting. I totally see what you mean now, even though it's a completely different way of looking at things. I think my brain likes to break things down into rational steps that can be analyzed individually, which doesn't work for this problem

>>	Anonymous Sun Mar 18 01:52:32 2018 No.9599943 >>9593548 Distinct

>>	Anonymous Sun Mar 18 13:50:49 2018 No.9601082 The problem can be equivalently rephrased as this: "You pulled a gold ball out of a box. What is the probability that the box you chose is the one with two gold balls in it?"

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/sci/ - Science & Math