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/sci/ - Science & Math

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9908326 No.9908326 [Reply] [Original]

This was posted on /g/ while ago and they mostly proved to be bunch of tards. Show that you are smarter /sci/.

>> No.9908329

>>9908326
50% ? i dont know im stupid

>> No.9908336

>>9908326
2/3

>>9908329
If you know you're a brainlet, why would you even post and ruin the reputation of /sci/?
fuck you

>> No.9908341

>>9908326
2/3. This is a classic -- counter intuitive problem though so it's understandable that people get it wrong.

>> No.9908348

Just started getting into math.
Id say 1/3, the silver box is eliminated from the possibilities and there are 3 balls left, two are gold. As there are two boxes and you may have selected the gold + silver one, the chances go from 2/3 to 1/3

>> No.9908351

>>9908329
>>9908348
it's 2/3

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.9908359
File: 14 KB, 728x441, bertrandBox.png [View same] [iqdb] [saucenao] [google]
9908359

stage #1: 3 ways to have picked the first gold
stage #2: Next ball, 1 silver, 2 golds

>> No.9908363
File: 3 KB, 125x118, bertrandBox_diagram.jpg [View same] [iqdb] [saucenao] [google]
9908363

>> No.9908389 [DELETED] 

Ok, /sci/ has proven really smarter than /g/. Now for a harder challenge, try to explain correct result to idiots in >>/sci/66958014

>> No.9908390

Ok, /sci/ has proven really smarter than /g/. Now for a harder challenge, try to explain correct result to idiots in >>>/g/66958014

>> No.9908406

>>9908326
it's a 2-step process with 6 different possible outcomes, each as likely to occur. (g1,g2) (g2,g1) (s1,s2) (s2,s1) (g3,s3) (s3,g3).
Since we know that we pick a gold ball in the first step, we're left with the following possible outcomes (g1,g2) (g2,g1) (g3,s3), again, each as likely to occur. Thus 2/3 chance that the second ball will also be gold.

>> No.9908427

It's 50%.
You have one gold ball so you are either in box 1 or 2 at random. So it's 50% chance you get gold(box1) or silver(box2).

>> No.9908431

>>9908326
2:1 ie 2/3 = 66.667%

>> No.9908508

>>9908427
>what is "three golden balls" leads to "three different 2nd ball"

>> No.9908514

You either do or you don't so 50%

>> No.9908515

can't be bothered to work out solution on paper but it looks like a trivial application of bayes formula

>> No.9908526 [DELETED] 

If you picked a gold ball, then by definition you picked box 1 or box 2 (as box 3 has no gold balls). This leaves 2 gold balls and 1 silver ball still in play. So the chance is 2/3.

>> No.9908531

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

>> No.9908543

This is posted every week

>> No.9908549

>>9908336
>ruin the reputation of /sci/

I've got some babnewsnfor ya, kid

>> No.9908569
File: 114 KB, 988x958, ball.jpg [View same] [iqdb] [saucenao] [google]
9908569

>> No.9908577 [DELETED] 

>>9908569
bob doesn't get to guess,
alice tell's him which box to pick from

>> No.9908582

>>9908531
The problem I have with this is that the question states you have already grabbed a gold ball. This means that the SS box is never considered within the problem, right?

>> No.9908589 [DELETED] 

>>9908582
>what is "favorable" for $750, Trebek

>> No.9908594 [DELETED] 

>what is "favourable" for $750, Trebek

>> No.9908597

>>9908582
>what is "favourable" for $750, Trebek

>> No.9908598

>>9908326
2/3

See it this way, you got one gold ball, because the first box have two balls and the other only one, its more likely that the ball you got came from the first box. So 50% its already discarded, do all the combinations and you get 2/3

>> No.9908603

>>9908569
bob doesn't get to guess,
alice tells him which box to pick from

>> No.9908616
File: 24 KB, 600x600, 1460240205248.jpg [View same] [iqdb] [saucenao] [google]
9908616

>>9908514

>> No.9908644

>>9908329
You are probably mistaking probability from the unaware participant's view (from not knowing which of the gold ball-containing boxes he picked) with the actual probability.

>> No.9908702
File: 556 KB, 1350x1800, 1532641746889.jpg [View same] [iqdb] [saucenao] [google]
9908702

>>9908326
x-posting from g reddit. 25% is the solution we went with. Can you prove /g/ wrong? Pro tip, you can't!

>>66968482

>> No.9908705

>>9908702
>*crosspost
>>>66968482

>> No.9908733

>>9908705
damn

>>>/g/66968482

>> No.9908759

>>9908326
There are 3 gold, the chance of you picking from the one with 2 gold is 2/3, so 2/3

>> No.9908792

>>9908702
it's 2/3 retard

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.9908808

>>9908589
Oh I see

>> No.9908823

>>9908792
check this >>9908733 tard. You are probably an engineer anyways

>> No.9908858

>>9908823
Wrong, consider the problem again for each case of choosing the 3 gold balls. Since it is within the premise of the problem that you already chose a box and took out one gold ball, then the probability at this point is if you chose the correct gold that would guarantee the other ball in the box to be a gold. From this, it is clear to observe that there are two gold balls that, when chosen, would guarantee a success of choosing another gold ball when reaching back into the box you already chose in the premise of the problem. The other gold ball, on the other hand, would result in immediate failure, so you have 1/3 of a chance of choosing the "losing" gold ball. Overall, there is a 2/3 chance your first gold ball was a "successful" gold ball, while there is a 1/3 chance your gold ball was a "losing" gold ball, making the answer to the problem 2/3. Remember, you cannot reach into another box, so the problem is whether or not you pulled out a gold ball from the RIGHT box, not if you will pull out a second gold ball.

>> No.9908866

>>9908823
in conditional probability order matters

>> No.9908872

>>9908582
Yes, the premise of the problem guarantees that you chose at least one gold ball from the box you chose. Because of this, choosing the 3rd box containing two silver balls cannot possibly happen, since that would violate the premise of the problem

>> No.9910105

>>9908326

Isn't this just the Monty Hall problem?

>> No.9910131
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9910131

>>9908616

>> No.9910174

>>9908326

Kek. I completely glossed over the second part of riddle and got 1/2 at first attempt. The conditional probability of the first draw being gold is essentially the whole problem.

(1/2)(1)+(1/2)(0+1)
2/3

>> No.9910237

>>9908531
so is the SS box irrelevant? It seems the final odds wouldn't change if we removed the SS box and you only had two box choices.

>> No.9910247

>>9908326
100%

>> No.9910250
File: 85 KB, 960x960, 1524011437231.jpg [View same] [iqdb] [saucenao] [google]
9910250

>>9910247
Why you ask? You dont need probability when you control the outcome.

>> No.9910263

>>9910237
>>9908363

>> No.9910272
File: 50 KB, 374x382, monty.png [View same] [iqdb] [saucenao] [google]
9910272

>>9910105
The two are the opposite in the way they mislead.
With Monty Hall, you think that the doors are independent, but you have to think of them as a group (image).
With Bertrand's Box, you think the gold balls aren't independent when actually they are (three ways to pick the first gold).

>> No.9910421

>>9908702
Gib milk.