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/sci/ - Science & Math

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11706900 No.11706900 [Reply] [Original]

Not a /pol/ user, but this was posted there yesreday. What is the right answer? Is there an answer?

>> No.11706914

>>11706900
possibilities are
G G
G S
S G
S S
eliminate the last one
G G
G S
S G
so its a 1/3rd chance

>> No.11706919

>>11706900
From the description, our space is:
1/4 chance two gold balls
1/4 chance two silver balls
1/2 chance one of each
The merchant will say what he did in the first and third situations (I am assuming his sniffer is 100% accurate regardless of how much gold there is), so our updated probabilities are
1/3 chance two gold balls
2/3 chance one of each

>> No.11706957

>>11706914
But isn't G S and S G - same thing? Does it matter which one comes first?

>> No.11706958

>>11706900
Assuming "randomly" means an equal chance of either choice.

P(2G|S) = P(2G) P(S|2G) / P(S) = (1/4) P(S|2G) / ((1/4) P(S|2G) + (1/2) P(S|1G) + (1/4) P(S|0G)) = P(S|2G) / ( P(S|2G) + 2P(S|1G) + P(S|0G))

So it depends on how confident we are that the Rabbi would smell a gold ball if there was two in the box or one in the box, and how confident we are that the Rabbi is not smelling something that isn't there. Assuming the Rabbi has a perfect sense of smell then these are 1, 1, and 0 respectively, meaning the chance there are two gold balls is 1/3.

>> No.11706961

either 100% or 0%

>> No.11706971

>>11706900
>trusting the Rabbi
0%
dumb goyim if there were actually gold in it he would have taken it for himself

>> No.11706982

>>11706957
The frog is essentially flipping two coins. This gives {TT,TH,HT,TT}. There is at least one heads in 3 of these cases, and one of the three has two heads.

>> No.11706988

>>11706982
That 2nd TT should be HH.

>> No.11706998

>>11706988
I got it Anyway. But still not sure if 1/3 100% correct

>> No.11707006

>>11706998
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

It's 1/3 but brainlets will say it's 1/2

>> No.11707007

>>11706900
it's 1/3
but why the anti-semitism?

>> No.11707013

>>11707007
/pol/ are obsessessed with jews. I just found this pic on Internet, Googled it - turns out /pol/ posted it yesterday. I didn't want to provocate or any of that

>> No.11707032

>>11706957
>Does it matter which one comes first?
It matters that there were two possible ways to gather a collection of 1 G and 1 S.

>> No.11707044

>>11707007
Are you black?

>> No.11707053

The box contains one gold orb and one tripwired grenade; the jew knows this and is waiting for you to open the box, so he can keep the gold after.

>> No.11707058

>>11706971
this desu

>> No.11707062

>>11706900
>not a /pol/ user
You'll have to leave, this is a /pol/ user exclusive board.

>> No.11707151

1/3 easy

>> No.11707289

50/50
Either both are gold or not

>> No.11707305

>>11707289
This is the answer. The identity of the first orb doesn't effect the second, so both have a 50% chance of being gold. This also means the order is arbitrary. We know one is gold, so essentially were asking for the 50/50 that the other one is gold. 50%.

>> No.11707332

>>11707289
>>11707305
tards

>> No.11707349

1/4 because I'm antisemite

>> No.11707356

>>11706900
>Not a /pol/ user, but this was posted there yesreday.
Why do you have pictures from yesterday's /pol/ saved if you don't use /pol/?

>> No.11707396

>>11707356
I saw it else where, but Googled and it brought me to pol

>> No.11707407

>>11707305
But isn't it was 1/4 chance (GG, GS, SG, SS) to get 2 golden in the beginning and then we got told there's at least one golden and then it became 1/3 (GG, GS, SG)? Why not? Can You explain?

>> No.11708331

>>11706971
This is the answer that studying Talmud gives you.

>> No.11709133

>>11706900
Use Bayes, im too lazy.
If frog chooses truly randomly p that chosen orb is s is g is 1/2.
We need p of both orbs are gold given at least one is gold.
So we have to compute p of at leastone orb is gold And two are gold/ p of at least one is gold= p of two are gold/at least one is gold.
Theese xan be modelled as bernoulli trials. 1/4 / 3/4 = 1/3.
Theese images are always the same fucking thing and for some reason people without undergrad course in prob and knowledge of this formula always fuck it up.

>> No.11709352

>>11706900
If you actually ran this for a large amount of trials and extracted how many times BOTH orbs were gold, you would realize the answer is 50%.

Remember the question is asking for the odds that BOTH orbs are gold and 1 of them will always be gold. It's not asking to enumerate the possibilities or any other bullshit.

>> No.11709382

>>11709133
I'm a highschool dropout but I get 1/3 without problems and I have no idea whats baeys or bernoulli
Maybe it's my experience as a poker pro that helps me understand probability

>> No.11709416

>>11709352
No. Inclusion of knowledge in the preceeding question implies that when you intepret the trials ypu propose you take into the account only the ones where at least one was gold. So you compute the proportion both are gold to at least one was gold so its 1/3.

>> No.11709437

>>11707305
Identity of the first orb affects the final result regardless of what the second is
If you got G first then you still can either have GS or GG after the second pick
If you got S first then you still can either have GS or SS after the second pick
It means that you can get GS from two different points, either from picking G after S or picking S after G, while both GG and SS has exactly one way - if you pick both G or both S
It means that probability of GS is twise as big as GG or SS, so system of equasiosn is like this GS+GG+SS = 1; GG=SS; GS = 2GG
Solving it will give you GG = SS = 1/4 and GS = 1/2
Other way to think about it is to realise that GS and SG are ALWAYS different states which exist under the surface regardless of what we think, and thus we can only think of them as the same by purposefully merging them into one, which means merging(summing) their probabilities too, thus we get SG1/4(true state) + GS1/4(true state) = GS1/2(merged state we pretend to be true)
And with those probabilities when you remove SS you will have a total of 3/4 instead of 1 and thus GG being 1/4 its 1/4 out of 3/4 which is 1/3

>> No.11709457

You have to be ESL or a "half an A press" pilpul extraordinare to think the answer is anything other than 50%.

>> No.11709458

>>11709382
The nuance is that without bernoulli you might want to model the anwsers as gg ss gs ie treating two trials as indistiguishable and equiprobable therefore 1/3 each. You would get different results than if you considered gg gs sg gg, middle ones as distinct outcomes. Bernoulli trials is just a formula that works for the latter case and counts all possibilities ie orderings of trials such thst you get k successes. In this case 1 success either in first trial or in second one. And the latter interpretation tends to be congruent to most long run frequencies of real life phenomena occurrencies such as coin toss.