If you pick a golden ball you either picked the box with 2 golden balls or the box with one golden one silver, but you dont know which, so either you pick the one with 2 golden balls -> 100% of next golden, or the one with 1/1 -> 0% of next golden, so in total 50% chance of next ball to be golden, other answer is just retarded
>Do the computation. P(both gold|first gold) = P(both gold and first gold) / P(first gold) = P(both gold) / P(first gold) = (1/3) / (1/2) = 2/3. Edit: or just do the usual trick of changing 2 to 1000. If one box has 1000 gold balls, one box has 1 gold and 999 black, and the third box has 1000 black, and you pull a ball at random and it's gold, which box did it come from?
>>720728328 It would have been this before you picked the first gold ball and eliminated the 2 silver ball box as a possibility. Now you're down to just the other ball being a binary choice of gold or silver. If you got to switch to the other box that would change the odds but as you are stuck with the box you already picked the only answer is 50/50. It is already either gold or silver and nothing changes that.
>>720728886 Who gives a shit about your thread, dumbass? I'm here to see how much of a pseudo-intellectual you are. Your shit is straight cringe, bro. Looking in the archives for that thread so I can paste shots here for people to see.
>>720729075 Only wrong if you misread the question. As it is written the only options a binary issue of the already selected ball. There is no choice and there were only 2 boxes 50/50. If you got to change or answer or had not yet selected the box it would get complicated but here there is no choice involved.
>>720729948 >wew lad >I'm going to link you to a different thread, talk shit, then expect not to be followed. >I was only pretending to be retarded >muh gf >y did you post muh gf >"I'm going to go troll on this other thread"
>>720729785 If you compute from the beginning then yes but that isn't what the question is asking. It stopped being that when the silver box became irrelevant to the question. Now all mention of that box is garbage data to what is being asked.
>>720729520 >pick random box: A: GG B: GS C: SS >reach in, grab random ball, which is gold >at this point, that ball could have come from: A: >GG A: GG< B: >GS >so there's a 2/3 chance that the ball you drew was from box A >this probability extends to the rest of the contents of the box you chose >since there is only one choice involved in the whole problem
Like another anon said, simply expand the example for clarity: >three boxes, contents as follows: A: GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG B: GSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS C: SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS >now, if you draw a gold coin, which box do you think it came from? Anyone that's not retarded would assume the gold coin came from box A.
>>720729785 Not hum but I'm going to ask you to explain why.
Since you originally choose one of two boxes (the third was never an option since you are to pull a golden ball) it seems equally likely that you pick either box. You can't change your choice, so you always pick the two balls in either box which means the only factor is your original choice, again, a 50/50 pick.
I'm going to need some solid reasoning to see it otherwise. If there are any factors I did not consider, state them.
>>720730852 >Not hum but I'm going to ask you to explain why.
>Since you originally choose one of two boxes (the third was never an option since you are to pull a golden ball) it seems equally likely that you pick either box. Yes, correct, although at this point the all silver box is still in play but that doesn't really matter.
>You can't change your choice, Correct.
However, when you randomly pick a ball from the box, if that ball is gold (it is), then you are twice as likely to have the all gold box from that point forward, because there are 2 ways to get a gold ball as your first ball from the all gold box and only 1 way to get a gold ball as the first ball from teh gold/silver box.
You could rewrite the question as there are two boxes and one has a silver ball. You have already selected your box. What are the odds that it has the silver ball. That is 50/50. The choice is already made and only one of the 2 is correct.
It matters not which gold ball was it, only the fact it was a gold ball. Sure, if you pick any gold ball there's a 66% chance its from the all-gold box, but that's only if picking the silver ball of GS first was ab option.
Since you are fated to pull a Gold ball first whichever box you choose, it becomes irrelevant what box is it more likely to come from.
For example, with a guaranteed gold draw even in the 1000 G box vs 1G, 999 S box it is just as likely to draw it from either.
>>720731607 A: GG B: GS >possibilities, if gold coin is chosen: A: >GG A: GG< B: >GS >still a 2/3 chance that the ball you chose was from box A You, like many people , have an inherently incorrect approach to probability. You are trying to solve the problem with common sense instead of logic. That approach usually leads to an incorrect answer. Seriously, if you doubt this, go look up one of the many mathematical resources that outlines the Bertrand's box paradox.
>>720732441 Who gives a shit if I'm samefagging? I'm going to keep reminding you and everyone else that you're under the age of 15 and that your whole intellectual bravado is bullshit. It's pissing you off isn't it? Get mad some more.
I love how you're frustrated to the point to where you can't even articulate a response. Its just gibberish at this point. Your life is a joke.
>>720731994 >It matters not which gold ball was it, only the fact it was a gold ball. Yes, and since you are only making one choice in the whole problem, that choice is the only one that matters. IF you choose a ball and it is gold, there is a 2/3 chance that irt came from the GG box. If you chose a silver ball, 2/3 chance it came from the SS box.
>Since you are fated to pull a Gold ball first There is no fate. The problem simply stated, as an example, you chose a gold ball. It could simply have been stated as: "if you choose a ball, what is the probability that the second ball in that box will be of the same color?"And the answer would be the same: 2/3. > it becomes irrelevant what box is it more likely to come from Incorrect. The only choice made in a question of probability is never irrelevant.
>with a guaranteed gold draw even in the 1000 G box vs 1G, 999 S box it is just as likely to draw it from either. This is proof that you have a severe lack of understanding of probability.
>>720732082 For the question you posted yes but not the question OP posted. You're answering the wrong question. We didn't start at that first draw to have the odds of the double silver box come into play like in your example. That extra sentence is what makes the difference and has reduced the odds from 2/3 to just being 1/2. Yes this is a classic logic equation but if the actual question in the picture from OP is just that you're stuck with 2 boxes that have gold and a gold and you need to know if yours also has silver. In your image it is what is the probability that I will pick a box with two coins of the same color or one with gold (which would be 2/3). They aren't the same question.
>>720731282 >there are 2 ways to get a gold ball as your first ball from the all gold box and only 1 way to get a gold ball as the first ball from teh gold/silver box.
But this is false as long as the first draw is guaranteed to be golden. There are infinite ways (or merely one) to pull a golden ball from either box if the first draw is 100% of the time gold. The paradox only applies when both your draws are truly random.
Thus, the problem is worded wrong. It should be "You pull a ball, then a second ball from the same box, what's the chance it's the same color as the first?".
If it isn't simultaneously valid for Silver and Gold then there the premise is invalid, there must be symmetry
>>720732305 The issue is that the choice is already made. Not that the probability is off. You're factoring in all of the steps probability. At this part of the equation it is now binary. It is not Bertrands box anymore.
>>720733152 >But this is false as long as the first draw is guaranteed to be golden There is no guarantee. The question would function precisely the same if the color of the ball was not revealed. The question is simply saying, "For example, if you chose a ball and it was X color, what's the probability that the other ball in the box is the same color?" Whether you chose silver or gold, that answer will always be 2/3: >A: GG >B: GS >C: SS >two out of three boxes has two balls of the same color >one out of three has balls of different colors >if you draw from a box, there is a 2/3 chance that the box has two of the same colored balls It really is that simple, I have no idea why you're so stuck on this. After all this explaining.
>>720733152 >there must be symmetry The only way there would be symmetry is the following: >GG >GS >SG >SS But there is no symmetry. THere are tow boxes with same colored balls, and one with different colored balls. Which is all you need to know to figure out the probability.
>>720733436 >The issue is that the choice is already made. That is as irrelevant. As is your choice to use common sense to solve a question of probability over logic or math. >At this part of the equation it is now binary. Are you this dense on purpose? I don't usually resort to ad hominems, but you're just stubbornly refusing to understand what people here are trying to explain.
>>720733076 >There is no fate. There is indeed. The problem precludes you pull out a gold ball first 100% of the time as its main premise. Your interpretation is nothing alike because if its "1st any ball > % of 2nd being of the same color" the all silver box becomes relevant again, which is what makes the paradox work as you describe.
As long as the first draw is any color guaranteed the choice becomes binary because one of the options becomes irrelevant. Only in true randomness does this paradox work.
So yeah, you are right in how does the problem work. But the problem shown in the OP has little to do with the one you talk about.
>>720733697 you forget that the SS box can NEVER be chosen on the first pick in accordance with the initial question. Yes it does matter that the color of the ball was revealed, because you can already cross off SS from being the box chosen. Out of the remaining two, GG and GS, there is a 50/50 chance of G or S being the second ball.
>>720734059 No it is not. It is precisely the same. The probability works the same. There is only one choice in Bertrand's box paradox, and there is only one choice in OP's question. The entirety of probability hangs on that one choice, and whether you say "I choose a random ball," or, "I choose a random ball from a random box," or, "I chose (past tense) a random ball," it amounts to the exact same thing. Whether the choice is already made or not, the probability remains exactly the same. It is still unknown what the second ball is, and whether you want to know the probability of the color of the second ball, or the probability of which box you chose, it is precisely the same.
you just cant read. this is not bertrand's box question. bertrand's box has 2 important points. first: there is no golden coin picked out of the box. (this is the troll of the pick - and all highschool learning nonintelligent faggas can fuck off) 2. if we would try to solve bertrand's quest: "you have to pick the same box again". so the "ss" box is out of game, only 2 options open: GG / GS. you choose one box at random. the chance to get one of these two boxes is 50%. GG is 100%, GS is 50% another 50% for the "SAME" box equal 2/3. still the riddle is not bertrand's box. go check your 2/3 iq fagot
The problem being generally discussed is 2/3s. The problem in the OP is 1/2. The OP picture is only concerned when you have picked a box that already has gold. The case where you can grab a box without it is ignored.
>>720734450 >It's not guaranteed. It simply happened by chance. The question makes this clear. Not so clear in my opinion but now thatI've re-read it I see it does not state your first draw HAS to be golden. Still, my point is that its 66% only when there is true randomness. As long as the first pick is precluded to be either color the choice becomes binary and it's not Bertrand's box any more.
>>720733697 You don't understand, what I'm saying is that if there's a 66% chance for the all-gold box, the effect must be mirrored for the silver box.
>>720734631 >you forget that the SS box can NEVER be chosen >read OP >OP says RANDOM twice How fucking dense can you get? I COULD have been chosen, it just wasn't. And it's irrelevant anyway. Even if you ruled out the SS box, here are all of your possibilities of draws, with named balls for clarity: >Box A: (G1) (G2) >Box B: (G3) (S1) >if chosen at random, which ball would be in the box with it? >(G1) -> (G2) >(G2) -> (G1) >(G3) -> (S1) That's a fucking 2/3 chance that, if you chose a gold ball at random, you would have another gold ball in the same box. if you don't understand this i'm gonna just assume you're pants-on-head retarded and give up.
>>720735150 Oh i get it, because the question is worded to say "randomly selected" yet still "you picked gold", you get to dispose of half the probability of selecting box 2. This is why mathematicians are retarded, zero useful application in reality.
>The OP picture is only concerned when you have picked a box that already has gold
Yes, that's Bertrand's Box Paradox. You pick a random box and then pick a random ball from that box. it's gold. It could be the all gold box or the gold/silver box. 2 choices, but 1 is twice as likely as the other.
>>720735672 >what I'm saying is that if there's a 66% chance for the all-gold box, the effect must be mirrored for the silver box. It is. If your first ball chosen is silver, there is a 2/3 chance that you c hose the SS box. Symmetry.
depends which box youre picking from, dont need to consider randomized event of picking from either box 1 or box 2 because it is given that you are picking from the same box the second time, you are picking from that one box, there is no random chance. So probability is 1 if the box you initially picked from is box 1 and probability is zero if the first box you picked from is box 2, and box 3 was not the first box you picked from. There are two distinct probabilities for the two distinct scenarios you simps.
>>720736155 This math is still assuming the question is about the total probability from the beginning, meaning you have that first choice. The question in OP is about your decision after the first ball has already been picked and is unquestionably gold. Two options, 50%.
>>720737540 It's not. the thing I missed that is crucial in understanding it isn't 50% is that you're picking randomly out of the boxes. So when you pick the one gold, one silver, you have a 50% chance of picking out the silver ball or the gold ball. There are three options where you can pick a gold ball. in two of those options there is another gold ball in the box. So 66%.
>>720737540 >The question in OP is about your decision THERE IS NO DECISION AFTER THE FIRST. You are retarded if you don't understand this. The question is determining the probability of which box you have chosen. Is there another gold, or a silver ball? There is NO SECOND DECISION.
Read this guy's response carefully, because it's precisely what you are saying: >>720737630
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