You're missing the words before it.
"You pick box at random. You put your hand in and take a ball from that box at random." Including the word "if" certainly helps for clarification, but it is not required. The randomness is explicitly stated in the question, it doesn't need to be implied.
Once again, if the first ball chosen from the box is not random, there is no way to solve this problem without further information. It is either 2/3 or it is impossible to solve, you can't get 50%.
Hitman, just save yourself the time. Humpy clearly likes to "stir the pot" just like the OP. All you're doing is encouraging him.
Stirring the pot has value, but only for so long. There is clearly enough information in this thread to get to the answer that people are searching for, including there being no answer or more than one answer, answers that don't make sense, and most importantly, the right answer .
Once again, if the first ball chosen from the box is not random, there is no way to solve this problem without further information.
Maybe it is easier to understand if you change the two gold balls to one gold and one bronze, and change the question to the chance of the second draw being gold or bronze after the first draw is either gold or bronze.
Then you have B+G and S+G, the chance of drawing S first from the second box is removed, so you are left with B first + G second as option, with G first + B second, and in the second box G first and S second.
So after you draw either a bronze or a gold ball from a chest, how high is the chance of drawing a gold or bronze one again?
This might make it easier to differentiate between the 2 balls in the 'gold box'.
The math demonstrates 2/3, but one can rationally argue 1/2 (yes, while taking all events into account). I started out arguing for 1/2, then accepted 2/3, but am presently agnostic.I need to know first if you're on the 1/2 side or the 2/3 side.
He's not stirring the pot, the posted question is.
You guys are using outside sources to answer a question that is worded differently then the one here. No where that I see does it say pick from 2 boxes. It only says 1 and you guys are ignoring this fact.
But the posted question is rigged as you don't know if you'rein a box with GG or GS.
If you do a reverse image search you'll see about 20 other forums went ape s$&@ over this problem including the body building forum from which the image is from.
If you feel the math demonstrates 2/3, then there is no way you should believe that one can rationally argue 1/2. That is part of the beauty of math, it is not refutable.The math demonstrates 2/3, but one can rationally argue 1/2 (yes, while taking all events into account). I started out arguing for 1/2, then accepted 2/3, but am presently agnostic.
You're missing the words before it.
"You pick box at random. You put your hand in and take a ball from that box at random." Including the word "if" certainly helps for clarification, but it is not required. The randomness is explicitly stated in the question, it doesn't need to be implied.
The math isnt wrong; its just math from a different experiment.If you feel the math demonstrates 2/3, then there is no way you should believe that one can rationally argue 1/2. That is part of the beauty of math, it is not refutable.
The math demonstrates 2/3, but one can rationally argue 1/2 (yes, while taking all events into account). I started out arguing for 1/2, then accepted 2/3, but am presently agnostic.
That's not what's being said at *all*.Bertrand: so, you have these boxes.
You: ok.
Bertrand: and one has a silver ball in it.
You: ok.
Bertrand: and you draw a ball.
You: ok i draw the silver ball.
Bertrand: no that cannot happen.
You: i am not allowed to draw the silver ball?
Bertrand: no, this experiment exists in a reality where you are not allowed to draw silver.
You: ok.
Bertrand: now calculate the possibilities.
You: this much.
Bertrand: Aha! You fool, you didnt consider the possibility of drawing silver!!
You: ...
The language is clear, it's just posing that conditional probability problem, which has a clearly defined solution. It's 2/3, end of story
Actually its sort of not, because it states you already pulled a gold ball from the box..
The odds of you pulling another gold ball from the same box would mean you got the first box with double gold balls to begin with.
So your doing P on the boxes itself and not the balls.
Ie... you have a 50% chance of selecting the box with the double gold balls, because there is a 0% chance of you getting a second gold ball from the other 2.
What further information would you need?
I'm pretty close to taking the L. I understand how the answer can be 2/3. I understand that the problem would be unsolvable if the silver ball was actually an unknown color.
you cant use bayes therom on this, because B = 0
The Second BOX has a 0 % chance of it being the solution.
So your looking at a straight probabilty of you selecting the First box when looking at all the boxes with gold.
The balls are whats confusing you, and tripping you out.
The answer to this question is only the box with GG..
Out of the 3, only 2 have G.
So your looking at out of the 2 which one has GG.
50% your got the box with GG, because the other box is 0% your going to get GG.
Here it was answered by Skorpio.
P(A) = P(both gold) = 1/3 -- one out of three boxes has both gold
you cant use bayes therom on this, because B = 0
The Second BOX has a 0 % chance of it being the solution.
So your looking at a straight probabilty of you selecting the First box when looking at all the boxes with gold.
The balls are whats confusing you, and tripping you out.
The answer to this question is only the box with GG..
Out of the 3, only 2 have G.
So your looking at out of the 2 which one has GG.
50% your got the box with GG, because the other box is 0% your going to get GG.
Here it was answered by Skorpio.
You do realize that Skorpio later in the thread stated that he was wrong in that answer and that it is 2/3, right? So you're using an answer the author already said was in error.
As far as solving the probability, you absolutely can use Baye's Theorem.
no because your not even looking at the third box.
You already pulled a gold ball from a box.
That means it was already determined you got either box 1 or 2 without probability.
So your looking at 2 boxes.
The odds of you getting the box with GG out of the 2 boxes is what?
50%?
lol... read the question carefully, your overthinking the problem.
A better title for this should be people with an IQ under 115 can answer this problem, because i honestly feel if your IQ is too high, you will overthink it.