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It's a gold coin and a silver coin at the same time, until we open the box.
Brendan (he who passeth judgement on the frequent changing of signatures): I don't do hentai anymore
Originally Posted by Scorpio View Post
something makes me think it's 2/3 since 2 out of 3 boxes has gold coins in them.

Yea, but if you had picked the box with gold-silver, and taken a gold coin, the other coin would have been silver, and we're looking for the odds that the other coin is gold too.

The correct answer, however, was indeed 2/3. Your logic was only flawed. Let me explain.

There are three boxes:
GG - gold/gold
GS - gold/silver
SS - silver/silver

Each box has an equal chance to be taken. 1/3. But that's not to the point: we can rule out the SS box, considering we have at least one gold coin.

The chance of the GS box and the SS box being taken is 1/2, right? Yes, BUT we have already opened the box and we see that it contains a gold coin. This is the clue of the riddle: the GS box has a lower chance of being the box we chose, BECAUSE THE GG BOX CONTAINS TWO GOLD COINS! The odds that the opened box WAS GG, are thus higher than the box being GS.

If the box was GS, we will not have a second gold coin.
If the box, however, was GG, we will have a second gold coin.

What are the odds? Well, let's look at just coins. There are six coins: 3S, 3G.

The GG box contains 2 of the three gold coins, the GS box contains only one of the three gold coins.

The riddle can be rephrased as "What are the odds that the first coin was one G coin from GG?". (because SS is impossible, and because GS would not have a second gold coin, and that's what we're looking for: a second gold coin. The only case that has a 'second gold coin', is the GG case.)

So there are three gold coins. GG amounts for 2 out of these three. We are looking for the GG box.

Table of possibilities
1/3: we chose G from GS
1/3: we chose G1 from GG
1/3: we chose G2 from GG

The sum of the last two is 2/3. Simple!



Bonus:
What are the odds that, by guessing randomly, you will answer this question correctly (completely unrelated to the riddle above)?
A: 25%
B: 50%
C: 75%
D: 25%
Last edited by Arglax; Jun 22, 2014 at 12:54 PM.
f=m*a syens
Well it is indeed conditional probability.

B- blind man picks a section with gold coin: P(B) 1/2 (3 sections out of 6). Boxes doesn't matter right now.
A- blind man picks a box with two gold coins: P(A) 1/3 (only one box out of 3 has two golden coins).

P(A joint B)- blind man picks a section with a golden coins IN A BOX WITH 2 GOLDEN COINS: It is still 1/3, as there is 100% probability that he picks section with golden coin when both coins are golden in this box (and 1/3 to get proper box).

Therefore, P(A|B) = P(A joint B)/P(B) = (1/3) / (1/2) = 2/3.

Also my logic was flawed first time, deleted it already as my P(A joint B) was dumb that time.
Originally Posted by Thrandir View Post
Well it is indeed conditional probability.

B- blind man picks a section with gold coin: P(B) 1/2 (3 sections out of 6). Boxes doesn't matter right now.
A- blind man picks a box with two gold coins: P(A) 1/3 (only one box out of 3 has two golden coins).

P(A joint B)- blind man picks a section with a golden coins IN A BOX WITH 2 GOLDEN COINS: It is still 1/3, as there is 100% probability that he picks section with golden coin when both coins are golden in this box (and 1/3 to get proper box).

Therefore, P(A|B) = P(A joint B)/P(B) = (1/3) / (1/2) = 2/3.

Also my logic was flawed first time, deleted it already as my P(A joint B) was dumb that time.

Yea, that's the mathematical way of explaining it. I know, however, that a lot of toribashians are too young to understand this/haven't seen probability in school yet, so I tried to explain it in a non-mathematical way.
f=m*a syens
Originally Posted by Arglax View Post
Yea, that's the mathematical way of explaining it. I know, however, that a lot of toribashians are too young to understand this/haven't seen probability in school yet, so I tried to explain it in a non-mathematical way.

Oh well, I just didn't read Your solution before posting mine, as I didn't want to lose this fun. Yeah Your solution is exactly the same as even in school conditional probability is explained this way (with probability tables). Thank You for this riddle, it was fun.

also AS JTANK LIKES NUMERICAL METHODS AND WANTED ME TO DO THIS: short console application simulating this case (yeah, I am C newb). Indeed, this probability is close to 2/3, even if rand() is shit.

http://pastebin.com/SeGzLWi8
Last edited by Thrandir; Jun 22, 2014 at 01:21 PM.
Lazors your referring to Schrodingers Cat am i right?
Originally Posted by Lazors View Post
It's a gold coin and a silver coin at the same time, until we open the box.

The Cat/Coin could be consider both Dead/Silver or Alive/Gold until we open the box.
I'd like to call this equation Schrodinger's Box due to the variables and probability and since the similarity is rather high between these 2 expirements/riddles.
Last edited by Merc; Jun 22, 2014 at 01:50 PM.
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Originally Posted by ZENBOY123 View Post
Lazors your referring to Schrodingers Cat am i right?

The Cat/Coin could be consider both Dead/Silver or Alive/Gold until we open the box.
I'd like to call this equation Schrodinger's Box due to the variables and probability and since the similarity is rather high between these 2 expirements/riddles.

Except we already know what is in the boxes. We don't know how the cat is doing, since we have no influence on how long it survives. There is no 'quantum state' of two coins: there are only two distinctly different coins, no coin that is both silver and gold at the same time.
f=m*a syens
Originally Posted by Arglax View Post
What are the odds that, by guessing randomly, you will answer this question correctly (completely unrelated to the riddle above)?
A: 25%
B: 50%
C: 75%
D: 25%

(there is probably a "more correct" answer to this but i am most likely too stupid to figure it out)

from what i understand, if i input an answer of 25% (options A and D) the correct answer would be 50%, because there are two options that say 25%, 2 out of 4 is obviously 50%.

but if i answered 50% (option B) the correct answer would be 25%, since there is only 1 out of four answers that allow you to select 50% as an answer.

same with option C, if I chose 75% then the correct answer would be 25% since there is only one "75%" option out of four.

there is no answer since the answer you input influences the correct answer, since this is a question that implies itself actually the answer is 0% but its not up there so there is no answer


EDIT: i read a post above about schrodingers cat, and i would like if someone convinced me that the theory isnt just plain stupid. granted i am not very intelligent, but just because we dont know what happened means that we have to take both options into account?? something did happen, and it was either the cat lived or the cat died, we just cant jump into any conclusions because we didnt see what happened. the logic should stop there due to the lack of information, right?
Last edited by pusga; Jun 22, 2014 at 02:39 PM.
oh yeah
Here is how the expirement worked:The cat was in a box with a ball of poison(Made of metal i think). It has motion or touch sensors so that when he touched it it would open and release the poison and kill him. Until they open the box they have no idea if the poison opened yet or not. Meaning he can be thaught of as both dead and alive Also this theory fits ur avatar XD
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Also will someone please attempt to solve my riddle on one of the other pages?
Last edited by Merc; Jun 22, 2014 at 02:49 PM. Reason: <24 hour edit/bump
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