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Jun 22, 2014 Old
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Jun 22, 2014 Old
Originally Posted by Arglax View Post
I know a better riddle. One that isn't retarded:

Imagine three identical boxes. Each box has two sections. In one box, each section has a gold coin. In another box, each section has a silver coin. In a third box, one section contains a gold coin, the other contains a silver coin.

A blind man picks a box, and retrieves one coin from one section. It is a gold coin.

What are the odds that the other coin in the chosen box is gold as well?


1/2?

edit: seems like solax already answered. It can't be this easy though
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Jun 22, 2014 Old
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Jun 22, 2014 Old
Originally Posted by Solax View Post
50%?
Here's one:
What is placed on a table and is cut but never eaten?

A deck of Fish´s cards?
h
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Jun 22, 2014 Old
Iron oh iron what you can do but when you are heated your stronger than glue. If it is done right you can get ten times stronger and if you are dense enough then you can be ten times longer. What am i?

Made by Me.
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Jun 22, 2014 Old
Originally Posted by Risk View Post
It's 1/3. Basically an inversion of the Monty Hall problem it seems.



I'd say it's 1/3 as well, tho I wouldn't associate it with Monty Hall problem anyhow. More like a simple case of conditional probability or however it's called in english. I may be wrong tho.
Also, monthy hall paradox is easier to explain due to this 100-gates-example.

how it should be solved in my opinion- YEAH I KNOW THAT I SAID IT'S 1/3 AS WELL BUT YEAH



If I am wrong, please correct me, I'd love to know how to solve it properly using probability (I am more into calculus and algebra on my uni, I didn't solve any probability tasks for over an year :/).
Last edited by Thrandir; Jun 22, 2014 at 12:45 PM.
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Jun 22, 2014 Old
Originally Posted by Risk View Post
It's 1/3. Basically an inversion of the Monty Hall problem it seems.

Originally Posted by Thrandir View Post
I'd say it's 1/3 as well, tho I wouldn't associate it with Monty Hall problem anyhow. More like a simple case of conditional probability or however it's called in english. I may be wrong tho.

Basically, we have 1/3 chance to get (gold | gold) box and a 2/3 chance to pick box with one gold coin. Getting a (gold | gold) when our 2/3 condition is already fulfilled is (1/3) / (2/3) = 1/2. If I am wrong, please correct me, I'd love to know how to solve it properly using probability (I am more into calculus and algebra on my uni, I didn't solve any probability tasks for over an year :/).

It's not 1/3 or 1/2. It has nothing to do with the Monty Hall problem, that's a completely different story. Your theory is also completely wrong

*edit: I re-read my post, and I think I sounded a bit harsh. Don't take it like that, I really applaud everyone's attempts! Keep it up! I never thought anyone would bother to respond.
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