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