There is a bag containing four coins, the first coin is red on both sides, the second is blue on both sides, the third coin has a red side and a green side, and the fourth coin has a red side and blue side. A randomly drawn coin from the bag and is placed on a table in front of you, You see a red side of the coin. What are the odds the other side of the coin is red?
My intuition is that this is a conditional probability. We are looking for the probability of a double red given that we see a red. The probability of seeing a red face is $\frac12$ and the probability of picking a double red is $\frac14$.
$\frac{\frac14}{\frac12} = \frac12$
Is this correct? If so, what is the intuition as to why the answer would not be $\frac13$, as there are $3$ coins with a red face where $1$ is double sided.
Alternative intuition:
You conduct an experiment $8$ times, so that each of the two sides of each of the $4$ coins is face-up one of the $8$ times.
Then, because the face-up side is red, you can presume that of the $8$ equally likely faces that might have appeared, the $4$ faces that are not red, did not appear.
So, you have $4$ equally likely faces, each of which is red, that represent that the face that you see is red. Of these $4$ faces, exactly $2$ of the $4$ faces have the color red on the opposite face.