Cards and probability There are four cards in a hat. Three cards have 0 on one side and 1 on the other, while the fourth card has a 0 on both sides. If I observe that one side of a card I have chosen at random has a 0 on it, what is the probability of the card I have chosen being the one with 0 on both sides? 
I think the answer's $\frac{2}{5}$ since we're accounting for sides only. We have five sides with a 0, and the two sided card has two of those 0s. Am I making a mistake?
 A: Your answer is correct. That is the more intuitive approach. If you want to brute force compute it, you can say that you are trying to solve P(draw 00 | see a 0). That is, the probability you draw the 0/0 card given that the side you draw shows a 0.
We can solve this breaking down P(draw 00) into all its conditions:
$
\text{P(draw 00)} = \text{P(draw 00 | see a 0)} \cdot \text{P(see a 0)} + \text{P(draw 00 | see a 1)} \cdot \text{P(see a 1)}
$
NOTE that P(draw 00 | see a 1) is 0 so the second term drops out (if you see a 1 you can't have the double zero card).
So we are left with:
$
\text{P(draw 00 | see a 0)} = \frac{\text{P(draw 00)}}{\text{P(see a 0)}} = \frac{\frac{1}{4}}{\frac{5}{8}} = \frac{2}{5}
$
The $\frac{1}{4}$ comes from 1 of 4 cards being a double zero, and the $\frac{5}{8}$ comes from 5 of 8 sides having a 0.
You realized the shortcut which is to say that there are only 5 zeroes and 2 of them belong to the double zero card.
A: Always when faced with such problems you should repeat the mantra:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
In words: The probability of $A$, given that $B$ happened, is the probability that both $A$ and $B$ happen, divided by the probability that $B$ happens.  
Here, $A$ is the event that you chose the double-zero card, and $B$ is the event that you see a zero. So $P(A \cap B)$ is equal to 1/4, and $P(B)$ is equal to 5/8. So your answer is correct.
