Is there a name for this particular bias estimating probability? I am struggling with this problem from An Introduction to Probability and Statistics by Sheldon Ross.
Chapter 3 Q23
Of three cards, one is painted red on both sides (RR), one is painted black on both sides (BB) and one is painted red on one side and black on the other (RB). A card is randomly chosen and placed on the table. If the side facing up is red, what is the probability that the other side is also red?
My first thinking was that the odds are 50:50 since we eliminated the black card and are now picking from the two remaining cards.
However the answer is 2/3
It has been explained to me that the card is more likely to be RR than RB because out of sides RR,RB R is twice as likely to be picked from RR than RB.
When I imagine drawing cards from a jar containing an equal number of RR and RB cards it does seem likely that I would have a larger pile of red cards.  Yet I can easily imagine myself falling back to my original thinking.
Is their a name for my original bias? and a technique to avoid it?
 A: The technique boils down to just thinking in terms of conditional probabilities! Although each card has a random probability of being chosen, they don't all have the same probability of being chosen and being red.
The key insight in this problem is to note that conditional on seeing one side is red, all three red sides from the 2 red cards have an equal probability of being shown, but only two of the sides have a red opposite side, hence the answer of 2/3.
We can see this using the definition of conditional probabilities. Note that
$$P(\text{down red} | \text{up red}) = \frac{P(\text{up red},\text{down red})}{P(\text{up red})} = \frac{(1/3)}{(1/2)} = \frac{2}{3},$$
where the numerator follows because only one of the 3 cards is RR, and the denominator follows from the fact that unconditional on other information, that a side is red is really picking 'randomly' from the 6 sides of the three cards and not from the three cards. Of these 6 sides, 3 are red.
I'd argue it's natural to fall for these types of questions... in many ways, this question is related to the Monty Hall problem, and the wikipedia link explains a bit about sources of confusion for these kind of problems if you are interested. But really they reduce to whether or not you approached the question carefully and to always think about what information you've learned after being first told that something was done 'randomly'. In this case, even though the first selection may have been random, you were then told more about the selection.
