Conditional Probability Problem (3 Cards, 1 red-red, 1 red-black, 1 black-black) Our class got into a heated debate over a question that even my professor couldn't answer with certainty. So here's the problem:
"Of three cards, one is painted red on both sides; one is painted black on both sides; and one is painted red on one side and black on the other. A card is randomly chosen and placed on a table. If the side facing up is red, what is the probability that the other side is also red?"
Our class is split into the "1/2" and the "2/3" camp. I lean more towards 1/2. My argument is that when you uncover the red card, the black-black card is no longer considered in the sample space. So there are only 2 cards remaining, and since only one of those cards will have red on its other side, the final probability is 1/2.
My professor leans more towards 2/3. He didn't have time to formulate his argument fully, however. I've been trying to think about an alternative explanation which would lead to 2/3 as an answer, but it always leads back to the same argument I used for 1/2. Which one would be correct, and why?


My argument is that when you uncover the red card, the black-black card is no longer considered in the sample space. 

It eliminates more than that.
The card is selected (1 from 3), and a side from that card is shown (1 from 2), without bias.   So this experiment may be represented using a sample space consisting of six unbiased outcomes.   Of these, witnessing a red side eliminates not just the two outcomes for the double black card, but also the outcome of "the two-colour card shows black".
$$\{R_1(R_2), R_2(R_1), R_3(B_3),\require{cancel} \cancel{B_3(R_3), B_2(B_1), B_1(B_2)}\}$$
So there are three red sides you could be looking at with equal probability.  Two of them have another red side on the flipside.   Thus giving the answer.
$$\dfrac 23$$
