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?
 A: You have randomly selected one of the three red sides to be showing.  Two of the three of them have a red side on the bottom, so the chance is $\frac 23$.  Yes, the black-black card is out of the picture, but the red-red card is twice as likely to be the card because you also eliminate half the times you choose the red-black card-the times the black face is up.
A: Imagine that, beforehand, you took the card that is red on both sides, and used invisible ink to number the sides $1$ and $2$. 

The card that is on the table is showing a red face. 
This red face can be one of three faces, each equally likely: 


*

*it could be the side with an invisible $1$ on it

*it could be the side with an invisible $2$ on it

*it could be the side with no invisible ink on it (the red side of the card with one side of each color)

Of these three equally likely possibilities, two of them imply that the facedown side of the card is red, so we get $$\frac{2}{3}$$
A: 
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.
$$\dfrac 23$$
