Your friend is right, because the expectation around $Y_2$ is computed before any information is known about what happended on the first draw.
To intuitively see this, consider the following thought experiment.
You have $6$ boxes, labeled 1-A, 1-B, 1-C, 2-A, 2-B, 2-C. You are blindfolded. You then make your first drawing from each of the $3$ boxes. However, instead of looking at the balls, you place the balls in boxes 1-A, 1-B, and 1-C, without ever having seen them. You now close the boxes, so the balls still haven't been viewed.
Continuing to stay blindfolded, you then make your second drawing from each of the $3$ boxes. Again, instead of looking at the balls, you place the balls in boxes 2-A, 2-B, and 2-C, without ever having seen them. You now close the boxes, so the balls still haven't been viewed.
Now, there is exactly $1$ ball left in box A, and you still don't know what color it is. Continuing to stay blindfolded, you draw the last remaining ball from Box A, place in a box labeled 3-A, and then close box 3-A.
Now, the first question is, which of the boxes 1-A, 2-A, or 3-A is more likely to have the lone red ball that was originally in box A.
Answer:
Since you have no information about the color of any of the balls in any of the boxes, there is no reason that the lone ball would be more likely than not to be in any of boxes 1-A, 2-A, and 3-A.
Now, suppose that instead of drawing the last remaining ball from box A, you had left it in box A. Would that have altered the color of each of the balls that is in box 1-A and 2-A.
Answer:
No. Therefore, the probability of a red ball in box 1-A is $(1/3)$ and the probability of a red ball in box 2-A is also $(1/3)$.
Now, consider boxes 1-B and 2-B. Here, besides those 2 balls, the other balls that were in box B, were left there. If the other balls that were in box B had each been drawn and placed in its own box, while you were still blindfolded, the color of the balls in each of boxes 1-B and 2-B would have remained unchanged.
Therefore, the fact that the remaining balls from box B were left there does not alter that the probability that the red ball from box B is in Box 1-B must still exactly equal the probability that the red ball from box B is in Box 2-B.
Analysis of the probabilities re the color of the balls in boxes 1-C and 2-C parallels the analysis of the probabilities re the color of the balls in boxes 1-B and 2-B.
Best Answer
There will be an urn in which the number $r$ of red balls will satisfy $r\in\left\{ 0,1,2\right\} $.
To be found is an answer to the question: "for which of the options $r=0,1,2$ will $P(S)$ be maximal?"
Let $E$ denote the event that this urn will be chosen at random and let $S$ denote the event that two balls are chosen that have the same color.
If $r=0$ then $S=E$ hence $P\left(S\right)=P\left(E\right)=\frac{1}{2}$.
If $r\in\left\{ 1,2\right\} $ then $$P\left(S\right)=P\left(E\right)P\left(S\mid E\right)+P\left(E^{\complement}\right)P\left(S\mid E^{\complement}\right)=\frac{1}{2}\left[P\left(S\mid E\right)+P\left(S\mid E^{\complement}\right)\right]$$
Here $$P\left(S\mid E\right)=\frac{r}{5}\frac{5-r}{5}+\frac{5-r}{5}\frac{4-r}{4}$$ and $$P\left(S\mid E^{\complement}\right)=\frac{5-r}{5}\frac{r}{5}+\frac{r}{5}\frac{r-1}{4}$$ (do you see why?).
Now substitute $r=1,2$ and draw conclusions.