For game where first player to N points wins, find the distribution of win probability and total number of points between players 
Two players, A and B, play a series of points in a game with player A winning each point with probability p and player B winning each point with probability
  q = 1 - p. The first player to win N points wins the game. Assume that
  N = 3. Let X be a random variable that has the value 1 if player A wins the
  series and 0 otherwise. Let Y be a random variable with value the number
  of points played in a game. Find the distribution of X and Y when p = 1/2.
  Are X and Y independent in this case? Answer the same questions for the
  case p = 2/3.

For the case of p = 1/2:
$P(Y=3)=(\frac{1}{2})^3+(\frac{1}{2})^3=0.25$
$P(Y=4)=3[(\frac{1}{2})^3(\frac{1}{2})+(\frac{1}{2})^3(\frac{1}{2})]=0.375$
$P(Y=5)=6[(\frac{1}{2})^3(\frac{1}{2})^2+(\frac{1}{2})^3(\frac{1}{2})^2]=0.375$
$P(X=1)=P(X=0)=0.5$
For the case of p = 2/3:
$P(Y=3)=(\frac{2}{3})^3+(\frac{1}{3})^3=0.333$
$P(Y=4)=3[(\frac{2}{3})^3(\frac{1}{3})+(\frac{1}{3})^3(\frac{2}{3})]=0.370$
$P(Y=5)=6[(\frac{2}{3})^3(\frac{1}{3})^2+(\frac{1}{3})^3(\frac{2}{3})^2]=0.296$
$P(X=1)=(\frac{2}{3})^3+3(\frac{2}{3})^3(\frac{1}{3})+6(\frac{2}{3})^3(\frac{1}{3})^2=0.79$
$P(X=0)=1-P(X=1)=0.21$
Is this correct?
I think for p = 1/2, X and Y are independent and for p = 2/3 X and Y are not?
I can't find anything wrong with it but I'm having trouble working through the logic of why it makes sense that $P(Y=3,4,5) = 1$.

Your calculation looks correct. As for why $\mathbb{P}(Y\in\{3,4,5\})=1,$ think in the following way.
$Y$ is the number of points at the end of the game.
Here $N=3,$ so  $Y$ to be at least $3$, since a player can win at most $k$ points in $k$ trials. Now suppose we have $5$ points at the end, but neither player $A$ nor player $B$ has $3$ points. Is this possible? No. That means $Y$ can be at most $5.$
