Is my reasoning for this probability question correct? Consider a game of three. You shuffle a deck of three cards: 1,2,3. You draw cards without replacement until total is 3 or more. You win if your total is 3. What is P[W], the probability that you'll win? 
I drew a tree diagram. Below is the number of the card along with its probability: 

(1/3)(1/2) + (1/3)(1/2) + 1/3 = 2/3. Is this right? 

Morning after.  Had a chance to go to photoshop.
This should be the tree.  Originally you overlooked that the last branch ($3$ is the first card) was a win.

Your new image is better but now it has the branch extending so that you are drawing after you've already won.  Obviously once you win, you stop playing.
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Can't draw but:
Branch 1a: draw 1 (prob $\frac 13$), go on; draw 2 (prob $\frac{1}{2}$) WIN
Branch 1b: draw 1 (prob $\frac 13$), go on; draw 3 (prob $\frac{1}{2}$) lose
Branch 2a: draw 2 (prob $\frac 13$), go on; draw 1 (prob $\frac{1}{2}$) WIN
Branch 2b: draw 2 (prob $\frac 13$), go on; draw 3 (prob $\frac{1}{2}$) lose
Branch 3: draw 3 (prob $\frac 13$) WIN
So you win if Branch 1a or 2a or 3.  So $\frac 16 +\frac 16 +\frac 13=\frac 23$
