Probability Interview Question - Brain teaser 
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?

So I have
$$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$
There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B
The answer for this question should be $0.35$, I am not sure where I did wrong. 
I just figured maybe I should do this instead 
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?

I've reasoned like this, as usual starting from the simplest case: 
A has a 3-sided dice, B a 2-sided one.
$P(\mathbf{B} wins)$ = $P(\mathbf{B}_{dice}=1)$ * $P(\mathbf{A}_{dice}=1)$ + 
              $P(\mathbf{B}_{dice}=2)$ * $P(\mathbf{A}_{dice}<=2)$ = 
= $\frac{1}{2}$ * $\frac{1}{3}$ + $\frac{1}{2}$ * $\frac{2}{3}$ = $\frac{1}{2}$
Generalizing ($n$ being the number of sides for B): 
$P(\mathbf{B} wins)$ = $\frac{1}{n} * \sum_{j=1}^{n}(\frac{j}{\frac{3}{2}*n})$
 = $\frac{2}{3n^2} * \frac{n*(n+1)}{2}$ = $\frac{1}{3n} * {(n+1)}$ = $\frac{1}{3} + \frac{1}{3n}$
So, with $n = 20$, $P(\mathbf{B} wins)$ = $\frac{1}{3} + \frac{1}{60}$ = $\frac{21}{60}$
