A d20, a d10 and a comparison

Question

You roll a 20-sided die and a 10-sided die. What's the probability that the 10- sided die shows a greater value?

The probability of the d20 showing 1-10 is $\frac12$. In this case, one has two d10s effectively. For one die to be greater than the other: $$P=\frac{1-\frac1{10}}2=\frac9{20}$$ So the result is $\frac9{20}\cdot\frac12=22.5\%$?

Answer 1

Yes, $22.5\%$ is right. Here's another approach using sums:

The probability that the $10$ sided die beats the $20$ sided die depends on the combinations of outcomes. Let $a$ be the outcome of the $10$-sided die, and $b$ be the outcome of the $20$-sided die. Then:

$$P = \dfrac{\sum_{b=1}^{9} \sum_{a=b+1}^{10} (1)}{20 \cdot 10} = \dfrac{\sum_{b=1}^{9}(10-b)}{200} =\dfrac{45}{200} = \dfrac{9}{40} = .225$$

Answer 2

The number shown on the 10 die is greater than the number shown on 20 die if the point $(i,j)$ lies above the line shown in the picture. There are $45$ such pairs and hence the probability is $\frac{45}{200} = \frac{9}{40}$