Is this a fair game (three players throw dice)? I found an interview question online summarized as follows:

Three players A, B, and C are playing a game. A throws three fair 6-sided dice (each one has face values from 1 to 6), while B and C each throws a fair 20-sided die (each one has face values from 1 to 20). Whoever gets a bigger number (or sum of numbers for A) wins the game. Is this a fair game? You can only calculate by hand and have only 10 minutes.

The method I thought of is calculate the probabilities of A wins and B wins respectively, and check whether they are equal. The probability of A wins can be calculated as follows:
$$
P(A>B, A>C)=\sum_{a}P(A>B,A>C|A=a)P(A=a)\\
=\sum_a P(A>B|A=a)P(A>C|A=a)P(A=a)
=\sum_a (\frac{a-1}{20})^2P(A=a)\\
=\mathop{\mathbb{E}}[(\frac{A-1}{20})^2]=\frac{1}{400}\mathop{\mathbb{E}}[A^2-2A+1]
=\frac{1}{400}(\mathop{\mathbb{E}}[A^2]-2\mathop{\mathbb{E}}[A]+1).
$$
If we denote $X_1,X_2,X_3$ as the three 6-sided dice of A, then we have
$$
\mathop{\mathbb{E}}[A]=\mathop{\mathbb{E}}[X_1+X_2+X_3]=3\cdot \frac{1+6}{2}=\frac{21}{2},
$$
and
$$
\mathop{\mathbb{E}}[A^2]=\mathop{\mathbb{E}}[(X_1+X_2+X_3)^2]=\mathop{\mathbb{E}}[X_1^2+X_2^2+X_3^2+2X_1X_2+2X_1X_3+2X_2X_3]\\
=3\mathop{\mathbb{E}}[X_1^2]+6\mathop{\mathbb{E}}[X_1X_2]=3\mathop{\mathbb{E}}[X_1^2]+6\mathop{\mathbb{E}}[X_1]\mathop{\mathbb{E}}[X_2]\\
=3 \cdot\frac{1}{6}\cdot \frac{6\cdot (6+1)\cdot (6*2+1)}{6}+6\cdot \frac{1+6}{2} \cdot \frac{1+6}{2}=119.
$$
Thus $$P(A>B,A>C)=\frac{1}{400}\cdot (119-2\cdot \frac{21}{2}+1)=\frac{99}{400}.$$
However, here comes the problem when I calculate $P(B>A,B>C)$ as follows:
$$
P(B>A, B>C)=\sum_{b}P(B>A,B>C|B=b)P(B=b)\\
=\sum_b P(B>A|B=b)P(B>C|B=b)P(B=b)
=\sum_b P(B>A|B=b)\cdot \frac{b-1}{20}\cdot P(B=b).
$$
I can't find an easy way to calculate $P(B>A|B=b)$ because the probability distribution of $A$ isn't as simple as $B$ or $C$. In fact, for $a$ from $3$ to $18$, $P(A=a)$ is $\frac{1}{216}\{1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1\}$. I can't find a general formula for these values or partial sum of these values. And if you calculate the probability with these values directly, it is very time-consuming and I don't think it is feasible when you are being interviewed. Thus I think there may be some more simple method to solve this problem (maybe by symmetry or whatever). For your convenience, with the help of my computer, I get that $P(B>A,B>C)=\frac{543}{1600}=\frac{135.75}{400}>P(A>B,A>C)$. Anyone has any ideas?

I can see how you can actually proceed with your own solution (or at least the same idea!) until you reach the conclusion. Still not sure it can be done within 10 min on an interview, though.
The probability for $A$ to win:
$$P(A\text{ wins})=P(A>B>C)+P(A>B=C)+P(A>C>B)$$
The probability for $B$ to win:
$$P(B\text{ wins})=P(B>A>C)+P(B>A=C)+P(B>C>A)$$
Now, because of overall symmetry:
$$P(A>B>C)=P(B>C>A)$$
$$P(A>B=C)=\frac{1}{20}P(A>B)=\frac{1}{20}P(B>A)=P(B>A=C)$$
so we only need to compare the remaining terms: $P(A>C>B)$ and $P(B>A>C)$. You can calculate those in the same way as you have already calculated $P(A>B\land A>C)$:
$$P(A>C>B)=\sum_{a}P(a>C>B)P(A=a)=\frac{1}{400}\sum_{a}\frac{(a-1)(a-2)}{2}P(A=a)=\frac{1}{400}\left[\frac{1}{2}E(A^2)-\frac{3}{2}E(A)+1\right]=\frac{1}{400}\left(\frac{119}{2}-\frac{3}{2}\frac{21}{2}+1\right)=\frac{179}{1600}$$
$$P(B>A>C)=\sum_{a}P(B>a>C)P(A=a)=\frac{1}{400}\sum_{a}(20-a)(a-1)P(A=a)=\frac{1}{400}(-E(A^2)+21E(A)-20)=\frac{1}{400}(-119+21\frac{21}{2}-20)=\frac{163}{800}=\frac{326}{1600}$$
and obviously $P(B>A>C)>P(A>C>B)$, thus $P(B\text{ wins})>P(A\text{wins})$.
Addendum/Check step: We can now additionally calculate $P(A=B)=\frac{1}{20}$ and so $P(A>B)=\frac{1}{2}P(A\ne B)=\frac{19}{40}$ so $P(A>B=C)=\frac{1}{20}P(A>B)=\frac{19}{800}=\frac{38}{1600}$. Thus, $P(A\text{ wins})=\frac{179}{1600}+\frac{38}{1600}+\frac{179}{1600}=\frac{396}{1600}=\frac{99}{400}$, which matches your result. Similarly, $P(B\text{ wins})=\frac{326}{1600}+\frac{38}{1600}+\frac{179}{1600}=\frac{543}{1600}$, which again matches your result.
