Betting game: $3$ $20$-sided dice versus $2$ $30$-sided dice Alice and Bob play a dice game. Alice rolls three $20$-sided dice and
  Bob rolls two $30$-sided dice. Whoever has the highest sum wins.
  
  Who has a better chance of winning?


This was a question asked in an interview. So I don't have much time to answer and I want to find the answer quickly. I did it the long way (enumerate all possibilities), and I figured out that Alice has a better chance of winning. But is there any way I can solve this problem very quickly?

I tried to make a "flipping over the dice" argument but got nowhere.

Thanks
Let the outcomes of the dice rolls be $(A_1, A_2, A_3, B_1, B_2)$ where the $A_i$ are uniform on $\{1, 2, \ldots, 20\}$ and the $B_j$ are uniform on $\{1, 2, \ldots, 30\}$.  You want to know whether $\Delta = A_1 + A_2 + A_3 - B_1 - B_2$ is more likely to be positive (Alice wins) or negative (Bob wins).  Of course it can be zero.

Consider the cases where these take on the values $(a_1, a_2, a_3, b_1, b_2)$ and ("flipping over" the dice) $(21-a_1, 21-a_2, 21-a_3, 31-b_1, 31-b_2)$.  

In the first case we get $\Delta = a_1 + a_2 + a_3 - b_1 - b_2$.

In the second case we get $\Delta = (21 - a_1) + (21 - a_2) + (21 - a_3) + (31 - b_1) + (31 - b_2) = 1 - (a_1 + a_2 + a_3 - b_1 - b_2)$.  

So this gives a bijection between outcomes where Alice wins by $\Delta$ and outcomes where Alice wins by $1-\Delta$.     In particular 0 and 1 are equally likely; -1 and 2 are equally likely; and so on.   So all the positive values of $\Delta$ have probability adding up to 1/2 - that is, Alice wins with probability 1/2.  There is a tie with some nonzero probability, and Bob wins with probability less than 1/2.
Twenty-sided dice roll 10.5 on average. Thirty-sided dice roll 15.5 on average. So Alice is going to roll 31.5 on average, while Bob only rolls 31 on average.

These are only averages, but the symmetry of the problem ensures Alice wins out (on average).