Betting game: $3$ $20$-sided dice versus $2$ $30$-sided dice

Question

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

Answer 1

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).

Answer 2

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.

Answer 3

The flipping dice argument is straight forward. When flipping a dice you have that $n$ turns into $21-n$ (or $31-n$ depending on which type of die).

What you do is note that if Alice win for some outcome Bob will win in the equally probable(*) outcome when all dice are flipped. This is because if Alice rolled $a$ and Bob $b$ in total and $a>b$ the result with dice flipped is Alice rolling $21-a$ and Bob $31-b$ and because $a>b$ we have $21-a < 21-b < 31-b$.

The outcomes in the preceding paragraph is as many wins for Bob as it's far Alice. For outcomes that are not covered by the preceding paragraph there is no win for Alice. If we just can find a single win for Bob amongst these there is more outcomes where Bob wins and he would be more likely to win than Alice. But there is, consider for example Alice rolling $21=11+10$ and Bob rolling $47=16+15+16$ which Bob wins and he wins even if the dice are flipped (Alice rolling $21=10+11$ and Bob $46=15+16+15$) so the case is not covered by the preceding paragraph.

(*) All outcomes will be equally probable if we consider each die uniquely identified (ie distingishing Alice rolling $10+11$ from Alice rolling $11+10$).