Whats the theoretical probability of winning when multiplying 2 dices vs summing 3 dices?

Question

So basically, I roll two $6$-sided dice and multiply the values. Then the other person would roll three $6$-sided dice and add up all of their values. What is the theoretical probability of me getting a higher value?

Answer 1

This seem to just be a case of counting outcomes, as mentioned in the comments this could be done with a simple scripts as there are only $6^5 = 7776$ possibilities. Such a script would reveal that there are $3635$ possibilities of winnig ($312$ draws and $3829$ losses). Since the outcomes are equally probable the probability of winning would be $3635/7776\approx 47\%$

It's somewhat doable to do the calculations by hand too. In order to do that one first start to count the outcomes for you and the corresponding scores and the outcomes for your opponent and the corresponding scores and from that count the number of combos where you win.

For your scores you take the multiplication up to $6$ and count the occurrence of distinct products. Then you get

$$\begin{align} P_1 &= 1 & P_8 &= 2 & P_{18} &= 2 & \\ P_2 &= 2 & P_9 &= 1 & P_{20} &= 2& \\ P_3 &= 2 & P_{10} &= 2 & P_{24} &= 2 & \\ P_4 &= 3 & P_{12} &= 4 & P_{25} &= 1 & \\ P_5 &= 2 & P_{15} &= 2 & P_{30} &= 2 & \\ P_6 &= 4 & P_{16} &= 1 & P_{36} &= 1 \\ \end{align}$$

(for example you have four ways to get $12$, by $2\times 6$, $3\times 4$, $4\times 3$ and $6\times 2$)

For counting the possibilities to get certain sums you have to do little more thinking. Up to the sum of $8$ it's triangle numbers as the ways to get those are the sum of ways to get the (six nearest) numbers below with only two dice which increases linearily until we get to $7$. After that we have to take into account that the number of ways to get a sum with two dice decreases again and that we must count only the six nearest below. We also have that the distribution of counts are symmetric. We get

$$\begin{align} S_3 &= 1 & S_7 &= 15 & S_{11} &= 27 & S_{15} &= 10 & \\ S_4 &= 3 & S_8 &= 21 & S_{12} &= 25 & S_{16} &= 6 & \\ S_5 &= 6 & S_9 &= 25 & S_{13} &= 21 & S_{17} &= 3 & \\ S_6 &= 10 & S_{10} &= 27 & S_{14} &= 15 & S_{18} &= 1 \\ \end{align}$$

Then we just have to take each product and sum up all sum counts that mean that you would win:

$$N_w = 1\times 0 + 2 \times 0 + 2 \times 0 + 3 \times 1 + 4 \times 10 + 2 \times 35 + 1 \times 56 + 2 \times 81 + 4 \times 135 + 2 \times 196 + 1 \times 206 + 1 \times 215 + 2 \times 216 + 2 \times 216 + 1 \times 216 + 2 \times 216 + 1 \times 216 = 3635 $$