Is it better to bet on a d12 or on the sum of two d6? The game is simple we throw a $d12$ and two $d6s$. We sum the value of the outcomes of the two $d6$ and compare it with the outcome of the $d12$.
If the sum of the two $d6$ is higher, $d6$ wins, if the $d12$ has a higher value, $d12$ wins. I have run a simulation and it is pretty clear that one should bet on the $d6s$. However, I it is also pretty clear that this should not be because of their higher expected value. (Expected value $d6s = 7$ and expected value $d12 = 6,5$). In fact, one can easily concoct scenarios with different dices in which even the expected value of a dice is higher, but the probability of winning is lower.
Am I right to think that the correct way to calculate the probability of winning this game is to compare every possible outcome of the two $d6$; i.e., $36$ outcomes, with every possible outcome of the $d12$; i.e., $12$ outcomes, and simply count those in which $d6 > d12$ over the total?
In this case there are $36 \times 12$ total possible outcomes of $d12$ and two $d6$s throws. In these, $180$ are won by the $d12$. Thus, the probability of winning with the $d12$ is $5/12$. Is this correct?
 A: Suppose you have two discrete random variables:

*

*$X$ is uniformly distributed over the set $1, 2, \ldots, 12$

*$Y$ is symmetrically distributed over the set $2, 3, \ldots, 12$
Because of the symmetry of $Y$'s distribution, we have

*

*$P(Y > 1) = 1-P(Y > 12)$

*$P(Y > 2) = 1-P(Y > 11)$

*$\cdots$

*$P(Y > 6) = 1-P(Y > 7)$
or in general, $P(Y > k) = 1-P(Y > 13-k)$. Then by the law of total probability,
\begin{align}
P(Y > X)
    & = \sum_{k = 1}^{12} P(Y > X \mid X = k) P(X = k) \\
    & = \sum_{k = 1}^{12} P(Y > k) P(X = k) \\
    & = \sum_{k = 1}^6 [P(Y > k) + P(Y > 13-k)] P(X = k) \\
    & = \sum_{k = 1}^6 P(X = k) \\
    & = \sum_{k = 1}^6 \frac{1}{12} \\
    & = \frac12
\end{align}
We also observe that
\begin{align}
P(Y = X)
    & = \sum_{k = 2}^{12} P(X = Y \mid Y = k) P(Y = k) \\
    & = \sum_{k = 2}^{12} P(X = k) P(Y = k) \\
    & = \sum_{k = 2}^{12} \frac{1}{12} P(Y = k) \\
    & = \frac{1}{12} \sum_{k = 2}^{12} P(Y = k) \\
    & = \frac{1}{12}
\end{align}
Therefore, $P(Y < X) = 1 - P(Y > X) - P(Y = X) = 5/12$. So yes:

tl;dr: You were right.

And also: You can just count the cases as (I think) you did.

There are various ways to generalize this, provided the symmetry of $Y$'s distribution is symmetrically covered by $X$'s distribution. I'll leave that as an exercise to the reader, or perhaps I'll come back and sketch it out, right around the time hell freezes over.
