Is it better to bet on a d12 or on the sum of two d6?

Question

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?

Answer 1

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.

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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.