# Best way to show the probability that when two dice are rolled, one the first die has a larger number than the other die

I am faced with the following problem:

There are three four-sided dice $D_1$, $D_2$, $D_3$, $D_4$ that have the following assignments of numbers to their sides:

$D_1$: (1, 8, 11, 14)

$D_2$: (2, 5, 12, 15)

$D_3$: (3, 6, 9, 16)

$D_4$: (4, 7, 10, 13)

It is said that die $D_i$ beats die $D_j$ when the probability that die $D_i$ comes up with a larger number than die $D_j$ is greater than $1/2$ in an experiment where each of these two dice is rolled once.

Now, the trivial way to solve this problem is to iteratively go through each pair of die and determine how many outcomes result in $D_i$ being larger than $D_j$ and dividing that by the total number of outcomes to find the probability.

This seems tedious. Is there a better way to show the probability of each pair of die, given the different numbers on each side?

• Do you want to find out which die beats which dies or do you want to find the probability of a die beating another die? – AlgorithmsX Sep 4 '16 at 21:38
• Your problem statement says that there are 3 dice, but then you seem to refer to 4 dice instead. – user84413 Sep 4 '16 at 21:44

Your dice $D_1, D_2, D_3, D_4$ have the following property. If we write $D_i$'s values as $a_i, b_i, c_i, d_i$, where $a_i < b_i < c_i < d_i$, then all $a_i$s are less than all $b_i$s which are less than all $c_i$s which are less than all $d_i$s. For example, $a_1 < b_2$, $b_4 < c_3$, and so forth.
It follows that if we roll dice $D_i$ and $D_j$, with probability $\frac{6}{16}$, $D_i$ will automatically beat $D_j$ and with probability $\frac{6}{16}$, $D_j$ will automatically beat $D_i$. The only case that is in dispute is the last $\frac{4}{16}$ of the time: when the rolls are $a_i$ and $a_j$, $b_i$ and $b_j$, $c_i$ and $c_j$, or $d_i$ and $d_j$.
You will further find in comparing $a_i$ with $a_j$, $b_i$ with $b_j$, and so on, that $D_2$ beats $D_1$ in three of the remaining four cases, $D_3$ beats $D_2$ in three of the remaining four cases, and $D_4$ beats $D_3$ in three of the remaining four cases. So with little actual calculation, you can see which dice beat which others.
Finally, you will want the observation that if $D_i$ beats $D_j$, and $D_j$ beats $D_k$, then $D_i$ beats $D_k$, to get in the end a total understanding of which dice beat which others.