Tetra Master Dice Rolls

Question

In the game Tetra Master, two players play a card with a number n between 0 and 15, inclusive. Then, both players roll a 16-sided die numbered from 0 to 15.

Both players than add up their dice rolls like so:

16 * n + dice(0, 15) = A

Then, both players roll another die from 0 to the number they previously rolled A inclusive (with an equally likely chance for every number) and subtract them:

A - dice(0, A) = B

The player with the higher B wins.

Given two different n, what is the probability that one n beats another? For instance, what is the probability that Player 1 with n=5 will beat Player 2 with n=2? Is there a formula that can be used to figure out the probability given two n without running simulations?

Answer 1

We assume final ties are split- $1/2$ to each player. Call the two players' $n$'s $n_1, n_2$, $A$ numbers $A_1, A_2$ and $B$ numbers $B_1, B_2$. $B_1$ is drawn uniformly from $[0,A_1]$ and likewise $B_2$. If we assume $A_1 \ge A_2$, the chance that player $1$ wins is the chance $B_1 \gt A_2+1/2$ because if $B_1 \gt A_2$ he wins and otherwise it is a tossup. The chance of that is $\frac {A_1-A_2}{A_1+1}+\frac 12$ For a large difference in $n$'s the first die is noise and you get the chance that player $1$ wins is about $\frac 12+\frac {16(n_1-n_2)}{16n_1+1}$. You can get the exact value following the same logic, but the variation in denominators makes me suspect that there will not be a nice form.