Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression?

Let the three numbers on the dice be $x, y, z$. The condition is that $x, y, z$ are in arithmetic progression and are distinct. Let the numbers be $a-d, a, a+d$ where $d > 0$.

Since the numbers are between $1$ and $6$, we have $1 \le a-d$ and $a+d \le 6$.

For $d=1$: $(a-1, a, a+1)$ can be $(1,2,3), (2,3,4), (3,4,5), (4,5,6)$. There are $4$ sets. Each set has $3! = 6$ permutations. Total outcomes = $4 \times 6 = 24$.

For $d=2$: $(a-2, a, a+2)$ can be $(1,3,5), (2,4,6)$. There are $2$ sets. Each set has $3! = 6$ permutations. Total outcomes = $2 \times 6 = 12$.

Total number of outcomes with three different numbers in arithmetic progression is $24 + 12 = 36$.

Now, we find the number of these outcomes that sum to $15$. The sum is $(a-d) + a + (a+d) = 3a = 15$, which implies $a = 5$.

For $a=5$, the possible sets are:
If $d=1$, the set is $(4, 5, 6)$. Permutations are $(4,5,6), (4,6,5), (5,4,6), (5,6,4), (6,4,5), (6,5,4)$. There are $6$ such outcomes.
If $d=2$, the set is $(3, 5, 7)$, but $7 > 6$, so this is not possible.

The number of favorable outcomes is $6$.

The probability is $\dfrac{6}{36} = \dfrac{1}{6}$.

Answer: $\dfrac{1}{6}$