3 Dice Game

If we can make a sum with $-$ , we can make one with $+$ instead ( $a=b+c$ instead of $a-b=c$ ). Similarly, if we can make a sum with $\div$ , we can make one with $\times$ . Thus we only have to count the number triples that give a sum with $+$ - $(1,a,a+1)$ for $1 \le a \le 5$ , $(2,a,a+2)$ for $2\le a \le 4$ and $(3,3,6)$ , and the number triples that give a sum with $\times$ - $(1,a,a)$ for $1 \le a \le 6$ and $(2,a,2a)$ for $2 \le a \le 3$ , if we only consider triples with digits in ascending order.

Thus there are (looking at digit triples in ascending order):

• one triple involving three identical digits - $(1,1,1)$ ,
• eight triples involving two different digits - $(1,2,2)$ , $(1,3,3)$ , $(1,4,4)$ , $(1,5,5)$ , $(1,6,6)$ , $(1,2,2)$ , $(2,2,4)$ , $(3,3,6)$ ,
• seven triples involving three distinct digits - $(1,2,3)$ , $(1,3,4)$ , $(1,4,5)$ , $(1,5,6)$ , $(2,3,5)$ , $(2,4,6)$ , $(2,3,6)$ .

Thus there are $1 \times 1 + 8 \times 3 + 7 \times 3! = 67$ different ordered triples of dice rolls that give a valid sum. The desired probability is $\tfrac{67}{216}$ , making the answer $67 + 216 = \boxed{283}$ .