When rolling a pair of standard six-sided dice, the possible sums range from 2 (1+1) to 12 (6+6). Since the maximum sum that can be rolled with two six-sided dice is indeed 12, the probability of rolling a sum less than 13 will cover all possible outcomes of the dice.
The total number of outcomes when rolling two dice can be calculated as follows:
\[ 6 \text{ (outcomes for the red die)} \times 6 \text{ (outcomes for the blue die)} = 36 \text{ total outcomes} \]
The sums that can be rolled are from 2 to 12. Thus:
- Sum = 2: (1,1)
- Sum = 3: (1,2), (2,1)
- Sum = 4: (1,3), (2,2), (3,1)
- Sum = 5: (1,4), (2,3), (3,2), (4,1)
- Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1)
- Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2)
- Sum = 9: (3,6), (4,5), (5,4), (6,3)
- Sum = 10: (4,6), (5,5), (6,4)
- Sum = 11: (5,6), (6,5)
- Sum = 12: (6,6)
Since the maximum achievable sum is 12, every single outcome will yield a sum of less than 13.
Thus, the number of successful outcomes (sum < 13) is all 36 possible outcomes from rolling the dice.
The probability of rolling a sum less than 13 is therefore:
\[ P(\text{sum} < 13) = \frac{36}{36} = 1 \]
Now, if we express the probability as a fraction in its simplest form:
\[ \frac{a}{b} = \frac{1}{1} \]
where \( a = 1 \) and \( b = 1 \).
Finally, to find \( a + b \):
\[ a + b = 1 + 1 = 2 \]
Thus, the answer is \(\boxed{2}\).