1. **State the problem:** We want to find the probability that the second die roll is exactly one greater than the first die roll after rolling a fair six-sided die twice. 2. **Understand the sample space:** Each die roll can result in one of 6 outcomes: 1, 2, 3, 4, 5, or 6. 3. **Define the event:** The event is that the second roll value is exactly one more than the first roll value. If the first roll is $x$, the second roll must be $x+1$. 4. **Possible first roll values:** Since the second roll must be $x+1$ and the maximum die value is 6, the first roll $x$ can only be from 1 to 5. 5. **Calculate the number of favorable outcomes:** For each first roll $x$ in $\{1,2,3,4,5\}$, there is exactly one second roll $x+1$ that satisfies the condition. So, there are 5 favorable pairs $(x, x+1)$. 6. **Calculate total possible outcomes:** Since each die roll is independent and has 6 outcomes, total outcomes for two rolls are $6 \times 6 = 36$. 7. **Calculate the probability:** $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{36}$$ **Final answer:** The probability that the second die roll is exactly one greater than the first die roll is $\boxed{\frac{5}{36}}$.