Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a
running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of 7. Play continues until either player wins, or else indenitely. If Nathaniel goes first, determine the probability that he ends up winning.
1. **Initial Setup and Understanding the Game:**
   - Nathaniel and Obediah take turns rolling a fair six-sided die.
   - The game continues until the running tally of the sum of all rolls is a multiple of 7.
   - Nathaniel goes first.

2. **Analyzing the First Roll:**
   - Nathaniel's first roll does not matter in terms of winning immediately because the sum of one roll (1 to 6) is not a multiple of 7.
   - Obediah's first roll has a \(\frac{1}{6}\) chance of making the sum a multiple of 7 (since he needs to roll a number that, when added to Nathaniel's roll, results in a multiple of 7).

3. **General Probability Analysis:**
   - For any given sum \(S\) that is not a multiple of 7, the next roll has a \(\frac{1}{6}\) chance of making the sum a multiple of 7.
   - This is because there is exactly one number on the die that, when added to \(S\), results in a multiple of 7.

4. **Calculating the Probability of Nathaniel Winning:**
   - The probability that Nathaniel wins on his first roll is \(0\) because the first roll cannot make the sum a multiple of 7.
   - The probability that Nathaniel wins on his second roll is \(\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)\). This is because Obediah must not win on his roll (\(\frac{5}{6}\) chance) and then Nathaniel must win on his next roll (\(\frac{1}{6}\) chance).
   - The probability that Nathaniel wins on his third roll is \(\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)\). This is because all previous rolls must not result in a win (\(\left(\frac{5}{6}\right)^3\)) and then Nathaniel must win on his roll (\(\frac{1}{6}\)).

5. **Generalizing the Probability for Nathaniel's \(n\)-th Roll:**
   - On his \(n\)-th roll, there are \(2n-2\) rolls before him, all of which must not result in a win (\(\left(\frac{5}{6}\right)^{2n-3}\)) and then Nathaniel must win on his roll (\(\frac{1}{6}\)).
   - Therefore, the probability that Nathaniel wins on his \(n\)-th roll is \(\left(\frac{5}{6}\right)^{2n-3}\left(\frac{1}{6}\right)\).

6. **Summing the Probabilities:**
   - We need to find the sum of the probabilities for all possible \(n\):
     \[
     \sum_{n=2}^{\infty} \left(\frac{5}{6}\right)^{2n-3} \left(\frac{1}{6}\right)
     \]
   - This is a geometric series with the first term \(a = \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \frac{5}{36}\) and the common ratio \(r = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\).

7. **Calculating the Sum of the Geometric Series:**
   - The sum of an infinite geometric series is given by:
     \[
     \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
     \]
   - Here, \(a = \frac{5}{36}\) and \(r = \frac{25}{36}\), so:
     \[
     \sum_{n=2}^{\infty} \left(\frac{5}{6}\right)^{2n-3} \left(\frac{1}{6}\right) = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}
     \]

The final answer is \(\boxed{\frac{5}{11}}\).