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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 indefinitely. If Nathaniel goes first, determine the probability that he ends up winning.

The solution for the same is posted here https://hmmt-archive.s3.amazonaws.com/tournaments/2011/feb/combgeo/solutions.pdf.

I have solved it using geometric series. want to know if the below is correct!!

Natheniel starts the game, and there is zero probability that he can have a multiple of 7 in the first roll. Then Obediah rolls,she has a prob of $\frac{1}{6}$ of winning. In the fourth roll, Obediah has a probability equal to $\frac{5}{6}.\frac{5}{6}.\frac{1}{6}$ of winning meaning, Obediah fails in the second roll, nathaniel (has the same change of failing to have a multiple of 7) fails in the third roll and Obediah wins in the fourth roll and sets off a series such as below.

Probability that Obediah wins $=\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+ (\frac{5}{6})^4\frac{1}{6}+\cdots = \frac{6}{11}$.

Thus Nathaniel has a $1-\frac{6}{11} = \frac{5}{11}$ chance of winning.

Is this a correct approach?

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Yes, it's a correct approach. Here's another, which you might find simpler.

Let $p$ be the probability that Obediah wins the game. She has a probability of $\frac 16$ of winning the game on her first roll. If she doesn't, then the players have switched positions so that Nathaniel now has a probability of $p$ of winning the game.

Thus, $p+ \frac 56 p =1$, so $p = \frac {6}{11}$ and Nathaniel's probability of winning is $1-p=\frac{5}{11}.$

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