<s>[INST] Probability that the eventually a six on a dice will appear. 
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m+n$.

If you roll a die $r$ times, the probability of getting a six on the $r$th try is:
$\frac{1}{6} \cdot \frac{5^{r-1}}{6^{r-1}}$
Now I believe casework for Dave so:
1 roll: $ \frac{1}{6} $, 2 rolls: $\frac{1}{6} \frac{5}{6}$, 3 rolls: $\frac{1}{6} \frac{5^2}{6^2}$, .... , $r$ rolls, $\frac{1}{6} \frac{5^{r-1}}{6^{r-1}}$
I would add this up, but still, we never found $r$?
HINTS ONLY
EDIT
As pointed by 5xum, $L=1$ isn't possible with $D = L - 1$. So I got:
$P = 1 \cdot (1/6)\sum_{r=2}^{\infty} (5/6)^{r-2} + 1  + ...$
But $P > 1$ already, which is impossible?
 [/INST] Let $D, L$ be the obvious random variables.  Then the desired probability is
\begin{align}
P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\
  & + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\
  & + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\
  & + \cdots
\end{align}
Each row constitutes an element in a geometric series.  Find the first row sum and the ratio, and the rest is trivial.
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