Slowpoke and Doubles probability -- does it converge as the numbers of laps increases? I posted this on puzzling stack exchange three months ago and it was immediately closed as off-topic, for being a "fairly straightforward probability calculation".
I have the solution to the problem, which I do not give here in case you might want to solve it yourself. My two questions for this forum are: 1) is this indeed an interesting problem to solve and not "fairly straightforward", and 2) how can I prove that the answer for the probability of N laps converges over time (i.e. as N increases)? (This second part I have been unable to solve.)
=== Here is the original post: ===
I created this dice and paper game for my kids years ago, and at the same time gave myself an interesting puzzle.
This game is a horse race between two horses: Slowpoke and Doubles.
With a sheet of lined writing paper, draw vertical lines to make six columns down the paper. The first five columns belong to Slowpoke, and the sixth column belongs to Doubles.
Each row across the paper represents one lap around a racetrack. Before the race starts, decide on the number of laps that will be run. You can then have fun predicting which horse will win the race.
To begin the race, roll a pair of dice. If you roll a double -- (1,1), (2,2), (3,3), (4,4), (5,5), or (6,6) -- then draw an X in Doubles' column in the first lap. Doubles has just completed the first lap!
If you roll anything other than doubles, then draw an X in Slowpoke's first column in the first lap. Slowpoke has made it one fifth of the way around the track.
Keep rolling the dice. Each time a doubles comes up, Doubles completes another lap.
Each time anything else comes up, Slowpoke advances across the current lap, left to write. Draw an X in the next free column. It will take five non-doubles rolls for Slowpoke to complete the lap.
In this way, roll after roll, one or the other horse advances with another X, until finally, one of the two horses completes their last lap and wins!
My puzzle for you is:
Is one of the two horses favored to win the race?
If so, what is the probability that a given horse wins a race of N laps?
Enjoy!
=== UPDATE:
@user2661923 I found solving the probability for Doubles is easier than for Slowpoke. It follows a simple pattern. (I'm using spoiler blocks in case others find it interesting to solve on their own.)
For one lap:

 $$ \frac {6^4 + (6^3\cdot 5) + (6^2\cdot 5^2) + (6\cdot 5^3) + 5^4} {6^5} $$

For two laps:

 $$ \frac {6^9 + (6^8\cdot 5) + (6^7\cdot 5^2) + (6^6\cdot 5^3) + (6^5\cdot 5^4) + (6^4\cdot 5^5) + (6^3\cdot 5^6) + (6^2\cdot 5^7) + (6\cdot 5^8) + 5^9} {6^{11}} $$

For n laps:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k\cdot 6^{(5n - 1) - k}} {6^{6n - 1}} $$

Which simplifies to:

 $$ \sum_{k=0}^{5n - 1} \frac{5^k} {6^{k + n}} $$

This last equation I would like to prove continually approaches, but never reaches, 1/2.
 A: Partial Answer Only

Let $p = (1/6), q = (5/6)$
If I understand the problem correctly, after exactly $(6N - 1)$ rolls there are
two mutually exclusive possibilities.  Either there have been at least $(N)$ doubles
or there have been fewer than $(N)$ doubles.
If there have been at least $N$ doubles, then have been no more than $(5N-1)$ non-doubles.
In this scenario, Doubles must have crossed the finish line, and Slowpoke could not
have crossed the finish line.  If there were fewer than $N$ doubles, then the
situation is reversed: Slowpoke had to have crossed the finish line, and Doubles
could not have crossed the finish line.
In R Bernoulli trial's of an independent event, with probability of success
$= p$, and $q = (1-p)$ the probability of exactly
$k$ successes, for $k \in \{0,1,2, \cdots, R\}$ is $\binom{R}{k} p^kq^{(R-k)}$.
Here, this means that the probability of at most $S$ successes is 
$\sum_{k=0}^S \binom{R}{k} p^kq^{(R-k)}$, 
while the probability of at least $(S+1)$ successes is 
$\sum_{k=(S+1)}^R \binom{R}{k} p^kq^{(R-k)}$.
Therefore, you can simply plug in the numbers of
$R = (6N-1)$, $S = (N-1)$, and $p = (1/6)$ 
to solve the general case of $N$ laps.
Examining the case of $N = 1$ gives Slowpoke's chances as 
$$\left(\frac{5}{6}\right)^5 \approx 0.40,$$
while $N = 2$ gives Slowpoke's chances as 
$$\left(\frac{5}{6}\right)^{11} + (11) \times (1/6) \times \left(\frac{5}{6}\right)^{10}
~\approx 0.43.$$

This suggests that as $N \to \infty$, Slowpokes chances are strictly increasing, and
approach $(1/2)$ from below.  This takes some proving however.  Shown below, is
analysis that goes as far as I can take it.

If you form the sequence of Slowpokes chances, based on the number of laps, as
$\{t_1, t_2, \cdots \}$, you can then simplify the analysis by expressing
$$t_n = \frac{a_n}{b_n} ~\text{where} ~b_n = 6^{(6n-1)}.$$
Then, the numerator, $a_n$ will equal
$$a_n = \binom{6n-1}{0} 5^{(6n-1)} + \binom{6n-1}{1} 5^{(6n-2)} + \cdots + \binom{6n-1}{n-1} 5^{(5n)}.$$

Somebody with a better knowledge of statistics than me, is going to have to establish that
as $n \to \infty$, $\frac{a_n}{b_n}$ is strictly increasing and approaches $(1/2)$ from below.
