Stuck at my current score

Let $x$ denote one round where $x=12+7=19$ darts

We have a total of $25\times 12 + 50 \times 7 =650$ points per round,

Maximum points that is possible = $50 \times 19 =950$

Now we can calculate average point gain per dart = $\dfrac{650}{x}$

and maximum points per dart = $\dfrac{950}{x}$

Ratio of points per dart to max points per dart = $\dfrac{650x}{950x}$

= $\dfrac{65}{95}$

$=0.684$

If you believe anything's missing or incorrect in the solution, please feel free to openly share your concerns.

The first paragraph introduces $X$ as a function of $n$ , and the second paragraph imposes "initial conditions" (coming from diffy-Q parlance), and both pieces of information are crucial to solving the problem.

After my $n$ -th throw, I've thrown $a$ -number of 50-point throws and $b$ number of 25-point throws. Here these numbers are reported so that their sum is equal to $n$ . The number of points I've gotten is their linear combination: $50a + 25b$ . The maximum number of points is when I've hit the 50-point mark every time: $50n$ . So my $X$ fraction looks like:

$X(n) = \dfrac{50a + 25b}{50n}$

The second paragraph relates the numbers $a$ and $b$ to my $n$ . Let's say there had been nineteen throws. I've hit seven 50-point marks and twelve 25-point marks. So the proportion of $a$ and $b$ inside every sample of $n$ is:

$\dfrac an = \dfrac{7}{19}$

$\dfrac bn = \dfrac{12}{19}$

I rewrite the set of equations:

$a=\dfrac{7}{19}n$

$b=\dfrac{12}{19}n$

I rewrite $X(n)$ as a function of just $n$ alone:

$X(n)=\dfrac{50\left(\frac{7}{19}\right)n + 25\left(\frac{12}{19}\right)n}{50n}$

The problem asks for the limit of $X$ as $n\to\infty$ , which I calculate using Wolfram|Alpha for free.

$X\to\dfrac{13}{19}$

Is what that machine tells me. I can use a calculator to get those decimals, which begin with the sequence 0.684. That is the answer I'll report here.

Thank you,

-YuJin

wolfram 0.684