Coin tossing run convergence. Show that $P(\vert \frac{Y_n}{n}-2p(1-p)\vert \ge\epsilon) \to 0$ A coin is tossed $n$-times. The probability of getting heads $H$ is $p\in(0,1)$. A run is defined  run as a maximum-length subsequence of equal values. For example, the sequence $\color{blue}{HHH}\color{green}{T} \color{purple}{HH}$ consists of three runs. Let $Y_n$ be the number of runs. Show that: $$P(\vert \frac{Y_n}{n}-2p(1-p)\vert \ge\epsilon) \to 0 \space \space \space \text{for} \space \space n\to\infty$$
Hint: Use the chebyshev inequality and write $Y_n$ using indicator functions.
My thoughts/what I have done so far:
First of all the Chebyshev inequality states that $$P(\vert X-\mu \vert \ge c )\le \frac{\sigma ^2}{c^2}$$
where $\mu$ is the mean and $\sigma^2$ the variance. I am not sure how to apply this inequality here though but I tried something assuming that the mean $\mu=E[X]$ of $\frac{Y_n}{n}$ is $2p(1-p)$.
$$P\left(\vert \frac{Y_n}{n}-2p(1-p)\vert \ge\epsilon\right)\le \frac{E\left[ \left(\frac{Y_n}{n}-2p(1-p) \right)^2 \right]}{\epsilon^2}$$
If I could show that the numerator goes to zero as $n$ goes to infinity then I could show this statement is true. $$\frac{E\left[ \left(\frac{Y_n}{n}-2p(1-p) \right)^2 \right]}{\epsilon^2}=\frac{1}{\epsilon^2}\left( E\left[ \frac{Y^2_n}{n^2}\right] \right)-\frac{1}{\epsilon^2} \left(E\left[2p(1-p) \right] \right)^2$$
Since the second experession is just the expected value of a number:
$$\implies \frac{1}{\epsilon^2}\left( E\left[ \frac{Y^2_n}{n^2}\right] \right)-\frac{1}{\epsilon^2}(2p-2p^2)^2$$
This doesn't really seem right so I would appreciate any help/hints/solutions.
A short off topic question: Is it normal for these questions to feel very difficult and to not know where to start/what to do considering this is my first course in probability (6 weeks in)? I am kind of starting to doubt myself because I have never had so many difficulties understanding concepts in math.
Edit: Even though E-A has helped me a lot, I am still not quite sure how to finish the problem (see comments below E-A's answer) so any help is still welcome.
 A: Hint: Let the sequence of coins be denoted $X_1, X_2, X_3, ...$ Note that $Y_n = 1+\sum_{i=2}^{n} 1_{X_i \neq X_{i-1}}$. (To see why, you can say that a round ends on turn $i-1$ if $i$ and $i-1$ don't match, and the $+1$ is the last run.) Can you finish from there?
As for your soft question, it is hard to establish what is normal and what is not. For a first course in probability though, if you have not taken real analysis, even just the idea of playing with epsilons and deltas should be really new, let alone stuff about different kinds of convergence, or what a random variable is (those are always hard). I left most of my final essentially blank for my first semester graduate level probability course, and through constant exposure, I got better (still a very long way to go). If you enjoy the flavor of problems, you should take more probability classes! I would advise you to not worry too much, but if you are concerned, I would talk to a human on your course staff. I am sure they would be happy to help.
