Convergence in probability and almost surely of number of occurences of yielding H twice in a row. A fair coin is tossed independently and infinitely  many times. Let $Y_n$ be the number of occurences of yielding H (heads) twice in a row during the first $n$ tosses.

*

*Does $\frac{Y_n}{n}$ converge in probability?

*Does $\frac{Y_n}{n}$ converge almost-surely?

I've managed to find the expected value and variance of $Y_n$ by constructing $Y_i=X_iX_{i+1}$ for all $1 \le i \le n-1$, with $X_i$ is $Ber(\frac{1}{2})$ (so $Y_n = \sum^{n-1}_{i=1}Y_i)$. I've come up with:
$$ \mathbb E\left[Y_n\right] = \frac{n-1}{4} $$
$$ Var(Y_n) = \frac{5n-7}{16}$$
But I'm at really not sure how to approach this. I think I'm supposed to use one of the known upper bounds (we've seen Markov's, Chebyshev's, Chernoff's and Hoeffding's inequalities) but I'm not sure how. Any help would be appreciated.
 A: It would be tempting to use SLLN, but unfortunately, we cannot since the random variables $X_i X_{i+1}$ are not independent. However, with a small trick, we can make this approach work. Let $U_i = X_{2i-1} X_{2i}$ and $V_i = X_{2i} X_{2i+1}$ for $i\in \mathbb{N}$. Now notice that we can write
$$Y_n = \sum_{i=1}^{\left\lceil (n-1)/2\right\rceil}U_i + \sum_{i=1}^{\lfloor (n-1)/2 \rfloor} V_i$$
Denote $S_n = \sum_{i=1}^n U_i$ and $T_n = \sum_{i=1}^n V_i$. From SLLN both $\frac{S_n}{n} \to \frac{1}{4}$ and $\frac{T_n}{n} \to \frac{1}{4}$ a.s. because the families $\{U_n\}$ and $\{V_n\}$ are i.i.d. Finally, write
$$\frac{Y_n}{n} = \frac{\lceil (n-1)/2 \rceil}{n}\cdot \frac{S_{\lceil (n-1)/2 \rceil}}{\lceil (n-1)/2 \rceil} + \frac{\lfloor (n-1)/2 \rfloor}{n}\cdot \frac{T_{\lfloor (n-1)/2 \rfloor}}{\lfloor (n-1)/2 \rfloor}$$ and take the limit as $n\to \infty$ to get
$$\lim_{n\to \infty} \frac{Y_n}{n}=\frac{1}{2}\cdot \frac{1}{4} + \frac{1}{2}\cdot \frac{1}{4} = \frac{1}{4} \;\; \text{a.s.}$$
Because almost sure convergence implies convergence in probability we also have $\frac{Y_n}{n} \xrightarrow{p}\frac{1}{4}$.
