Law of large numbers for sum

Question

Liz is standing on the real number line at position 0. She rolls a die repeatedly. If the roll is 1 or 2, she takes one step to the right (in the positive direction). If the roll is 3, 4, 5 or 6, she takes two steps to the right. Let $X_n$ be Liz's position after n flips. Find $$\lim_{n \rightarrow \infty} P(X_n>1,6n)$$

I determined a random variable N as the number of rolls that give 1,2. Then $N\sim Bin(n,\frac{1}{3})$. $X_n=N+2(n−N)$. My question is: what is the distribution of $X_n$, how to apply Law of Large Numbers?

Answer 1

HINT To apply the law of large numbers to a series $\sum_n A_n$, you don't need the distribution of $A_n$, just variance and expectation. Here too, $$ X_n = \sum_{k=1}^n A_k, \quad \text{with} \quad A_k = \begin{cases} 1, & \text{ with probability } 1/3,\\ 2, & \text{ with probability } 2/3. \end{cases} $$ Can you finish this?

Answer 2

$$X_n = 2n -N$$

\begin{align} \lim_{n \to \infty}P(X_n > 1.6n) &= \lim_{n \to \infty}P(2n-N>1.6n) \\ &=\lim_{n \to \infty}P(N < 0.4n)\\ &= \lim_{n \to \infty} P\left(N - \frac{n}3 < \frac{n}{15}\right) \\ &= \lim_{n \to \infty} \left[1- P\left(N - \frac{n}3 \ge \frac{n}{15}\right) \right]\\ &= \lim_{n \to \infty} \left[1- P\left(\frac{N - \frac{n}3}{\frac{\sqrt{2n}}{3}} \ge \frac{\frac{n}{15}}{\frac{\sqrt{2n}}{3}}\right) \right]\\ &\ge \lim_{n \to \infty} \left[ 1- \left(\frac{\frac{\sqrt{2n}}{3}}{\frac{n}{15}}\right)^2\right] \\ &= \lim_{n \to \infty} [1- \frac{50}{n}] \end{align}