Probability problem regarding limit theorems

Question

Let's consider this problem:
We have a slot machine that generates a number $k$ with probability: $$P_k=\frac{2^{k-1}}{3^k},k\in \mathbb{N}$$ If the generated number is divisible by 3, the player gets $0$ points. If the remainder is $1$, the player gets 10 points and if the remainder is $2$, the player loses 10 points.
What is the probability that after a 1000 generated numbers the player has between 50 and 100 points?

What bothers me is how we can identify what type of distribution this is? If I'm guessing right, this might be a Binomial distribution. How would you approach this problem?

Answer 1

An incomplete sketch:

If we assume $\mathbb{N} = \{1, 2, \dots\}$, we have $$\sum_{k=1}^{\infty}\dfrac{1}{2}\cdot\left(\dfrac{2}{3}\right)^k=\dfrac{1}{2}[(2/3)+(2/3)^2+\cdots]=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot3=1\text{.}$$ The function$$P_k=\left(\dfrac{2}{3}\right)^k\cdot\dfrac{1}{2}=\left(\dfrac{2}{3}\right)^{k-1}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}=\left(1-\dfrac{1}{3}\right)^{k-1}\cdot\dfrac{1}{3}$$ for $k = 1, 2, \dots$ corresponds to the probability mass function of a geometric distribution (in the "trials" form) with $p = \dfrac{1}{3}$. I will use $\text{Geom}(1/3)$ to denote this distribution.

Let $X_1, \dots, X_{1000} \sim \text{Geom}(1/3)$. Define $$Y_i=\begin{cases} 0, & X_i \equiv0\mod 3 \\ 10, & X_i \equiv 1 \mod 3 \\ -10, & X_i \equiv 2 \mod 3\text{,} \end{cases}\qquad i = 1, \dots, 1000\text{.}$$ Find the probability mass function of $Y_i$, and what you're essentially looking for is $$\mathbb{P}\left(50\leq\sum_{i=1}^{1000}Y_i\leq100 \right)\text{.}$$

Answer 2

Encode using probability generating functions. Let $X_i$ be the score of the $i$th round (where $X_i=0, 10, -10$ with probability one). Put $X=\sum_{i=1}^{1000}X_i$. Suppose that the $X_i$ are independent and note the $X_i$ are identically distributed. Further $$ \sum_{k}P(X_1=k)t^k=Et^{X_1}=\frac{4}{19}t^0+\frac{9}{19}t^{10}+\frac{6}{19}t^{-10}=t^{-10}\left( \frac{4}{19}t^{10}+\frac{9}{19}t^{20}+\frac{6}{19} \right)\tag{0} $$ since $$ P(X_1=0)=\frac{1}{2}\sum_{j=1}^\infty \left(\frac{2}{3}\right)^{3j}=\frac{4}{19}; \quad P(X_1=10)=\frac{1}{2}\sum_{j=0}^\infty \left(\frac{2}{3}\right)^{3j+1}=\frac{1}{2}\times\frac{2}{3}\times\frac{27}{19}=\frac{9}{19} $$ and $$ \quad P(X_1=-10)=\frac{1}{2}\sum_{j=0}^\infty \left(\frac{2}{3}\right)^{3j+2}=\frac{1}{2}\times\frac{4}{9}\times\frac{27}{21}=\frac{6}{19} . $$ Because the $X_i$ are independent and identically distributed it follows at once that $$ \sum_{k}P(X=k)t^k=Et^{X}=(Et^{X_1})^{1000}=t^{-10^4}\left( \frac{4}{19}t^{10}+\frac{9}{19}t^{20}+\frac{6}{19}\right)^{1000} $$ We want $$ \sum_{k=50}^{100}P(X=k)=\sum_{k=50}^{100}[t^{k+10^4}]\left( \frac{4}{19}t^{10}+\frac{9}{19}t^{20}+\frac{6}{19}\right)^{1000}\tag{1} $$ where $[t^m]$ extracts the coefficient of $t^m$ in the polynomial. Presumably the sum in (1) can be computed using a computer algebra system.