Probability of earning money on game slot. Let's say you are playing a game on the game slot where probability of getting
a number $k$ is $p_k=\frac{2^{k-1}}{3^k}$, if number you got happens to have a reminder of one when dividing by three you earn 10 dollars, in case you got a number that is divisible by three, you earn nothing, but you don't lose anything either (we could say that you earn zero dollars in that case), if it happens that you got the number that has reminder two when dividing by three you lose ten dollars. Find the probability that amount of money player is going to earn is 50 to 100 dollars after 1000 attepmts.
This is my attempt:
probability of earning 10 dollars in one attempt is:
$P(X_j=10)=\sum_{k=0}^{\infty} \frac{2^{3k}}{3^{3k+1}}=\frac{9}{19}$
probability of "earning" zero dollars in one attempt is:
$P(X_j)=\sum_{k=0}^{\infty} \frac{2^{3k-1}}{3^{3k}}=\frac{9}{38}$
probability of losing 10 dollars in one attempt is:
$P(X_j)=\sum_{k=0}^{\infty} \frac{2^{3k+1}}{3^{3k+2}}=\frac{11}{38}$
Now i am supposed to analyze what happens when i have thousand attempts, money earned then is sum of all money player earned in every single attempt so we have $$Y_{1000}=\sum_{k=1}^{1000}X_k$$
Now i could simply solve this by implementing central limit theorem , however, i need to find $E(Y_{1000})$ and $Var(Y_{1000})$
As far as i know, expectation should be 
$E(Y_{1000})=\sum_{k=1}^{1000}X_kp_k$
but i don't know how to determine this, so i cannot progress any further, any help appreciated!
 A: Your probabilities for the individual outcomes are wrong. The probability for $X_j=10$ is right. The probability for $X_j=-10$ should be $\frac23$ of that; here you wrote the right sum but got the wrong result. The probability for $X_j=0$ should again be $\frac23$ of that; here you wrote the wrong sum, as $k$ should start at $1$. (The $p_k$ are only normalized if we let $k$ start at $1$.)
Thus
\begin{eqnarray*}
\mathsf P(X_j=10)&=&\frac9{19}\;,\\
\mathsf P(X_j=0)&=&\frac4{19}\;,\\
\mathsf P(X_j=-10)&=&\frac6{19}\;.\\
\end{eqnarray*}
The expectation of a sum of random variables is the sum of the individual expectations. (Here no independence is required.) The variance of a sum of independent random variables is the variance of the individual variances. (Here independence is required.)
The expected gain from one round is
$$
\mathsf E[X_j]=\frac9{19}\cdot10+\frac6{19}\cdot(-10)=\frac{30}{19}\;.
$$
The expected squared gain from one round is
$$
\mathsf E[X_j^2]=\frac9{19}\cdot10^2+\frac6{19}\cdot(-10)^2=\frac{1500}{19}\;.
$$
Thus the variance of the gain from one round is
$$
\mathsf E[X_j^2]-\mathsf E[X_j]^2=\frac{1500}{19}-\left(\frac{30}{19}\right)^2=\frac{27600}{361}\;.
$$
The expectation and variance for $1000$ rounds are therefore $\frac{30000}{19}$ and $\frac{27600000}{361}\approx\left(\frac{5254}{19}\right)^2$, respectively. The range $[50,100]$ is almost six standard deviations away from the mean, so the desired probability is very low.
