Application of the weak law of large numbers (Roulette) I am currently working on the following problem. Imagine the following situation: A player bets 1 dollar, and looses his bet with the probability of 19/37, but is given his bet and an extra dollar back with the probability 18/37. What is the probability that the casino looses money with this game on a (very) long run?
I wanted to use the weak law of large numbers. If the random variable $X_i$ is the money the player has after the $i$th game, then $X_i-X_{i-1}$ is the money he won in the $i$th game. So I need to look for $\lim_{n \to \infty} \mathbb P(\frac{1}{n}\sum_{i = 1}^\infty X_i - X_{i-1}> \epsilon)$ for some $\epsilon > 0$ and the $X_i$ are Binomial distributed with parameters $p = 19/37$ and $q = 18/37$. Is this correct and how can I go on?
 A: Let us debunk several misconceptions that plague your analysis. First, the money $X_n$ that the player has after $n$ games is not binomially distributed since $X_n$ can be negative. However, $X_n=2Y_n-n$ where $Y_n$ is binomial $(n,p)$ with $p=18/37$ (and not $19/37$). Thus, $Y_n=Z_1+\cdots+Z_n$ where $(Z_n)$ is i.i.d. Bernoulli with parameter $p$, that is $P[Z_n=1]=p$ and $P[Z_n=0]=1-p$. 
The law of large numbers (strong or weak) asserts that $Y_n/n\to E[Z_1]$ (almost surely or in probability), thus $X_n/n\to2E[Z_1]-1$. Now, $E[Z_1]=1\cdot p+0\cdot(1-p)=p$ hence $X_n/n\to2p-1$. Since $2p-1=-1/37\lt0$, this implies that $\lim\limits_{n\to\infty}P[X_n/n\gt0]=0$.
Actual betting models take into account the fact that the fortunes of both players (the casino and the player) are finite, then one can be sure that the game ends with the ruin of either the player or the casino. The casino's fortune being immensely larger then the player's makes the ruin of the player overwhelmingly more probable. For $p\lt1/2$ as above, and for a player with initial fortune $f$ and a casino with initial fortune $F$, the probability of the player ruining the casino is
$R=(r^f-1)/(r^F-1)$, where $r=(1-p)/(1-2p)$. If $p=18/37$, then $r=19$ and $R\approx1/r^{F-f}$, which is very small.
