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Suppose a European style roulette wheel has the following probabilities: a red number appears with probability $\frac{18}{37}$ , a black number appears with probability $\frac{18}{37}$, and a green number appears with probability $\frac{1}{37}$. Jon bets exactly $\$1$ on black each round. Explain why this is not a good long-term strategy.

Outline:

Let $S_n:=$ net wins after $n$ games. Then after "many" games $P(S_n>0)$ goes to zero. We have, $$ \begin{align} \lim_{n\to\infty}P\left(S_n>0\right)&=\lim_{n\to\infty}(1-P\left(S_n\leq0\right))\\ &=\lim_{n\to\infty}[1-(P\left(S_n=0\right)+P\left(S_n<0\right))]\\ &=\lim_{n\to\infty}[1-0-P\left(S_n<0\right)]\\ &=\lim_{n\to\infty}1-\lim_{n\to\infty}P\left(S_n<0\right)\\ &\leq1-\lim_{n\to\infty}P\left(|\frac{S_n}{n}-\mu|<0\right)\\ &=1-1 \enspace(\text{by LLN})\\ &=0 \end{align} $$

I looked around online and it seems to me that this problem is related to the gambler's ruin problem. Is my approach correct? Or at least on the right track?

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The expected value of betting on blacks is $1*\frac{18}{37}+0*\frac{18}{37}+0*\frac{1}{37}=\frac{18}{37}$. Now, the expected value of profit of each round is $-1+\frac{18}{37}=-\frac{19}{37}$. Thus, the expected value of the profit in $n$ truly random rounds is $-\frac{19n}{37}$. As it is negative, it means that the gambler is more likely to lose money than not to. That's one of the meanings of the gambler's ruin problem - if you play a fair game with negative expected value, you'll eventually go bankrupt.

TL;DR the casino always wins

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