# European Gambler's ruin problem?

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?

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.