Originally Posted by

**Laurent** Let's let the numbers talk by themselves...

Suppose you use the following stategy:

Bet 1. If you win, you win 1 and you stop. If you lose, you lose 1 and then you bet 2.

If you win, you win 2-1=1, and you stop. If you lose, you lose 2+1=3 and then you bet 4.

If you win, you win 4-3=1, and you stop. If you lose, you lose 4+3=7 and then you bet 8...

Doing this, you always win 1, except if your next bet exceeds the maximum bet allowed in the casino ...in which case you lose *much money*. This happens with probability $\displaystyle \frac{1}{2^n}$ ($\displaystyle n$ losses), and you lose $\displaystyle 2^n-1$. The expected amount won is therefore $\displaystyle 1\left(1-\frac{1}{2^n}\right)-(2^n-1)\frac{1}{2^n}=0$. This is not surprising, because the game is fair: at each bet, your expected win is 0. With whatever strategy (with bounded amounts), it would have been the same. Fair games are not a good deal for casinos, and I suspect you pay a tax when you buy token or when you convert them into dollars back again. Anyway, what is for sure is that you can't have positive (>0) averaged winnings...

You can be "almost" sure to win something. But in the other case, however rare it is, you lose so much that it compensates for your winnings.

Laurent.