Betting strategy for roulette

Peter's strategy only works if it has an infinite amount of capital. However, since his funds are finite, eventually comes the point, at which he can not double the bet and thus loses a large portion of his money.

The probability of winning a spin is $p = \frac{13}{37} \approx 48.65 \,\%$ whereas the probability of losing gives $q = 1 - p$ . We assume, that Peter has a monetary assets of $x$ dollar from the beginning with $x = \$ \, \left(2^{N} - 1 \right) = \$ \,\sum_{n = 0}^{N-1} 2^n$ This means that Peter would go bankrupt after a series of $N$ lost spins with a probability of $Q = q^N$ (The profit in this case is equal to $-x$ ) If he loses only a small series of $k$ games with $k < N$ , Peter's profit results to $y = \$ \, \left(2^k - \sum_{n = 0}^{k-1} 2^n \right) = \$ \, 1$ with a win probability $P = 1 - q^{N}$ . Therefore, the expected value for the profit yields $\begin{aligned} E &= y \cdot P + (- x) \cdot Q \\ &= \$ \, (1 - (1 + x) q^N ) \\ &= \$ \, (1 - 2^N q^N ) \\ &< \$ \, (1 - 2^N \frac{1}{2^N} ) = \$ \, 0 \end{aligned}$ since $q > \frac{1}{2}$ . Thus, the expected value is negative and Peter will lose his money in the long run. The expectation value is even lower than for a single roulette spin with $E_0 = (\$ \, 1) \cdot p + (- \$ \,1) \cdot q = \$ \, (1 - 2 q) > \$ \, (1 - (2 q)^N) = E$ although you get the same profit of one dollar if you win. Paradoxically, the win probability $P$ can be close to 100% for large numbers for $N$ . Let's give a numeric example $\begin{aligned} N &= 10\\ x &= \$ \, 1023 \\ P &\approx 99.87 \% \\ E &\approx - \$ \, 0.306 \\ E_0 &\approx -\$ \, 0.027 \end{aligned}$ In this example, Peter loses an average of 31 cents when using his betting strategy, while losing only 3 cents for a single spin.