# Expected gain or loss in roulette

There are $37$ numbers, from $0$ to $36$. Each number has an equal chance of turning up. Zero is green in color and odd numbers are in black and even numbers are in red. If you place $\$1$on red (black) you get$2$dollars if you are right and$\$0$ if you are wrong. For each bet there are $19$ losing numbers and $18$ winning numbers.

Suppose you place $\$1$bets$18$times on red. What is your expected gain or loss after$18$tries? My working: Expected value of one game =$2(18/37) + -$1(19/37) Expected value of 18 games = ($2(18/37) + -$1(19/37))^18 Is this right? Part 2: Suppose I use a dollar for each color bet? How many games can I play if I have an initial capital of$18?

Not sure how to start on this...

• The value of a win is $1$, not $2$.
• The expected outcome after $18$ rounds is $18$ times the expected expected outcome after $1$ round, by linearity of expectation.
• Are you asking about the expected stopping time? I'll need to think a bit about that. My guess would be that it's the minimium $k$ such that $k\times \mathbb{E}[\text{outcome after 1 game}] \leq -18$. – Rebecca J. Stones Sep 23 '13 at 14:07
• I'm not sure what you mean, but the change in the total money will either be +1 or -1 after betting $1. – Rebecca J. Stones Sep 24 '13 at 17:26 We can think of the game as a$1$-dimensional random walk where we take a step$+1$(when we win$\$1$) with probability $p := \frac{18}{37}$ and a step $-1$ (when we lose $\$1$) with probability$1-p = \frac{19}{37}$. The expected profit after$n$plays is $$(2 p - 1) \, n = \left(\frac{36}{37} - 1\right) n = -\frac{n}{37}$$ which is negative, as expected. If$n = 18$, then the expected loss is$\frac{18}{37}$dollars, which is approximately$50\$ cents.