Roulette with extraordinary strategy I am struggeling with the following excercise. 
A gambler plays roulette(we neglect 0) and bets always the same amount on red.
If he has won three times or has played 8 rounds, he bails out. 
The question is: How much profit does he make and how likely is it that he plays less than 8 rounds?
I am not familar with this type of question, so I don't really know how to describe this.
If anything is unclear, please let me know.
 A: We first find the probability that the game lasts fewer than $8$ rounds. We describe two approaches.  
(i) We find the probability that the game lasts $3$ rounds, $4$ rounds, $5$ rounds, $6$ rounds, $7$  rounds, and add up. 
The game is over at the end of round $k$, where $3\le k\le 7$, if there is a win in round $k$, and exactly $2$ wins in rounds $1$ to $k-1$. The probability of $2$ wins in the first $k-1$ rounds is $\binom{k-1}{2}\frac{1}{2^{k-1}}$. Given this has happened, the probability of a win in the $k$-th round is $\frac{1}{k}$. So for $k\le 7$, the probability the game lasts $k$ rounds is $\binom{k-1}{2}\frac{1}{2^k}$. 
The probability the game lasts $7$ or fewer rounds is therefore
$$\sum_{k=3}^7 \binom{k-1}{2}\frac{1}{2^k}.$$
(ii) Alternately, we find the probability $p$ that the game lasts $8$ rounds. Then the required probability is $1-p$. 
The game lasts $8$ rounds if there are $2$ or fewer wins in the first $7$ rounds. We have 
$$p=\binom{7}{0}\frac{1}{2^7}+\binom{7}{1}\frac{1}{2^7}+\binom{7}{2}\frac{1}{2^7}.$$

We now attack the first problem. Let random variable $X$ be the gambler's profit. One answer to the question of how much profit he makes is to give an expression for $\Pr(X=n)$ for all possible values of $n$. Assume that $1$ dollar is bet each time. 
It is easy to find $\Pr(X=3)$. This only happens if the first $3$ games result in wins, which has probability $\frac{1}{2^3}$.
The profit is $2$ if the game ends on the $4$-th round. By an earlier calculation, this has probability $\binom{3}{2}\frac{1}{2^4}$.
The profit is $1$ if the game ends on the $5$-th round. By an earlier calculation, this has probability $\binom{4}{2}\frac{1}{2^5}$.
Similarly, $\Pr(X=0)= \binom{5}{2}\frac{1}{2^6}$, and $\Pr(X=-1)= \binom{6}{2}\frac{1}{2^7}$, and $\Pr(X=-2)= \binom{7}{2}\frac{1}{2^8}$.
The only other possibility is $X=-8$. We have $\Pr(X=-8)=\frac{1}{2^8}$.
A related question we might want to answer is the expected profit. Since we have the probability distribution of $X$, we can use the values just computed to find $E(X)$.
Compute. You will find that $E(X)=0$. 
Remark: The fact that $E(X)=0$ is also a consequence of a general result, that no "strategy" can affect expectation. In the more realistic situation where there is a $0$ (and, in American roulette, a $00$), on average the more you play the more you lose. The mean loss depends only on the mean number of games played. A strategy that makes the mean number of games small is best. For the usual roulette wheel, the optimal strategy is to stay away. 
Note that by suitable strategies we can easily arrange to have a probability $\gt \frac{1}{2}$ of ending up with a positive net gain. But this is at the cost of having a significant probability of a large net loss. 
