Roulette and Discrete Distribution A roulette wheel has 38 numbers. Eighteen of the numbers are black, eighteen are red, and two are green. When the wheel is spun, the ball is equally likely to land on any of the 38 numbers. Each spin of the wheel is independent of all other spins of the wheel. One roulette bet is a bet on black - that the ball will stop on one of the black numbers.
The payoff for winning a bet on black is 2 dollars for every 1 dollar bet. That is, if you win, you get the dollar ante back and an additional dollar, for a net gain of 1 dollar; if you lose, you get nothing back, for a net loss of 1 dollar. Each 1 dollar bet thus results in the gain or loss of 1 dollar.
Suppose one repeatedly places 1 dollar bets on black, and plays until either winning 7 dollars more than he has lost, or losing 7 dollars more than he has won.
What is the chance that one places exactly 9 bets before stopping?
I supposed 9 bets consist of eight wins(loses) and one lose(win). I realized that $${}_{9}C_8 \times (\frac{18}{38})^8 \times\frac{20}{38}+ {}_{9}C_1 \times (\frac{20}{38})^8  \times \frac{18}{38}$$ doesn't work because it cannot be winning or losing 8 times consecutively.  Any help is appreciated.
 A: We find the probability of first being $7$ dollars up on the ninth bet. So we must lose exactly once in the first $7$ bets, then win, then win. If $p=18/38$, the probability is $\binom{7}{1}p^6(1-p)p^2$. 
One can obtain a similar expression for the probability of first being $7$ dollars down on the ninth bet. Add.
A: "What is the chance that one places exactly 9 bets before stopping?"
The last two bets have to both be wins or both be loses otherwise stopping would happen after 7 bets, not 9 bets. That makes 7 bet sequence results
possible for a final, 7 bet ahead, win and 7 bet sequence results possible for a final, 7 bet behind, loss.
They are:
Lwwwwwwww, Wllllllll,
wLwwwwwww, lWlllllll,
wwLwwwwww, llWllllll,
wwwLwwwww, lllWlllll,
wwwwLwwww, llllWllll,
wwwwwLwww, lllllWlll,
wwwwwwLww, llllllWll,
Let p be the probability "that one repeatedly places 1 dollar bets on black, and plays until either winning 7 dollars more than he has lost, or losing 7 dollars more than he has won." and "places exactly 9 bets before stopping?"
Here is the equation that gives the answer p:
p=((7 choose 6)(18^8)(20)+(7 choose 1)(20^8)(18))/38^9
p=(7(9^8)(10)+7(10^8)(9))/19^9
answer:
p=490172130/16983563041=0.02886156036967
The chance that one places exactly 9 bets before stopping is about 2.89 percent or about 1 in 34.65.
