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.
Best Answer
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.