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 had $p = \dfrac{18}{38}, q = \dfrac{20}{38}$. Thus, my calculation was: $9(qp^8 + pq^8) – (p^3 + q^3)$ which was incorrect.
I would really appreciate any help.
Best Answer
You will be $7$ ahead in exactly $9$ bets by winning $8$ bets, with $1$ loss in the first $7$ bets,
and a mirror image for being $7$ behind,
thus $\binom71\cdot[p^8q + pq^8]$