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.

  • $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $LaTeX$ syntax $\endgroup$ – Shailesh Jul 19 '16 at 8:50
  • $\begingroup$ $\binom92 \cdot \left(\frac{16}{38}\right)^{2} \cdot \left(1-\frac{16}{38}\right)^{9-2} + \binom92 \cdot \left(1-\frac{16}{38}\right)^{2} \cdot \left(\frac{16}{38}\right)^{9-2}$ $\endgroup$ – barak manos Jul 19 '16 at 8:56
  • $\begingroup$ @barakmanos: OP has given correct figures for $p$ and $q$, and you have to be $7$ ahead or behind after exactly $9$ bets. $\endgroup$ – true blue anil Jul 19 '16 at 17:12
  • $\begingroup$ @LHoang: Pl. note the altered approach ! $\endgroup$ – true blue anil Jul 19 '16 at 17:13
  • 1
    $\begingroup$ @barakmanos: Well, congrats ! :) $\endgroup$ – true blue anil Jul 20 '16 at 9:02

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 $\binom71\cdot[p^8q + pq^8]$


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