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.

  • $\begingroup$ Don't forget that he could stop because he was on a losing streak as well. There are $\binom{9}{8}=9$ ways to arrange eight w's and one l where you don't care. How many of those are "bad" and actually would have ended in seven turns instead of nine? (if the first seven are w's and the last two are either wl or lw). How many of them then are good (and actually end in nine turns)? Do so similarly for the case of him having a bad losing streak. $\endgroup$ – JMoravitz Jul 11 '16 at 21:33
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    $\begingroup$ Also, for the record, "Russian Roulette" commonly refers not to the casino game you describe, but something much more grim: using a revolver with only one loaded chamber, spinning the chamber, and placing the gun to your head and pulling the trigger, under the expectation that the other players will do the same, running the risk of shooting yourself dead. $\endgroup$ – JMoravitz Jul 11 '16 at 21:37
  • $\begingroup$ That's right. Thanks for editing. $\endgroup$ – MakiK Jul 11 '16 at 21:46

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.


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