In American casinos, the roulette wheels have the integers between 1 and 36, together with 0 and 00. Half of the non-zero numbers are red, the other half are black, and 0 and 00 are green. A common bet in this game is to bet a dollar on red. If a red number comes up, the bettor gets her dollar back, and also gets another dollar. If a black or green number comes up, she loses her dollar.
There are 18 red spaces on the wheel and 18 black spaces along with 2 green spaces for 38 total spaces.
\[ P(R) = \frac{18}{38} = 0.4736842 \\ P(!R) = \frac{20}{38} = 0.5263158 \]
| results | $ +1 | $ -1 |
|---|---|---|
| x | 0.4736842 | 0.5263158 |
I terms of the text book set up for the Gambler’s ruin problem, \(p = 0.4736842\) and \(q = 0.5263158\).
\[ s = 40 \\ M = 50 \\ \frac{q}{p} = \frac{0.5263158}{0.4736842} = 1.111111 \]
\[ P = \frac{1-(\frac{q}{p})^s}{1-(\frac{q}{p})^M} \\ P = \frac{1-1.111111^{40}}{1-1.111111^{50}} \]
using R:
(1.111111^40-1)/(1.111111^50-1)## [1] 0.3453046There is 34.5% chance of making $10.
We need to solve for s, given M =$10 and P = 0.95
\[ P = \frac{1-(\frac{q}{p})^s}{1-(\frac{q}{p})^M} \\ P({1-(\frac{q}{p})^M}) = 1-(\frac{q}{p})^s \\ (\frac{q}{p})^s = 1- P({1-(\frac{q}{p})^M}) \\ s(log(\frac{q}{p})) = log(1- P({1-(\frac{q}{p})^M})) \\ s = \frac{log(1- P({1-(\frac{q}{p})^M}))}{log(\frac{q}{p})} \\ s = \frac{log(1- 0.95({1-(1.111111)^{10}}))}{log(1.111111)} \\ \]
Using r to calculate
log10(1-(1-1.111111^10)*0.95)/log10(1.111111)## [1] 9.685763The person would have to have $9.69 to have a 95% chance of winning $10.
If the odds were 50-50 you would have an equal number of people making money off the casino and losing money to the casino. The overall result would be the casino breaking even on the roulette wheel. Given that the casino has operating expenses, they would be operating at a net loss and would not remain in business.