Setup: There are 18 red, 18 black, and 2 green.
Therefore there are 38 possible outcomes. Betting a dollar on red… \[
\begin{aligned}
p &= \frac{18}{38} = 0.4736842 \\
q &= \frac{20}{38} = 0.5263158
\end{aligned}
\]
Using formula.. \[ \begin{aligned} P &= \frac{1 - (\frac{q}{p})^s}{1 - (\frac{q}{p})^M}\\ where, \\ s &= 40 \\ M & = 50 \end{aligned} \]
Therefore; \[ \begin{aligned} P &= \frac{1 - (\frac{0.5263158}{0.4736842})^{40}}{1 - (\frac{0.5263158}{0.4736842})^{50}} \\ &= \frac{1 - 1.111111^{40}}{1 - 1.111111^{50}} \\ &= 0.3453046 \end{aligned} \]
(1-1.111111^40)/(1-1.111111^50)## [1] 0.3453046Therefore the probability to getting to $50 is 34.53%
Solve for s, given M = $10, and P = 95% \[ \begin{aligned} P &= \frac{1 - (\frac{q}{p})^s}{1 - (\frac{q}{p})^M} \\ 1 - (\frac{q}{p})^s &= P(1 - (\frac{q}{p})^M) \\ (\frac{q}{p})^s &= 1 - P(1 - (\frac{q}{p})^M) \\ s(log(\frac{q}{p})) &= log(1 - P(1 - (\frac{q}{p})^M)) \\ s &= \frac{log(1 - P(1 - (\frac{q}{p})^M))}{log(\frac{q}{p})} \\ &= \frac{log(1 - 0.95(1 - 1.111111^{10}))}{log(1.111111)} \\ &= 9.685763 \end{aligned} \]
(log(1 - 0.95 * (1 - 1.111111^10)))/(log(1.111111))## [1] 9.685763Therefore to have a 95% chance to win $10 she must start with $9.69
Removing 0 and 00 would make the possibility to be 50-50, and although it is still possible for the casino to make a profit this case will make it harder because they will need to rely less on the chances of the game and the fact that they are getting more customers who are not just playing but eating, drinking, and engaging in other profitable activities.