Problem: At roulette, you bet a dollar on red $30$ times in a row. Each time you win a dollar with prob. $18/38$ and lose a dollar with prob. $20/38$. Find approximately the probability that after $30$ bets you have at least as much money as you started with. (Use normal approximation)
Attempt: Let $p = 18/38, q = 20/38$. So for me to win a dollar is a binomial distribution $(30, 18/38)$. I don't know exactly how to set up the probability
Anything helps! Thanks!
You will end up with at least as much money as you started if and only if you win $15$ times or more. The number of wins $W$ has a binomial distribution with $n=30$ and $p=\frac{18}{38}=\frac{9}{19}$. This distribution has mean and variance $$\mu=np=\frac{270}{19}\ ,\quad \sigma^2=npq=\frac{2700}{19^2}\ .$$ We approximate by using a normal distribution with the same mean and variance, $$X\sim N\Bigl(\frac{270}{19},\,\frac{2700}{19^2}\Bigr)\ ,$$ and use the continuity correction: since $W$ can only take integer values, $W\ge15$ is the same as $W\ge14.5$. So the required probability is $$P(W\ge15)\approx P(X\ge14.5)\ ,$$ and as $X$ is a normally distributed variable with known mean and variance, I am sure that you can evaluate this.