You go to Las Vegas with $\$1000$ and play roulette $100$ times by betting $\$10$ on the number 36 each time. Compute the probability of leaving the Casino with at least $\$1010$. Hint: Each bet you have chance $1/38$ of winning $\$350$ and chance $37/38$ of loosing $\$10$.
So in order to have at least $\$1010$ when you leave you have to win at least $3$ times. $$3\cdot 1050+1000=2050-(97\cdot 10)=1080$$
I am assuming that I am looking for the probability of $97$ losses or less.
I interpreted this as a Binomial Distribution, which I plugged into Excel with the following calculation:
=Binom.dist(97,100,(37/38),TRUE), however, it does not get me the right answer. I am unsure on how to fix the program.
Thank you!
I don't know what casino you've been going to but on standard Roulette there are $37$ numbers, numbered $0-36$. So I doubt it's a maths error.
By your method, plugging
into Excel gives $\approxeq0.5093$
Well, anybody knows your expected winnings from any casino bet are less than what you took in so the answer must be $\approxeq1-0.5093=0.4907$