I am stuck on the following problem: You go to Las Vegas with $\$1000$ and play roulette $100$ times by betting $\$10$ on red each time. Compute the probability of losing more than $\$100$. Hint: Each bet you have chance 18/38 of winning $\$10$ and chance 20/38 of losing $\$10$.
So essentially this is recurring 100 times. Now each time you do it you have an 18/38 chance of winning and 20/38 chance of losing. You have a greater chance of losing than winning, so based on speculation you need to lose 56/100 in order to lose more than $\$100$, to counterbalance your winnings . How would I set up an equation to solve for this?
Thank you!
This is a classical Bernoulli experiment: You already mentioned that we need to loose at least $56$ times to loose more than $100\$$. Let $X$ describe the number of times you loose. We want to compute $P(X > 56) = 1 - P(X \leq 56)$. We also have: $$P(X \leq 56) = \sum\limits_{k = 0}^{\ 56}{100 \choose k}\left(\frac{18}{38}\right)^{100-k} \left(\frac{20}{38}\right)^{k}\space\space\approx 0.7805$$ which is the probability that we loose at most 56 times. Thus we finally obtain $1-0.7805 = 0.2195 = 21.95\%$ as our end result.