Jaylene Hunter

2022-07-23

100 Gambles of Roulette
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?

yatangije62

Expert

Step 1
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 $P\left(X>56\right)=1-P\left(X\le 56\right)$.
Step 2
We also have:

which is the probability that we loose at most 56 times. Thus we finally obtain $1-0.7805=0.2195=21.95\mathrm{%}$ as our end result.

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