Jaylene Hunter

Answered

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?

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?

Answer & Explanation

yatangije62

Expert

2022-07-24Added 16 answers

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 $P(X>56)=1-P(X\le 56)$.

Step 2

We also have:

$P(X\le 56)=\sum _{k=0}^{\text{}56}{\textstyle (}\genfrac{}{}{0ex}{}{100}{k}{\textstyle )}{\left(\frac{18}{38}\right)}^{100-k}{\left(\frac{20}{38}\right)}^{k}\text{}\text{}\approx 0.7805$

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

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 $P(X>56)=1-P(X\le 56)$.

Step 2

We also have:

$P(X\le 56)=\sum _{k=0}^{\text{}56}{\textstyle (}\genfrac{}{}{0ex}{}{100}{k}{\textstyle )}{\left(\frac{18}{38}\right)}^{100-k}{\left(\frac{20}{38}\right)}^{k}\text{}\text{}\approx 0.7805$

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

Most Popular Questions