100 Gambles of RouletteI am stuck on the following problem: You go to Las Vegas...

Jaylene Hunter

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?

Answer & Explanation

yatangije62

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 56 ).
Step 2
We also have:
P ( X 56 ) = k = 0   56 ( 100 k ) ( 18 38 ) 100 k ( 20 38 ) k     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.

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