100 Gambles of RouletteI 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?

Question

yatangije62 · Accepted Answer

Step 1This 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  &amp;gt;  56  )  =  1  −  P  (  X  ≤  56  ).Step 2We also have:  P  (  X  ≤  56  )  =      ∑          k      =      0                     56                                    (                            100        k                              )                                (              18        38            )              100      −      k                  (              20        38            )              k              ≈  0.7805which 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.