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roulette win probabilityIn the game of roulette the probability of winning is 18/37. Suppose that in each game a player earns 1 lira or loses 1 lira.(a) (5 pts) How many games must be played in a casino in order that the casino earns, with probability 1/2, at least 1000 liras daily? (b) (5 pts) Find the percentage of days on which the casino has a loss. |

Re: roulette win probabilityHi; Old post but interesting question. In European roulette the odds that the player winds is 18 / 37, and that the house wins 19 / 37 Quote:
n = 37000 games. This makes sense if 37000 games were played the joint would expect to win 19000 and lose 18000 (from p and q ). This provides the 1000L profit and this is the 50% point. Quote:
To dip below the 18500 mark requires 5.2 standard deviations from the mean 19000 on the left. This will occur with a probability of .00000009927. This is about one chance in 10073134. Virtually the joint has no chance of booking a bad day at 1L per spin. This too is obvious, gamblers know that when you have an advantage and want to insure a win keep the wagers small. |

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