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May 9th, 2009, 04:09 AM  #1 
Newbie Joined: Dec 2008 Posts: 9 Thanks: 0  roulette win probability In 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. 
December 21st, 2009, 04:50 PM  #2  
Senior Member Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0  Re: roulette win probability
Hi; 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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