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 bogazichili May 9th, 2009 03:09 AM

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.

 bobbym December 21st, 2009 03:50 PM

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:
 How many games must be played in a casino in order that the casino earns, with probability 1/2, at least 1000 liras daily?
To find how many games have to be played for the casino to expect to earn 1000L is

$n \ \frac{19}{37}-\frac{18}{37}n=1000$

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:
 Find the percentage of days on which the casino has a loss.
In order for the casino to lose it would have to drop below the 18500 win mark out of the total 37000 spins. The standard deviation computes this:

$\text{sd}= \sqrt{37000*\frac{18}{37}*\frac{19}{37}}= 96.1418$

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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