# roulette win probability

• May 9th 2009, 03:07 AM
bogazichili
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.
• May 9th 2009, 09:37 PM
cl85
(a) The earning of each game can be modeled as a Bernoulli random variable with mean 1/37 and variance 1368/1369. If N games are played, then the casino earnings are the sum of N iid Bernoulli random variable, which has a binomial distribution, or since N is large, by Central Limit Theorem, it is approximately normal.

So if X denotes the casino earnings, then P(X > 1000) = 0.5 means, by symmetry of the normal distribution, that X has a mean of 1000. Since the Bernoulli random variable has a mean of 1/37, the casino needs to play 37000 games.

(b) Assuming that N games are played and N is large enough such that CLT is applicable, then
$P(X < 0) = P(\dfrac{X-\dfrac{N}{37}}{\sqrt{\frac{1368N}{1369}}} < \dfrac{0-\dfrac{N}{37}}{\sqrt{\frac{1368N}{1369}}}) = P(Z < \dfrac{0-\dfrac{N}{37}}{\sqrt{\frac{1368N}{1369}}})$
where $Z$ is standard normal
• May 10th 2009, 06:42 AM
bogazichili
i wanna see the calculations.
how can u find 1/37 and variance 1368/1369. can u be more clear in part of a. im trying to understand. anyway, thank u.