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October 23rd, 2017, 06:45 PM  #1 
Member Joined: Apr 2017 From: PA Posts: 45 Thanks: 0  What is the probability that she lost no money?
A tourist in Las Vegas was attracted by a certain gambling game in which the customer stakes $1$ dollar on each play; a win then pays the customer $2$ dollars plus the return of her stake, although a loss costs her only her stake. Las Vegas insiders, and alert students of probability theory, know that the probability of winning at this game is $\frac{1}{4}$. When driven from the tables by hunger, the tourist had played this game $240$ times. What is the probability that she lost no money? For this, I found $E(X)=60$. Afterwards, I knew I needed to find $V(X)$, so I got $\frac{9}{4}$ as my variance, then I got the standard deviation and tried to find the area of the normal distribution when $X>0$, but couldn't solve it. Last edited by poopeyey2; October 23rd, 2017 at 06:47 PM. 
October 23rd, 2017, 07:02 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 197 Thanks: 105 Math Focus: Dynamical systems, analytic function theory, numerics 
Why would a normal distribution be helpful here? The game clearly has a binomial distribution.

October 23rd, 2017, 11:16 PM  #3  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817  Quote:
First off we need to determine how many wins out of 240 she needs to make to break even. $w(n,k) = 2k  (nk) = 3kn$ $3k240 \geq 0 \Rightarrow k \geq 80$ now to approximate this hard to compute binomial distribution we find the mean and standard deviation as $\mu = n p = 240 \cdot \dfrac 1 4 = 60$ $\sigma = \sqrt{n p (1p)} = \sqrt{240 \cdot \dfrac 1 4 \cdot \dfrac 3 4} = \sqrt{45}$ $P[\text{breaking even or better}]=P[k\geq 80] = 1  P[k\leq 79] = 1  \Phi\left(\dfrac{7960}{\sqrt{45}}\right) \approx 0.00231$ The exact probability of $P[k \leq 79]$ using the binomial distribution can be computed with software. Doing this we find that $P[\text{breaking even or better}] = 1P[k\leq 79] = 1  \displaystyle \sum_{k=0}^{79}\dbinom{n}{k} \left(\dfrac 1 4\right)^k \left(\dfrac 3 4\right)^{nk} = 0.00234$  

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