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October 23rd, 2017, 06:45 PM   #1
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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.
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October 23rd, 2017, 07:02 PM   #2
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Why would a normal distribution be helpful here? The game clearly has a binomial distribution.
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October 23rd, 2017, 11:16 PM   #3
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Quote:
Originally Posted by poopeyey2 View Post
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.
Ok, it looks like you want to use the Normal approximation to the binomial distribution.

First off we need to determine how many wins out of 240 she needs to make to break even.

$w(n,k) = 2k - (n-k) = 3k-n$

$3k-240 \geq 0 \Rightarrow k \geq 80$

now to approximate this hard to compute binomial distribution we find the mean and standard deviation as

$\mu = n p = 240 \cdot \dfrac 1 4 = 60$

$\sigma = \sqrt{n p (1-p)} = \sqrt{240 \cdot \dfrac 1 4 \cdot \dfrac 3 4} = \sqrt{45}$

$P[\text{breaking even or better}]=P[k\geq 80] = 1 - P[k\leq 79] = 1 - \Phi\left(\dfrac{79-60}{\sqrt{45}}\right) \approx 0.00231$

The exact probability of $P[k \leq 79]$ using the binomial distribution can be computed with software. Doing this we find that

$P[\text{breaking even or better}] = 1-P[k\leq 79] = 1 - \displaystyle \sum_{k=0}^{79}\dbinom{n}{k} \left(\dfrac 1 4\right)^k \left(\dfrac 3 4\right)^{n-k} = 0.00234$
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