Question: HW Problem here, not sure where I'm messing up.

Let $X$ be the amount won or lost in betting \$5 on red in roulette. Then $P(5) = \frac{18}{38}$ and $P(-5) = \frac{20}{38}$. If a gambler bets on red one hundred times, use the central limit theorem to estimate the probability that those wagers result in less than $50 in losses.

So I calculated:

$E(x) = 100\frac{18}{38} = \frac{900}{19}$

$var(x) = \frac{900}{19}\times\frac{20}{38} = \frac{9000}{361}$

$\sigma = \sqrt{\frac{9000}{361}} = 4.993...$

Then this is what I was thinking:

$P(-\infty \leq x \lt 50) \rightarrow P(\frac{x - 50}{4.993} \lt 0)$

Which would give

$\int\_{-\infty}^{0}{\left(\frac{1}{\sqrt{2\pi}}\right)\left(e^{\tfrac{-z^2}{2}}\right)}$

Which using the table should be $(0.0 - 3.09) = (.5 - .001) = .499$

However I know from the back of the book it should be $.6808$. So where am I going wrong?

EDIT: I see from the comments and from one answer that I have my $E(x)$ wrong, but I'm still stuck on how to finish it out. I would normally go into my teacher's office hours for this, but due to Thanksgiving, there won't be any.

Answer: 
By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\mbox{sgn}{\left(
\begin{array}{ccccc}
 1 & 2 & 3 & 4 & 5 \\
 3 & 1 & 5 & 2 & 4 \\
\end{array}
\right)}(1000)^5=\mbox{sgn}\left((13542)\right)(1000)^5=+(1000)^5$.

Each of the other terms has absolute value at most $9\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\cdot9\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\cdot9\cdot1000^4$, which is positive.
You can use 1000 in every row to eliminate the other element to get an upper triangular matrix, and in the process for 1000 is large enough, you can reckon the element in other column won't change. Then you can get the elements on the diagonal all are1000, which shows the determinant is positive.`