Approximating Coins Flips Problem 
Approximate the probability of getting 500 heads out of a 1000 coin
  flip of unbiased coins to be within 5% of its true value (without the use of a calculator).

I know that an exact probability is $$\binom{1000}{500}(.5)^{1000} = .02522...$$
I am unsure how one could simplify this problem through estimation to get an approximate answer however. 
Thanks for any help. 
 A: An alternative approach would be to use the central limit theorem. Let $X_n$ be i.i.d. binary random variables, taking 1 if a head takes place and 0 otherwise. Having exactly 500 heads translates to $0 \le \sum_{n=1}^{1000} X_n  - 500 < 1$. Let $$Z = \dfrac{\sum_{n=1}^{1000} X_n - 500}{\sqrt{0.5 \times 1000}}.$$ By the central limit theorem, $Z$ is asymptotically a standard normal. The probability of getting 500 heads can be approximated by $$P\Big(0\le Z < \frac{1}{5\sqrt{10}}\Big)\approx \frac{1}{\sqrt{2\pi}} \int_0^{\frac{1}{5\sqrt{10}}} e^{-\frac{x^2}{2}} dx.$$ Approximating $e^{-\frac{x^2}{2}}$ by 1 gives 0.025.
A: To avoid a calculator, you certainly need Stirling's approximation for the factorials.  So $P=\binom {1000}{500}2^{-1000} \approx \frac {1000^{1000}\exp(500)\exp(500)}{500^{500}500^{500}\exp(1000)}\frac {\sqrt{2\pi 1000}}{\sqrt{2\pi 500}\sqrt{2 \pi 500}}2^{-1000}=\frac 1{\sqrt{\pi 500}}\approx \frac 1{\sqrt{1550}} \approx \frac 1{40}=0.025$
Stirling's approximation is within a factor $\frac 1{12n}$, so the error is negligible.  Using $\pi \approx 3.1$ is within 2%.  The last we were within 4% under the square root sign, so the root is within 2%, which means the calculation is within 4%.  In fact this is within 1% of your exact value.
