# 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.

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Are you familiar with Stirling's approximation for factorials? Applied to middle binomial coefficients, you get lots of cancellation, what's left can probably be well-approximated without a calculator. – Gerry Myerson Jul 2 '12 at 3:53
Ahhh ok yes I think that might work let me go try! – Matt Jul 2 '12 at 3:55

To avoid a calculator, you certainly need Stirling's approximation for the factorials. So $P=\binom {1000}{500}2^{-1000} \approx \frac {1000^{1000}\exp(-500)\exp(-500)}{500^{500}500^{500}\exp(-1000)}\frac {\sqrt{2\pi 1000}}{\sqrt{2\pi 500}\sqrt{2 \pi 500}}2^{-1000}=\frac 1{\sqrt{2 \pi 250}}\approx \frac 1{\sqrt{1550}} \approx \frac 1{40}=0.025$ 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.
@Matt: The first $\approx$ is plugging in Stirling's. The first = is clearing the exp's, and the powers of 2,500, and 1000. The second $\approx$ is $\pi \approx 3.1$, so I could do $500 \pi \approx 1550$. The last is that I know $40^2=1600$ and 1600 isn't very different from 1550. My last paragraph puts values on these to show we are within the 5% error budget. – Ross Millikan Jul 2 '12 at 4:19