What is the correct way to use Stirling's approximation in this example?

[leonne]

Homework Statement

I don't really understand how to use Stirling's approximation. here's an example
you flip 1000 coins, whts the probability of getting exactly 500 head and 500tails

Homework Equations

N!=N^Ne^-N(2pieN)^1/2

The Attempt at a Solution

wht they did was
2¹⁰⁰⁰ total number outcome
\Omega(500)=1000!/500!² multiplicity
than used the Stirling's approximation to get this which I don't get how. What do I plug in for N
\Omega(500)=2¹⁰⁰⁰/(500pie)^1/2
P=\Omega(500)/2¹⁰⁰⁰= 1/(500pie)^1/2= .025

thanks

 

[sgd37]

you divide the stirling approximation for N=1000 by the square of the stirling approximation for N=500. And it's Pi not pie silly

 

[leonne]

lol thanks,
so it would be 1000¹⁰⁰⁰e^-1000(2pi1000)^1/2/ (500⁵⁰⁰e^-500(2pi500)^1/2)²

 

[sgd37]

yes and that simplifies to what you want

 

[leonne]

um I don't think that is right i keep getting overflow error.
nvm after typing it in wolfam alpha it worked but how would i do it on my calculator, o well lol

 

[Andrew Mason]

um I don't think that is right i keep getting overflow error.
nvm after typing it in wolfam alpha it worked but how would i do it on my calculator, o well lol

These are very big numbers. You might find it easier to reduce the fraction as a whole rather than work out the denominator and numerator separately.

AM

 

[fzero]

We're going to need a bigger calculator.

 

[Andrew Mason]

We're going to need a bigger calculator.

Not really. The expression reduces to:

\frac{\Omega(500)}{2^{1000}} = \frac{(2\pi *1000)^{1/2}}{2\pi*500} = \frac{1}{(500\pi)^{1/2}} = .025

AM

 

[mavanhel]

The Stirling's approximation you want to use for this is

\ln(n) \simeq n\ln(n) - n

This is applicable for n \gg 1. So, for your example

\omega = \frac{1000!}{500!500!}

Take the logarithm of both sides and we find

\omega = \ln(1000!) - 2\ln(500!)

Which, using Stirling's approximation, gives us

\omega = 1000\ln(1000) - 1000 - 2(500\ln(500) - 500)

At this point it is easy to find that \omega = 10^{300}