Question

Consider flipping N fair coins (assume N is even) (a) 1 point] What is the probability of getting half heads and half tails, in no particular order? 1, simplify your answer, and show the limiting behavior of the (b) [1.5 points] Assume that N probability as N -> co.

Answer 1

Answer with Explanation:

The problem can be simplified as follows

The number of possible outcomes after tossing a coin N times is
`2^N` since for each toss 2 outcomes are possible

Since we need equal heads and equal tails the no of cases amont the
`2^n` cases are

`\binom{N}{N/2}=(N!)/((N-N/2)!(N/2)!)=(n!^2)/((n/2)!^2)`

Thus the required probability is

`P(E)=((n!)/((n/2)!^2))/(2^n)=(n!)/((n/2)!^2\cdot 2^n)`

Part 2)

For the limit as N approaches infinity we have

`P(E')=\lim_(n\rightarrow \infty )(n!)/((n/2)!^2\cdot 2^n)`

Using Stirling's approximation and solving we get

`\lim_(n\rightarrow \infty )\binom{n}{i}\approx \sqrt{(n)/(2\pi i(n-i))}* (n^n)/(i^i(n-i)^i)\\\\\lim_(n\rightarrow \infty )\binom{n}{n/2}\approx \sqrt{(2n)/(\pi n^2)}* (n^n)/((n/2)^(n/2)\cdot (n/2)^(n/2))\\\\\lim_(n\rightarrow \infty )\binom{n}{i}\approx \sqrt{(2n)/(\pi n^2)}* 2^n\\\\P(E')=\frac{\sqrt{(2n)/(\pi n^2)}* 2^n}{2^n}=\sqrt{(2)/(\pi N)}`