MHB Coin Toss Probability: Comparing $P_n$ and $P_{2n}$ for $2n$ and $4n$ Tosses

[anemone]

Let $n$ be a positive integer. Let $P_n$ be the probability that in $2n$ tosses of a fair coin exactly $n$ heads occur, and $P_{2n}$ the probability that in $4n$ tosses of a fair coin exactly $2n$ heads occur. Which is larger, $P_n$ or $P_{2n}$?

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[anemone]

Congratulations to lfdahl for his correct solution!:)

Suggested Solution:

Spoiler

Note that $P_n$ is always larger. Thus, it suffices to show that $P_n$ is a decreasing function of $n$, and for this it suffices to show that $P_n>P_{n+1}$ for all $n$.

We have $P_n={2n \choose n}\left(\dfrac{1}{2} \right)^{2n}$ and $P_{n+1}={2n+2 \choose n+1}\left(\dfrac{1}{2} \right)^{2n+2}$.

So,

$\begin{align*}\dfrac{P_n}{P_{n+1}}&=\dfrac{{2n \choose n}}{{2n+2 \choose n+1}}\dfrac{\left(\dfrac{1}{2} \right)^{2n}}{\left(\dfrac{1}{2} \right)^{2n+2}}\\&=\dfrac{(2n)!}{n!n!}\cdot\dfrac{4(n+1)!(n+1)!}{(2n+2)!}\\&=\dfrac{4(n+1)^2}{(2n+1)(2n+2)}\\&=\dfrac{2n+2}{2n+1}\\&>1\end{align*}$

Thus $P_n>P_{n+1}$ for all $n$. It follows that $Pn>P_m$ whenever $m>n$. In particular, $P_n>P_{2n}$.