Let $X$ and $Y$ be independent and identically distributed random variables, find $P(X \geq 2Y)$ 
Let $X$ and $Y$ be independent and identically distributed random variables with probability mass function $p(n)=\frac{1}{2^{n}}$. Find $P(X \geq 2Y)$.

My attempt: \begin{align} P(X \geq 2Y) & = \sum\limits_{y=1}^{\infty}\sum\limits_{x=2y}^{\infty}P(X) \\ & =  \sum\limits_{y=1}^{\infty} \left( \frac{1}{2^{2y}}+\frac{1}{2^{2y+1}}+\frac{1}{2^{2y+2}} + \ldots \right)  \\ & = \sum\limits_{y=1}^{\infty}\frac{1}{2^{2y}}  \left( 2 \right) \\ & = \frac{2}{3}.\end{align}
$\frac{2}{3}$ however does not match the answer at the back. What am I doing wrong? Help!
(The answer is given to be in the interval $[0.27,0.3]$.)

We have $$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \mathbb{P}(X \geq 2Y, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \mathbb{P}(X=x, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \frac{1}{2^x} \frac{1}{2^y}.$$
Thus,
$$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \frac{1}{2^y} \frac{2}{2^{2y}} = 2 \sum_{y \geq 1} \frac{1}{2^{3y}}= \frac{2}{7}.$$
