Simple counting exercise. The number of subsets of $\{1,2,\ldots ,2n \}$ such that the smallest element is equal to $2i+1$ is $2^{2n-2i-1}$, and so in total we have $2^1+2^3+\ldots+2^{2n-1}=\dfrac{2(4^n-1)}{3}$ such subsets, and so the answer is $\dfrac{2(4^n-1)}{3 \cdot 4^n}$

The ratio is $\frac{2}{3}$. We proceed by induction. The case $n=1$ is obvious with the subsets in question being $\{ 1 \}, \{ 2 \}, \{ 1, 2 \}$. We now consider the case for $n+1$. Let $S$ be the set for $n$. We note that the set of subsets including $2n+1$ but not $2n+2$ has a bijection to the powerset $\{ 1, 2, \cdots , 2n \}$. Analagous holds for $2n+2$ and $\{ 2n+1, 2n+2 \}$. Note that in any of these the minimum does not change. We then consider the subsets $\{ 2n+1 \}, \{ 2n+2 \}, \{ 2n+1, 2n+2 \}$. We can now compute the new ration using our incuction argument: $ \frac{OddMIN}{Subs} = \frac{4\cdot \frac{2}{3}(2^{2n}-1) + 2 }{2^{2n+2}-1} = \frac{2}{3}\cdot \frac{2^{2n+2}-4+3}{2^{2n+2}-1} = \frac{2}{3} $