smc

Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation $$P={\left(\frac{1}{3}\right)}^{50}+\left(\genfrac{}{}{0ex}{}{50}{2}\right){\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^{48}+\cdots +{\left(\frac{2}{3}\right)}^{50}$$ This is essentially the expansion of ${\left(\frac{2}{3}+\frac{1}{3}\right)}^{50}$ but without the odd power terms. To get rid of the odd power terms in ${\left(\frac{2}{3}+\frac{1}{3}\right)}^{50}$, we add ${\left(\frac{2}{3}-\frac{1}{3}\right)}^{50}$ and then divide by $2$ because the even power terms that were not canceled were expressed twice. Thus, we have $$P=\frac{1}{2}\cdot \left({\left(\frac{1}{3}+\frac{2}{3}\right)}^{50}+{\left(\frac{2}{3}-\frac{1}{3}\right)}^{50}\right)$$ Or $$\frac{1}{2}\left(1+{\left(\frac{1}{3}\right)}^{50}\right)$$ which is equivalent to answer choice $\boxed{\frac{1}{2}(1+\frac{1}{3^{50}})}$.