Taking $4$ shots or $6$? 
Suppose your skill (the ability to make a shot) follows a binomial distribution (with some unknown $p$). You need to make at least half of the shots. Would you rather take $4$ or $6$? More precisely, what should be the $p$ which is the threshold of your decision?

My Try: assuming $0 < p < 1$.
$${{6}\choose{3}}p^3(1-p)^3 ={{4}\choose{2}}p^2(1-p)^2 \\
\implies p(1-p) = 6/20
$$
but apparently the answer is: $3/5$. 
What's wrong with my calculation?

Probability of making AT LEAST half shots out of 6 is 
$$
\binom{6}{3}p^3(1-p)^3 +\binom{6}{4}p^4(1-p)^2 +\binom{6}{5}p^5(1-p) +\binom{6}{6}p^3 
$$
and for making at least 2 out of 4 is
$$
\binom{4}{2}p^2(1-p)^2 +\binom{4}{3}p^3(1-p) +\binom{4}{4}p^4
$$
For threshold $p$, equating both, we get:
$$
10p^4-36p^3+48p^2-28p+6 = 0
$$
which reduces to 
$$
2(x-1)^3(5x-3) = 0
$$
which gives $p = {\frac 35}$ since $0 \lt p \lt 1$.
