On page 19 http://www.iecn.u-nancy.fr/~chassain/GDT/documents/SpencerStFlour.pdf All in the first Paragraph.
it gives an estimate of (they use equal instead of approximation) $(1-p^{6})^{\binom{n}{4}}=e^{-n^{4+o(1)}p^{6}}$. I am familiar with the estimate $(1-p)\approx e^{-p}$ and so i understand $(1-p^{6})^{\binom{n}{4}} \approx e^{-\binom{n}{4} p^{6}}$. Im wondering how we can show that $\binom{n}{4}=n^{4+o(1)}$?
Secondly it says this estimate works well for small $p$ but for $p=1/2$ it gives the estimate $e^{-n^{4+o(1)}}$ surely that should be $e^{-n^{4+o(1)}(0.015625)}$? Where has $(1/2)^{6}$ gone?
In both cases you must realize that any nonzero constant is in the class $n^{o(1)}$. In the first case, $$ {n\choose4}=\frac{n(n-1)(n-2)(n-3)}{24}\sim\frac{n^4}{24}=n^{4-\log(24)/\log(n)}=n^{4+o(1)}, $$ and in the second case, at least when $p$ is positive and does not depend on $n$, $$ n^{4+o(1)}p^6=n^{4+o(1)+6\log(p)/\log(n)}=n^{4+o(1)}. $$ Note finally that the class $n^{o(1)}$ contains many more functions than the constants, for example every power of $\log n$ is in this class.