 \leqslant \epsilon) \geqslant 1 - 4 \exp\left(-\frac{2\epsilon^2}{N}\right).
\end{equation*}\]</span> Now set <span class="math inline">\(\epsilon = N^{0.55}\)</span> and let <span class="math inline">\(N=M \cdot 2^P\)</span> (so that <span class="math inline">\(N\)</span> is some multiple <span class="math inline">\(M\)</span> of <span class="math inline">\(2^P\)</span> where we shall choose <span class="math inline">\(M\)</span> large enough). The last inequality says that with probability at least <span