User blog comment:P進大好きbot/Large Number in Probability Theory/@comment-36957202-20181222171428/@comment-35470197-20181223121827

I guess that "\(n\) iterations" in your context means the \(n\)-th loop. The average value of the return under the condition that you have already won the flips \(n\) times is also \(\infty\). The conditional expected value \(E_n\) can be computed in the following way: \begin{eqnarray*} E_n = \lim_{N \to \infty} \sum_{i=1}^{N} \left( \frac{1}{2} \right)^i 2^{n+i-1} = \lim_{N \to \infty} 2^{n-1} N = \infty \end{eqnarray*}