User blog comment:P進大好きbot/Large Number in Probability Theory/@comment-36957202-20181222171428/@comment-35470197-20181222233510

> after n steps the expected value for E anfter n

Are you considering the expected value of the expected value? Sorry, I could not understand what you precisely mean. Instead, I add a description.

\(E\) is the expected value of the return of this process. The probalibity of the event "the process halts at \(n\)-th loop" (i.e. \(i = 2^{n-1}\)) is \((\frac{1}{2})^{n-1} \times \frac{1}{2} = (\frac{1}{2})^n\) (\(n-1\) times lose and once win). Therefore \(E\) is the sum of \((\frac{1}{2})^n \times 2^{n-1} = \frac{1}{2}\).