How can I calculate the mean number of attempts needed for success?

Question

Given that

$$P(\text{First success on the }n\text{th attempt}) =\left( 1 - 0.8(1/2)^{n-1}\right) * P(\text{failure of all previous attempts})$$ for $n = 1, 2, 3, \dots$

How can I calculate the mean number of attempts needed for success?

Answer 1

Let $X$ denote the number of attempts needed for success. Then $$ \mathbb P(X\geq k)=\prod_{n=1}^{k-1}0.8(1/2)^{n-1}=0.8^{k-1}(1/2)^{0+1+\ldots+(k-2)}=0.8^{k-1}\left(\frac12\right)^{\frac{(k-2)(k-1)}{2}} $$

The mean value of $X$ can be found as $$ \mathbb E[X]=\sum_{k=1}^\infty \mathbb P(X\geq k). $$ For a proof, look at wiki https://en.wikipedia.org/wiki/Expected_value#Finite_and_countably_infinite_case. Then $$ \mathbb E[X]=\sum_{k=1}^\infty \mathbb P(X\geq k)=\sum_{k=1}^\infty 0.8^{k-1}\left(\frac12\right)^{\frac{(k-2)(k-1)}{2}} $$ Wolframalpha says that this sum equal to 2.19073.