If I have a coin with $\ 0.6$ probability of getting $\ H $and I throw it until I get $\ H $ for the second time. If $\ Y $ is the number of $\ T$ I get and $\ 2^Y $ is my revenue. how do I calculate my revenue expectancy ?
If I set $\ X $ as my total number of throws then $\ Y = X-2 $
I know $\ X \sim NB(2,0.6) $ so $\ E[X] = \frac{2}{0.6} = \frac{10}{3} $
$$\ E[2^Y] = E[2^{x-2}]$$
but now I need some algebraic manipulation so I'll have something like $\ E[aX + b] $ ?
I also tried different direction based on the probability function of negative binomial $\ P\{X = i\} = { i-1 \choose r-1 } (1-p)^{i-r} \cdot p^r$ $$\ E[2^Y] = E[2^{x-2}] = \sum_{x} g(x)\cdot p_x(x) = \sum_{i=2}^\infty 2^i \cdot {i-1\choose r-1} \cdot 0.4^{i-2} \cdot 0.6^2 = \sum_{i=2}^\infty 2^i \cdot { i-1 \choose 1 }\cdot 0.4^{i-2} \cdot 0.6^2 = \sum_{i=2}^\infty 2^i \cdot(i-1) \cdot 0.4^{i-2}\cdot 0.6^2$$
Any suggestions?