probability and expected value Hey I am not sure if I thinking correctly on this question? In a carnival, there is game which charges you $3$ dollars to play a game. You win $1$ dollar for every consecutive head you get and you you can play till you get tail. if you get head head tail you get back two dollars. What is expected profit?

my theory is your probability of t,ht,hht,hhht and so on will be $1/2,1/4,1/8,1/16$ and so on, and income will be $0,1,2,3$ and so on.

so average income will be 
$$ 0 * \frac{1}{2} + 1 * \frac{1}{4} + 2 * \frac{1}{8} + 3 *\frac{1}{16} + ....$$

which will come to a dollar?

so average profit will be, $1-3 = -2$??
okay I tried to solve this one,

$$ [a+(n-1)d] r^{n-1} $$ is term used to show nth term of Arithmetico-geometric sequence

if I take $1/2$ common from the whole series I get

$$ 0 * 1  + 1 * \frac{1}{2} + 2 * \frac{1}{4} + 3 *\frac{1}{8} + ....$$

which gives me$$ a = 0, d=1, r=1/2$$

now the sum is

$$\lim_{n \to \infty}S_{n} = \frac{a}{1-r}+\frac{rd}{(1-r)^2}$$

which is $2$.

and after multiplying with previous $1/2$ we took out, we get $1$.

so average profit is $$1-3 = -2$$

reference Wikipedia https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence
The problem states that you CAN play until you get tails, so according to my understanding you can stop even if you don't get tails. So your calculations are correct but you should add in the factor of stopping of getting multiple consecutive heads. So the probability of getting 1 head is 1/2 and you get $1, two heads is 1/4 and $2, and so on. Adding this all up gets $2, and adding this to your calculation gets $3, which means the expected profit is $0-- according to my understanding.