Poisson distribution and price reduction A store owner has certain existence of an item and he decides to use the following scheme to sell it:
The item has a price of \$100. The owner will reduce the price in half for every customer that buys the item on a given day. That way, the first customer will pay \$50, the next one will pay $25, and so on... 
Suppose that the number of customers who buy the item during a day follow a Poisson distribution, with $\mu = 2$ Find the expected price of the item at the end of a day.

I defined a "final price" function like this:
$$g(x) = \frac{100}{2^x}$$
where $x$ is the random variable of the Poisson distribution:
$$f(x)=\frac{\lambda^xe^{-x}}{x!}$$
Since $\mu = \lambda = 2$, I can rewrite $f(x)$ like this:
$$f(x) = \frac{2^x}{e^2x!}$$
And there's where I'm stuck. I see two possible roads:
Assuming the final price is the price reduction after the last customer buys it. So for example if only 1 customer buys the item, then the final price is 50, which gives formula $P(X)=100/2^X$. Then since $X$ is Poisson with parameter $\mu=2$,
$$E[P(X)]=\sum_{x\geq 0} \frac{100}{2^x}\frac{1}{e^\mu}\frac{\mu^x}{x!}=\frac{100}{e^2}\sum_{x\geq 0} \frac{1}{x!}=100/e.$$
