Solution:

By linearity of expectation, we can treat the number of turns each card contributes to the total independently. Let $f(x)$ be the expected number of turns a card of value $x$ contributes (we want $f(0.9)$). If we have a card of value $x$, we lose it after $1$ turn with probability $1-x$. If we don't lose it after the first turn, which happens with probability $x$, then given this, the expected number of turns this card contributes is $f(x) + \frac{1}{x} \int_{0}^{x} f(t) dt$. Thus, we can write the equation
$$f(x) = 1 + x f(x) + \int_{0}^{x} f(t) dt$$
Differentiating both sides gives us
$$f'(x) = x f'(x) + f(x) + f(x) \Longrightarrow \frac{f'(x)}{f(x)} = \frac{2}{1-x}$$
Integrating gives us $\ln f(x) = -2 \ln (1-x) + C \Longrightarrow f(x) = \frac{e^{C}}{(1-x)^{2}}$. Since $f(0) = 1$, we know that $C = 0$, so $f(x) = (1-x)^{-2}$. Thus, we have $f(0.9) = (1-0.9)^{-2} = 100$.