Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is
$$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$
$$
\begin{array}{lc}
\hline
\text{compound} & K_\mathrm{eq} \\
\hline
\text{acetaldehyde} & 6 \times 10^{-7} \\
\text{acetone} & 5 \times 10^{-9} \\
\hline
\end{array}
$$
This means that acetone exists predominantly (if not wholly) in its keto form. For every one molecule of the enol form observed, we can expect $\pu{2E8}$ molecules of the keto form.
Because of this, any deprotonation that takes place in acetone will have to be via breaking a $\ce{C-H}$ bond, and that is not a simple thing to do when you compare it to deprotonating an $\ce{O-H}$ bond in ethanol. Thus ethanol is more acidic than acetone.