For quantumnumber, I don't want to dispute mjc's explanation. If this were on an exam and taught, it is the answer.
Try drawing resonance structures for an acetal anion where the negative charge is stabilised by delocalisation onto the oxygen. You can't; the O would have too many electrons. But you can with a thioacetal, because the S has vacant 3d orbitals to accept the electrons. (I should say my edition of March from the 1980s states that the reasons for carbanion stabilisation by S and P "are in dispute", but the fact is that S atoms stabilise carbanions.)
However, one might wonder what was learned after 1980 and how it was proven? If you look at the energy levels shown here,
http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html, you might think it should be the 4s orbitals that should fill. If you read the page, perhaps the filling doesn't follow a simplified filling order.
H2S is a lot more acidic than water. It could seem as though all of those protons in the nucleus can simply attract electrons better. If so, then HI should be the strongest haloacid and iodide a good leaving group.
I can understand the d-orbital argument may seem attractive, but I am skeptical that it stands on an iron clad proof.
Back to the original question. Acetal anions can be found if you look hard enough, however just as you might have predicted, they are much less acidic. You may find an analogy in the haloforms, CHF3 is the least acidic, then CHCl3, CHBr3, and CHI3 the most acidic. (If anyone has read this far, I've searched for the pKa of iodoform, but unable to find it. If anyone could send me a calculated value, I'd appreciate it. Some interesting things happen with bond lengths.)