It's interesting. I'm being a bit circumspect, because my old Cotton & Wilkinson (3rd ed.) says as follows (p.535: "It is also well to point out that the interelectronic forces and variations in total nuclear charge play a large part in determining the configurations of ions. We cannot say that because 4s orbitals became occupied before 3d orbitals they are always more stable. If this were so, then we should expect the elements of the 1st transition series to ionise by loss of 3d electrons, whereas in fact they ionise by loss of 4s electrons first. Thus it is the net effect of all forces -- nuclear-electronic attraction, shielding of one electron by others, interelectronic repulsions and the exchange forces -- that determines the stability of an electronic configuration; and, unfortunately, there are many cases in which the interplay of these forces and their sensitivity to changes in nuclear charge cannot be simply described." So yes indeed it seems it is the 4s that get stripped off first. I suppose this makes sense in that taking out a 4s reduces the shielding of the 3d, which then drop in energy, leading to an overall lower energy ground state than you would get by taking out a 3d electron and leaving the 4s in situ. But in general my recollection of the transition elements is that the ability to predict their chemistry from their electronic configuration becomes rather weak, compared with what you can do in the s and p block. There are just too many interacting factors.