Both the above contain elements of the truth. The first is the more "traditional" mathsy way of doing such a thing. To be honest, although integrating is the proper way to work such a measurement out, on a reasonably sized coin the thickness is much smaller than the diameter so the small angle approximation will be appropriate - A~sinA~tanA - and you could approximate the length of the arc at the top of the coin to the width of the coin itself. Hence the surface area of the band around the middle is approximately 2pi*r*thickness of coin. The surface area of the sphere is 4pi*r^2, and hence the ratio of the two is t/2r, which interestingly is independent of pi. Thus a 10p piece at about 1mm thick and 12mm radius would have a chance of 1/24 ~ 4%.
Clearly, however, this is a gross overestimate because there are many complicating factors to be taken into account. One such complication is touched upon by bilm_ks, namely the spin. However, he/she effectively assumes that the coin is rotating on an axis parallel to the table and that the centre of gravity, when the coin stops, will determine whether it tips onto a face or onto the edge. However, this does not account for the fact that this is not how a coin hits the table. A real (flipped) coin will hit the table and bounce, end over end, until its angular and linear momenta are sufficiently small that it cannot flip to the next face. At this point, does it have just enough momentum to become vertical and not fall back (or forwards), or will it settle on the final face? I would argue for the latter, on the grounds that the coin has a lot more "spin" to lose than this - it will rock round and round on the table until it settles, although always on that final face (try it, to see what I mean). If it were to tip up onto the edge, it would have sufficient momentum to keep going as the frictional force on the average table is not enough to carry it away. Hence I think that on the average table the chance of actually landing on one edge from a flip is so small as to be practically zero.
bilm_ks's approach would be useful if you actually dropped the coin on edge or rolled it down a slide, to calculate how flat the table would have to be and how perfect the edge of the coin such that the protrusions were not enough to tip it over.
Alternatively, if you covered the table in a thick shag rug or frog saliva, then it may well cause the coin to stand in an upright position much more easily.
Incidentally, John von Neumann was once asked how thick a coin would have to be to have a 1/3 chance of landing on edge. He used the same approach as in Mike Hildner's link and in 20 seconds in his head calculated ~35% of the diameter of the coin.