This is a pretty mathematical question, so I will answer it in pieces as I get the time to work it out. To find the time it takes to fall, you start with Newton's second law, ma=m[d2r/dt2]=-GMm/r2=dv/dt and integrate it once to get v=dr/dt=√[2GM(R-r)/(Rr)] where, for your question, R is the distance from the center of the earth to the point from which it was dropped, r is where it is at the time t(r), G is the universal gravitation constant, and M is the mass of the earth. (The mass m of the ship does not matter.) (You can more easily get v(r) from energy conservation.) Next integrate v(r)=dr/dt to get t(r)=-2.69x105{0.5tan-1((2x-1)/(2√[x(1-x)]-√[x(1-x)]-π/4} where x=r/R; this is not a simple integral to perform, but you can find good integrators on the web, for example Wolfram Alpha. Once you have gotten the indefinite integral you are not finished, you need to find the integration constant which will put the ship at rest at x=r/R=1; that is where the π/4 comes from. Putting in your numbers, I find t=4.22x105 s=117 hr=4.89 days. The figure shows the graph of t(x).
Say the cube has sides numbered 1-6 that means there are 6 possible outcomes. 3 and 4 are only two possible outcomes so the probability of that is 2:6. Simply 1:3 or 33%
