impressive idea.

you can get the probabilities v1 < u2 < v2 etc using convolutions, its a triple integral though (ouch! should be perfectly doable though).

alternatively, can you make a symmetry argument you for the order of v1 , v2 , u2? since they are i.i.d it seems to me that they all must have the same chance of being in the middle, ie 1/3.