A bit of probability theory: if X has the Uniform(0,1) distribution, the support (range) is [0,1]. If the support is actually [0,1), then the distribution is actually Uniform(0, 1-epsilon) for some small epsilon. In this case, if you knew what epsilon is, you could scale by the width of the support to get Y = X/(1 - epsilon - 0) = X/(1 - epsilon) and Y would have true Uniform(0,1) distribution.

As others have pointed out, there are a number of factors involved and epsilon likely isn't just the machine epsilon or unit roundoff. Besides, even if it were, see pryrt's reply Re^6: inclusive rand. The figure in question is too small; on normal floating point hardware, dividing by 1-epsilon would be no different from dividing by 1.

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