For a single word the distribution of R is a geometric distribution with p = 1/2, and its standard deviation is sqrt(2) ≈ 1.41.
So for a word with hash ending in 100000b the algorithm will, indeed, yield 25/0.77351 = 41.37. But the probability of that is only 1/64, which is consistent with the statement that the standard deviation of R is close to 1.