I am afraid that it won't do. With your method, the probability of a random text token having k letters would be roughly p*(1-p)**(k-1) where p is the probability of inserting a space (1/10 or 1/6 in your examples).
One way to test "Nullspace" theories is to remove all spaces from the VMS text, then re-insert them according to the proposed method. If the theory is correct, the resulting text should have the same word statistics and structure as the original. The above space-insertion methods would definitely fail this test.