If we replace the random variables with Gaussian variables having the same variances, namely formula_17 then the average at any point will also be normally distributed. The width of the distribution of the average will tend toward zero (standard deviation asymptotic to formula_18), but for a given ε, there is probability which does not go to zero with "n," while the average sometime after the "n"th trial will come back up to ε. Since the width of the distribution of the average is not zero, it must have a positive lower bound "p"(ε), which means there is a probability of at least "p"(ε) that the average will attain ε after "n" trials. It will happen with probability "p"(ε)/2 before some "m" which depends on "n". But even after "m", there is still a probability of at least "p"(ε) that it will happen. (This seems to indicate that "p"(ε)=1 and the average will attain ε an infinite number of times.)