Let's assume that final digits are uniformly distributed in such measurements (Benford's law says this isn't the case for first digits, but that's beside the po

Let's assume that final digits are uniformly distributed in such measurements (Benford's law says this isn't the case for first digits, but that's beside the point). Then P(last_digit in size 5 set) =1/2 for a single measure. So the probability they all are is 1/(2^30), which is on the order of one in a billion as the other reply says (good heuristics: 2^10 ≈ 1000, 2^20 ≈ 1,000,000, 2^30 ≈ 1,000,000,000)