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)