ujm, I think my logic would work for more pairs:
the first pair is least reliable, as I described before, because it is easier for them to cheat selecting a target and time. The rest are more reliable than first and all equally reliable to each other, as they all need to testify about given people and time.
Let’s for simplicity say value (1) = 1, value (2) and later = 10.
For 2nd pair, easy 10 > 1.
For 3rd pair, their value is 10 + 1 from first pair, 11>10
for 4th pair 10 + 10 (4th, 2nd) > 10 + 1 (3rd, 1st)