Given that it's arbitrary which questions they choose, call test A (the test that No.1 takes), say that A=(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
Now, we've defined person 1 as choosing (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1), so the prob of that happening is =1.
The probability of person 2 choosing (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) , however, is 1/(129,140,163), as there are 129,140,163-1 different tests they could have chosen