Six athletes of equal ability run a race, so that all finishing orders are equally likely and there are no ties. What is the probability that Arun finishes ahead of BOTH Bina and Chetan?

The other three runners are irrelevant, so look only at the order in which Arun, Bina and Chetan finish among themselves. Those three can be ordered in 3! = 6 equally likely ways, and Arun comes first in 2 of them — Arun, Bina, Chetan and Arun, Chetan, Bina. So the probability is 2/6 = 1/3. The same follows from symmetry: each of the three is equally likely to be the first of the three home. The value 1/2 answers the head-to-head question 'is Arun ahead of Bina', which ignores Chetan altogether. The value 1/4 comes from multiplying 1/2 by 1/2 as though 'ahead of Bina' and 'ahead of Chetan' were independent — they are not, since beating Bina makes it likelier that Arun ran a fast race and beat Chetan too. The value 1/6 is the probability that Arun wins the whole race against all five rivals, which is far more than was asked.