Uniform Distribution: Order Statistics 3

You are practicing your dart throwing skills. You aim for the center of the dartboard and throw two darts in succession. You observe that the second dart lands farther from the center than the first. If you decide to throw a third dart, aiming once again for the center, what is the probability that this third dart lands farther from the center than your first dart?

Answer

There are $$3!$$ (6) possible permutations of the distances at which the three darts can land. However, we are given the condition that the second dart lands farther from the center than the first dart, i.e., $$D_2 > D_1$$ . This condition reduces the number of possible permutations to 3:

$$D_1 < D_2 < D_3$$ , $$D_1 < D_3 < D_2$$ , $$D_3 < D_1 < D_2$$ ,

Out of these 3 permutations, the third dart lands farther than the first dart in 2 cases:

$$D_1 < D_2 < D_3$$ , $$D_1 < D_3 < D_2$$

Thus, the probability that the third dart lands farther from the center than the first dart is:

P(D_3 > D_1 \mid D_2 > D_1) = \frac{2}{3}

Notes and comments

Comment 1: Conditioning on the event $$D_2 > D_1$$ reduces the outcome space by half because, for any two darts, there are two possible orderings ( $$D_1 > D_2$$ and $$D_2 > D_1$$ ). By fixing $$D_2 > D_1$$ , we eliminate one of these possibilities, effectively reducing the total number of permutations from 6 to 3.

When dealing with 5 darts, conditioning on a similar event (e.g., $$D_2 > D_1$$ ) would reduce the outcome space by half, leaving $\frac{5!}{2} = 60$ possible permutations instead of $$5! = 120$$ .