The blind archer's third shot

Question

A blind archer shoots at a target. His shots are independent and identically distributed: each lands at a random distance from the centre drawn from the same (continuous) distribution, with no memory of earlier shots. He cannot see where they land.

He fires two arrows and is told that the first landed closer to the centre than the second. He then fires a third arrow.

What is the probability that the third arrow is the closest to the centre of all three?

Accepted Answer

One third. The symmetry. The three arrows are independent draws from the same distribution, so before any information, each of them is equally likely to be the best of the three: probability 1/3 each (ties have probability zero for a continuous distribution). Does the information change anything? We are told the first arrow beat the second. That is information about the relative order of arrows one and two. The question is about whether arrow three beats both of them. The event "arrow three is best" is independent of the event "arrow one beats arrow two": if you list the 6 equally likely orde…

The blind archer's third shot

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