Dart Game

Jason throws darts at a board with constant skill. His second dart lands farther from the centre than his first. If he throws a third, what is the probability that it lands farther from the centre than his first?

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#Rank, do not measure

Let $d_1, d_2, d_3$ be the three distances from the centre. Constant skill makes them independent draws from one distribution, so all $3! = 6$ orderings of their sizes are equally likely and the exact distribution never matters. I want $\PP(d_3 > d_1 \mid d_2 > d_1)$ , and both pieces fall to symmetry.

#Two symmetries

The joint event $d_2 > d_1$ and $d_3 > d_1$ says $d_1$ is the smallest of the three, the closest dart, which is one of three darts equally likely to be nearest,

\PP(d_2 > d_1 \text{ and } d_3 > d_1) = \PP(d_1 = \min) = \tfrac{1}{3}. \tag{1}

The condition $d_2 > d_1$ holds half the time. Dividing,

\PP(d_3 > d_1 \mid d_2 > d_1) = \frac{1/3}{1/2} = \frac{2}{3}. \tag{2}

nearer the centrefarther out

d1d2d3d3 farther than d1

d1d3d2d3 farther than d1

d3d1d2d3 nearer than d1

Knowing the second dart landed farther than the first leaves three equally likely orderings of the three distances. The third dart is farther than the first in two of them, so the probability is two thirds.

#The intuition

The naive guess is one half, treating the third dart as a fresh coin against the first. But learning that the first beat the second is evidence that the first was a good throw, close to the centre, so a later dart clears it more often than not. The information pulls the answer up to two thirds.