Dart Distance Probability
Jason throws three independent, identically distributed darts. We are told that the second dart landed farther from the center than the first. What is the probability that the third dart also lands farther than the first?
2/3
≈ 66.7%
The key idea: only the ranking matters
Let D₁, D₂, D₃ be the three distances from the center. Since the throws are i.i.d. and ties have probability zero for a continuous distribution, all six orderings are equally likely.
Orderings satisfying D₂ > D₁
3
Also satisfying D₃ > D₁
2
Conditional probability
2/3
Condition on what we know
Once we know D₂ > D₁, exactly three equally likely rankings remain:
1
D₁ < D₂ < D₃
✓ D₃ > D₁
2
D₁ < D₃ < D₂
✓ D₃ > D₁
3
D₃ < D₁ < D₂
✗ D₃ < D₁
Two of the three possible rankings work, so
P(D₃ > D₁ | D₂ > D₁) = 2/3.
Monte Carlo check
Simulate 100,000 triples, keep only trials where D₂ > D₁, and measure how often D₃ > D₁.
Expected result: about 66.7%.
The exact shape of the dart-distance distribution does not matter, as long as the three
throws are independent, identically distributed, and ties occur with probability zero.