Jason plays a game involving darts thrown at a board, all aimed at the center. Here’s the puzzle setup:

The Problem

  • Jason throws two darts.
  • The second dart lands farther from the center than the first.
  • Now, Jason throws a third dart.

Question: What is the probability that the third dart lands farther from the center than the first?

All throws are independent and identically distributed around the center.

Reasoning

Let the distances from the center be:

  • \( X_1 \): distance of the first dart
  • \( X_2 \): distance of the second dart
  • \( X_3 \): distance of the third dart

We are told that:

\[ X_2 > X_1 \]

We are asked to compute:

\[ P(X_3 > X_1 \mid X_2 > X_1) \]

Key Insight

The distances \( X_1, X_2, X_3 \) are i.i.d. continuous random variables, so no two distances are exactly equal.

We condition on the event \( X_2 > X_1 \) and ask for the probability that \( X_3 > X_1 \).

Symmetry implies that all 6 permutations of the values \( X_1, X_2, X_3 \) are equally likely.

Given \( X_2 > X_1 \), we are equally likely to be in any of the 3 permutations of \( X_1, X_2, X_3 \) that satisfy \( X_2 > X_1 \). In two of these three, \( X_3 > X_1 \).

Conclusion

Thus, the probability that the third dart is farther than the first, given that the second dart was farther than the first, is:

\[ \boxed{\frac{2}{3}} \]

Reference