Dart Game: Will the Next Throw Also Be Worse Than the First?
Jason throws darts at a board. Each throw's distance from the bullseye is an independent draw from the same continuous distribution (his skill neither improves nor deteriorates, and ties have probability zero).
(a) Jason's second throw lands farther from the bullseye than his first. He now throws a third dart. What is the probability that the third throw also lands farther from the bullseye than the first?
(b) Generalize: Jason's throws $2, 3, \ldots, n$ all land farther from the bullseye than his first throw. What is the probability that throw $n + 1$ also lands farther from the bullseye than the first?
Hints
- The temptation is to say 1/2 because the third throw is independent of the first two. But the information that throw 2 was worse than throw 1 tells you something about how good throw 1 was.
- Rephrase: the event that throws $2, \ldots, n$ are all worse than throw 1 is exactly the event that throw 1 is the best of the first $n$ throws. By exchangeability, that has probability $1/n$.
- $P(\text{throw 1 best of } n+1 \mid \text{throw 1 best of } n) = \dfrac{1/(n+1)}{1/n} = \dfrac{n}{n+1}$.
Worked Solution
How to Think About It: Every statement of the form "throw $k$ is farther than throw 1" is a statement about the ranking of the throws, and all $n!$ rankings of $n$ i.i.d. continuous draws are equally likely. Translate the conditioning event into a statement about ranks and the answer falls out.
Quick Estimate: After seeing that throw 2 was worse, throw 1 is "the better of two," which is like a draw from the upper half of the skill distribution. A fresh throw beats a typical upper-half throw well under half the time; the exact answer is $1/3$, so the third throw is worse with probability $2/3$.
Formal Solution:
Part (a): Three throws
*Step 1 -- Translate to ranks.* Let $D_1, D_2, D_3$ be the distances. "Throw 2 is worse than throw 1" is the event $D_1 < D_2$, and "throw 3 is worse than throw 1" is $D_1 < D_3$.
*Step 2 -- Use exchangeability.* All $3! = 6$ orderings of $(D_1, D_2, D_3)$ are equally likely. $D_1 < D_2$ holds in 3 of them, and $D_1 < D_2$ together with $D_1 < D_3$ (i.e. $D_1$ is the minimum) holds in 2 of them. So
$$P(D_1 < D_3 \mid D_1 < D_2) = \frac{2/6}{3/6} = \frac{2}{3}.$$
Part (b): $n$ throws, then one more
*Step 1 -- Restate the events.* $A_n = \{D_1 < D_k \text{ for all } 2 \le k \le n\}$ is the event that throw 1 is the best of the first $n$ throws. By exchangeability each throw is equally likely to be the best, so $P(A_n) = 1/n$.
*Step 2 -- Condition.* $A_n \cap \{D_1 < D_{n+1}\} = A_{n+1}$, the event that throw 1 is the best of $n+1$ throws, with probability $1/(n+1)$. Hence
$$P(D_1 < D_{n+1} \mid A_n) = \frac{P(A_{n+1})}{P(A_n)} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1}.$$
For $n = 2$ this recovers $2/3$. As $n$ grows the probability tends to 1: a throw that has beaten many others is almost surely an excellent throw.
*Alternative (integral) check for (a).* Let $F$ be the CDF of the distance. Conditional on $D_1 = d$, the other throws are worse independently with probability $1 - F(d)$, so
$$P(D_1 < D_2, D_1 < D_3) = \int (1 - F(d))^2 \, dF(d) = \int_0^1 (1-u)^2\,du = \frac{1}{3},$$
and dividing by $P(D_1 < D_2) = 1/2$ gives $2/3$.
Answer: (a) $\dfrac{2}{3}$. (b) $\dfrac{n}{n+1}$.
Intuition
Independence of the throws does not make the answer $1/2$, because conditioning on "throw 2 was worse" tells you throw 1 was probably a good throw, and good throws are hard to beat. The clean way to see it is through ranks: "all later throws worse than the first" just means "the first throw is the best so far," and the best of $n$ exchangeable draws stays the best after one more draw with probability $n/(n+1)$. This is the record-value calculation that underlies inspection of running maxima, "how often does a new high occur" questions, and Bayesian intuition about why a signal that has kept winning is likely to keep winning.