2025-12 ~ Jane Street Puzzle ~ Robot Javelin

Puzzle Prompt

It’s coming to the end of the year, which can only mean one thing: time for this year’s Robot Javelin finals! Whoa wait, you’ve never heard of Robot Javelin? Well then! Allow me to explain the rules:

• It’s head-to-head. Each of two robots makes their first throw, whose distance is a real number drawn uniformly from [0, 1].
• Then, without knowledge of their competitor’s result, each robot decides whether to keep their current distance or erase it and go for a second throw, whose distance they must keep (it is also drawn uniformly from [0, 1]).
• The robot with the larger final distance wins.

This year’s finals pits your robot, Java-lin, against the challenger, Spears Robot. Now, robots have been competing honorably for years and have settled into the Nash equilibrium for this game. However, you have just learned that Spears Robot has found and exploited a leak in the protocol of the game. They can receive a single bit of information telling them whether their opponent’s first throw (distance) was above or below some threshold d of their choosing before deciding whether to go for a second throw. Spears has presumably chosen d to maximize their chance of winning — no wonder they made it to the finals!

Spears Robot isn’t aware that you’ve learned this fact; they are assuming Java-lin is using the Nash equilibrium. If you were to adjust Java-lin’s strategy to maximize its odds of winning given this, what would be Java-lin’s updated probability of winning? Please give the answer in exact terms, or as a decimal rounded to 10 places.

My Solution

Robot Javelin Puzzle – Final Proof

1. Nash Equilibrium of the Original Game

Assume both players use a threshold strategy with threshold $t$. Let the opponent’s final score be $Y$, and define

Distribution of $Y$

• If $y \le t$, the opponent must have rerolled:

• If $y \ge t$, either:
  • the first throw is kept ($Y_1 \in [t,y]$), or
  • the first throw is rerolled and $Y_2 \le y$

Thus,

So

Expected Value of $Y$

Using the tail integral formula,

Compute:

Indifference Condition

At equilibrium, keeping or rerolling at $x=t$ must give equal win probability:

This yields:

2. Optimal Strategy of the Cheating Player (Stage II)

The cheating player (Spears) observes whether

Case 1: $J_1 < c$

Java must reroll, so her final score is uniform on $[0,1]$. Spears compares:

• Keep $s$: win probability $= s$
• Reroll: win probability $= \frac12$

Thus,

Case 2: $J_1 \ge c$

Java keeps her first throw, so

CDF:

Reroll win probability:

Indifference gives:

3. Java-lin’s Optimal Counter-Strategy

Focus on the branch $J_1 < c$, where Spears uses $t_0 = \frac12$.

Spears’ Final Distribution

Reroll Win Probability

Indifference Condition for Java

Let $a_0$ be Java’s threshold in this branch. Indifference:

Since $a_0 > 0.5$,

Thus Java keeps iff $J_1 \ge \frac{7}{12}$.

4. Java-lin’s Final Winning Probability

Partition $J_1 \in [0,1]$:

A. $J_1 \in [0,\frac{7}{12})$

Java rerolls:

B. $J_1 \in [\frac{7}{12}, c)$

Java keeps, Spears uses $t_0 = \frac12$:

C. $J_1 \in [c,1]$

Spears uses $t_1 \approx 0.690983$:

Final Answer

That is,

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