I apologize for this post being a bit late! It was December: a time for festivities, my birthday, vacation, ...and finals. I also almost didn't get my solution in on time, as I attempted the problem on the very last day of the year. Anyway, here is Jane Street's December 2025 puzzle, Robot Javelin :
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.
Before we attempt to exploit the "leak" in the game, let's first figure out how the fair version of the game works. We know the robots have settled into Nash equilibrium, so we should figure out the threshold for going for a second throw. Let's call this threshold $k$.
To find the optimal $k$, the probability of winning by keeping the first throw must equal the probability of winning by going for a second throw at the threshold $k$. So we have
If we keep a throw $\geq k$, then we only win if Spears throws lower than $k$. For Spears to throw lower than $k$, their first throw must be less than $k$ (probability $k$), and then their second throw must also be less than $k$ (probability $k$). Thus,
If we go for a second throw, we win if our second throw is greater than Spears' final throw. There are two scenarios: either they keep their first throw or go for a second throw.
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If Spears keeps their first throw, they must have rolled higher than $k$ (probability $1-k$). The probability that our second throw in $[0, 1]$ is greater than their first throw in $[k, 1]$ is $\frac{1 - k}{2}$. Therefore, the probability of us winning in this case is
\[ \text{Pr[We Win } |~(\text{We Throw Again } \cap \text{ Spears Keeps})] = (1-k) \cdot \frac{1-k}{2} = \frac{(1-k)^2}{2} \] -
If Spears goes for a second throw, they must have thrown less than $k$ (probability $k$). The probability that our second throw in $[0, 1]$ is greater than their second throw in $[0, 1]$ is $\frac{1}{2}$. Therefore, the probability of us winning in this case is
\[ \text{Pr[We Win } |~(\text{We Throw Again } \cap \text{ Spears Throws Again})] = k \cdot \frac{1}{2} = \frac{k}{2} \]
Summing these two cases gives us the total probability of winning by going for a second throw:
Setting the two probabilities equal to each other, we have
We take the positive root since $k$ must be in $[0, 1]$, so $k=\frac{\sqrt{5} - 1}{2} = \varphi^{-1}$, where $\varphi$ is the golden ratio! So in a standard game of Robot Javelin, the robots will go for a second throw if their first throw is less than $\varphi^{-1}$. Note that this value is approximately $0.618$.
Now let's consider the modified game where Spears can receive a bit of information if our first throw is below or above a certain threshold $d$. To maximize their chances of winning, Spears should choose a threshold of $d = k = \varphi^{-1}$. This is because if our first throw is below $\varphi^{-1}$, they know we will go for a second throw, and if our first throw is above $\varphi^{-1}$, they know we will keep our first throw. This choice of $d$ maximizes the information gained for Spears.
Spears can adjust their strategy accordingly by changing their original threshold of $\varphi^{-1}$ to a new threshold $t$ based on the bit they receive. If Spears receives a bit indicating our first throw is below $\varphi^{-1}$, they believe we will go for a second throw that will be uniformly distributed in $[0, 1]$. In this case, Spears should make $t$ equal to $\frac{1}{2}$ to maximize their probability of winning, as it is the median of the uniform distribution.
If Spears receives a bit indicating our first throw is above $\varphi^{-1}$, they know we have thrown a value in $[\varphi^{-1}, 1]$ and believe we will keep it. Spears can now do one of two things:
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Spears can rethrow if their first throw is below $t$. Regardless of $t$, they will win with probability $\frac{1 - \varphi^{-1}}{2}$, as the probability that their second throw in $[0, 1]$ is greater than our first throw in $[ \varphi^{-1}, 1]$ is $\frac{1 - \varphi^{-1}}{2}$.
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Spears can keep their first throw as long as it's above $t$. If Spears keeps a throw $\geq t$, they win only if our throw is lower than $t$. Since our throw is uniformly distributed in $[\varphi^{-1}, 1]$, the probability that Spears wins by keeping their first throw is $\frac{t - \varphi^{-1}}{1 - \varphi^{-1}}$.
Like before, to find Spears' new threshold $t$, the probability of them winning by keeping their first throw must equal the probability of them winning by going for a second throw at the threshold $t$. Thus, we have
which was simplified using the identity $(\varphi^{-1})^2 = 1 - \varphi^{-1}$. Therefore, Spears' new threshold is $t = 1 - \frac{\varphi^{-1}}{2}$.
So to recap, Spears' new strategy is to use the following conditional threshold:
Now we can calculate our updated probability of winning against Spears using their new strategy. There are two cases to consider based on our first throw:
If our first throw is above $\varphi^{-1}$ (probability $1-\varphi^{-1}$), we will keep our first throw. If we rethrow, then we reset to a random throw in $[0, 1]$, which has an average value of $\frac{1}{2}$. This is worse than keeping our first throw, so we will keep it.
If our first throw is below $\varphi^{-1}$ (probability $\varphi^{-1}$), something interesting happens. Spears assumes that if our throw is below $\varphi^{-1}$, we will always rethrow. They then lower their threshold to $\frac{1}{2}$, which creates an opening. In the standard game, a throw of $0.60$ is considered "bad" (because it's below $\varphi^{-1}$). But against an opponent who is willing to keep a $0.51$, a throw of $0.60$ is actually quite strong! Therefore, we should calculate a new threshold $j$ for ourselves to decide whether to go for a second throw.
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If we rethrow, we get a random value in $[0, 1]$. With a threshold of $t = \frac{1}{2}$, Spears has a $50\%$ chance of throwing below $t$, in which case they will rethrow. We have a $50\%$ chance of beating their second throw in $[0, 1]$. So there is a $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ chance of us winning by rethrowing if Spears rethrows. If Spears keeps their first throw, they must have thrown above $t$, which occurs with probability $\frac{1}{2}$. The probability that our random throw in $[0, 1]$ then beats their first throw in $[\frac{1}{2}, 1]$ is $\frac{1 - 1/2}{2} = \frac{1}{4}$. Thus, there is a $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$ chance of us winning by rethrowing if Spears keeps. So in total, we have
\[ \text{Pr[We Win } | \text{ We Throw Again}] = \frac{1}{4} + \frac{1}{8} = \frac{3}{8} \]