Jane Street’s May 2022 puzzle asks: in a robot swimming tournament played at Nash equilibrium, what is the probability $p$ that a robot devotes all its fuel to a single race?

The answer is p ≈ 0.999560 — robots almost always go all-in on one race, but not quite.

The Setup

24 robots compete in 8 races. Each robot has a fixed fuel budget and allocates it across the races however it chooses. In each race, the robot that spends the most fuel wins and advances to the finals. Ties are broken uniformly at random.

8 out of 24 robots advance, so each robot’s baseline win probability is $1/3$.

Nash Equilibrium and the Key Insight

A Nash equilibrium is a set of strategies where no player can improve their outcome by unilaterally changing their own strategy. In a symmetric game like this, the equilibrium is the same for every robot.

Why pure discrete fails: If every robot put all its fuel into one randomly-chosen race, a clever robot could deviate: spread fuel thinly across all 8 races, show up in every race, and win any race that nobody else targeted. That’s a profitable deviation — so pure discrete is not an equilibrium.

The mixed strategy: The equilibrium must mix between two strategies:

  • Discrete: put all fuel into a single randomly-chosen race.
  • Continuous: spread fuel across multiple races.

Each robot plays discrete with probability $p$ and continuous with probability $1 - p$. At the equilibrium $p$, a robot is completely indifferent between the two strategies — both yield the same win probability.

Since 8 of 24 robots advance, that win probability is exactly $1/3$. So the Nash equilibrium condition is simply:

\[W_C(p) = \frac{1}{3}\]

where $W_C(p)$ is the win probability of a continuous robot when all opponents play the mixed strategy. We don’t need to characterize the continuous strategy in detail — we just need to compute $W_C(p)$ and solve for $p$.

The f(i, j) Function

Before deriving $W_C$, we need a counting tool.

$f(i, j)$ is the number of ways to assign $i$ labeled robots to $j$ labeled races such that every race gets at least one robot — a surjection (onto function).

Example — $f(2, 2)$: Two robots (A, B) assigned to two races (1, 2), every race covered:

Robot A Robot B Race 1 covered? Race 2 covered?
Race 1 Race 2 ✓ ✓
Race 2 Race 1 ✓ ✓
Race 1 Race 1 ✓ ✗
Race 2 Race 2 ✗ ✓

Only the first two are valid, so $f(2, 2) = 2$.

More values:

  • $f(1, 1) = 1$ — one robot, one race, trivially covered.
  • $f(2, 1) = 1$ — both robots go to the only race.
  • $f(3, 2) = 6$ — equals $2^3 - 2$ (all assignments minus those where everyone picks the same race).
  • $f(3, 3) = 6$ — equals $3!$ (one robot per race, in any order).

Recurrence: For each new robot $i$, it either joins a race already covered ($j$ choices, leaving $f(i-1,j)$ ways for the rest) or opens a brand-new race ($j$ choices for which race it opens, leaving $f(i-1,j-1)$ ways for the rest):

\[f(0,0) = 1, \quad f(n,0) = f(0,n) = 0 \text{ for } n > 0\] \[f(i,j) = j \cdot \bigl(f(i-1,j) + f(i-1,j-1)\bigr) \quad \text{for } i,j > 0\]

This is the Stirling numbers of the second kind scaled by $j!$: $f(i,j) = j!\cdot S(i,j)$.

Computing $W_C(p)$

Our robot plays continuous. The 23 opponents each independently play discrete with probability $p$. We assume a continuous robot always places nonzero fuel in every race (an assumption the official solution takes as given).

The only thing that matters about the discrete robots is which races they cover — they outspend us in those races, so our only opportunities are the uncovered ones. We condition on:

  • $i$: how many opponents play discrete
  • $j$: how many distinct races those $i$ robots collectively cover
\[W_C = \sum_{i=0}^{23} \left[ \underbrace{\binom{23}{i} p^i (1-p)^{23-i}}_{\text{prob. of } i \text{ discrete opponents}} \cdot \sum_{j=0}^{7} \underbrace{\binom{8}{j} f(i,j) \left(\frac{1}{8}\right)^i}_{\text{prob. they cover } j \text{ races}} \cdot \underbrace{\min\!\left(1,\, \frac{8-j}{24-i}\right)}_{\text{our win prob.}} \right]\]

Outer sum — probability of exactly $i$ discrete opponents:

Each of the 23 opponents independently plays discrete with probability $p$, so the count is binomial:

\[\binom{23}{i} p^i (1-p)^{23-i}\]

Inner sum — probability that those $i$ robots cover exactly $j$ distinct races:

Each discrete robot picks a race uniformly from 8. We want the probability they collectively land on exactly $j$ distinct races. To count this:

  1. Choose which $j$ races get covered: $\binom{8}{j}$ ways.
  2. Assign $i$ robots to those $j$ races so every chosen race gets at least one robot: $f(i,j)$ ways (the surjection count from the previous section).
  3. Each robot picks its race independently and uniformly: probability $(1/8)^i$.

So the probability that $i$ discrete robots cover exactly $j$ specific races is $\binom{8}{j} \cdot f(i,j) \cdot (1/8)^i$.

Win probability given $(i, j)$:

The $j$ covered races are locked — discrete robots outspend us there. We only have a chance in the $8 - j$ uncovered races, competed for by all $24 - i$ continuous robots (us plus $23 - i$ others). By symmetry among continuous robots, our expected win probability is:

\[\frac{8-j}{24-i}\]

The $\min(1, \cdot)$ caps this at 1 for edge cases where continuous robots are outnumbered by uncovered races, guaranteeing we win at least one.

Worked example — $i=1$, $j=1$:

One opponent plays discrete and claims one race. We and $22$ other continuous robots compete for the remaining $7$ uncovered races. Our win probability is $7/23 \approx 0.304 < 1/3$ — the discrete robot consumed a race without us getting credit for it.

Solving for p

Setting $W_C(p) = 1/3$ numerically gives:

\[\boxed{p \approx 0.999560}\]

At Nash equilibrium, robots play discrete with ~99.96% probability. The continuous strategy gets a tiny but nonzero weight — just enough to make everyone indifferent.

Why is $p$ so close to 1? When $p$ is large, almost all opponents are discrete. With 23 robots each independently picking from 8 races, they cover on average $8\left(1-(7/8)^{23}\right) \approx 7.6$ of 8 races — leaving roughly 0.4 uncovered on average. A lone continuous robot wins those free races, giving $W_C > 1/3$. Equilibrium requires just a hair of continuous mixing to bring this back down to $1/3$.

Implementation

Note: This site is fully static — there is no server. The solver below runs entirely in your browser as JavaScript.

Step 1 — Precompute the surjection table

Storing $f(i,j)$ directly overflows: $f(23,8) > 10^{20}$, which exceeds 64-bit float precision. Instead define $g(i,j) = f(i,j) \cdot (1/8)^i$. Dividing the recurrence through by $8^i$:

\[g(i,j) = \frac{j}{8}\left(g(i-1,j) + g(i-1,j-1)\right)\]

All values of $g$ stay in $[0,1]$, and the $(1/8)^i$ factor drops out of the formula.

const G = Array.from({length: 24}, () => new Float64Array(9));
G[0][0] = 1;
for (let i = 1; i <= 23; i++) {
  for (let j = 1; j <= Math.min(i, 8); j++) {
    G[i][j] = (j / 8) * (G[i - 1][j] + G[i - 1][j - 1]);
  }
}

Step 2 — Evaluate $W_C(p)$

Direct translation of the formula above, using G[i][j] in place of f(i,j) * (1/8)^i.

function wc(p) {
  let total = 0;
  for (let i = 0; i <= 23; i++) {
    const pI = binom(23, i) * Math.pow(p, i) * Math.pow(1 - p, 23 - i);
    if (pI < 1e-300) continue;
    let inner = 0;
    for (let j = 0; j <= Math.min(i, 7); j++) {
      inner += binom(8, j) * G[i][j] * Math.min(1, (8 - j) / (24 - i));
    }
    total += pI * inner;
  }
  return total;
}

Step 3 — Bisect for the root

$W_C(0) = 1/3$ by symmetry (all continuous). As $p$ increases, $W_C$ dips well below $1/3$ as discrete robots cluster and cover most races. Then near $p=1$, $W_C$ climbs back above $1/3$ as opponents are nearly all discrete and leave uncovered races. We bisect in $[0.99, 1)$ to find where $W_C$ crosses $1/3$ again.

let lo = 0.99, hi = 1 - 1e-12;
for (let iter = 0; iter < 100; iter++) {
  const mid = (lo + hi) / 2;
  if (wc(mid) < 1 / 3) lo = mid; else hi = mid;
}
const p = (lo + hi) / 2;  // ≈ 0.99956044

100 iterations gives ~30 digits of precision — far more than the 6 significant figures required.











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