The large-board limit

1The model

The board has n cells, with n growing without limit, and time t counts shots as a share of the board. The fleet is {3,3,2}. A ship counts as found at its first hit: finishing it costs a few shots, the same for both racers, which vanish next to n. T is the moment the last ship is first hit. Edges, corners and ships touching each other affect a vanishing share of layouts and are ignored.

Two choices of the hunter's make every fixed plan's finishing time exactly computable. Before the game it draws a random offset: a diagonal direction, one of that direction's three sets of every-third diagonals, and which half of that set is A (section 2). The hider cannot tell which diagonals are which, so only the ships' positions relative to each other matter. Inside each part of the board (A, B and C) it fires in uniformly random order. A cell's place in a random order is equally likely to be anywhere, so each ship's cell in that part comes up at a uniformly random moment of the part's sweep, independently across ships. The hunter uses only what the rules announce: hit, miss, and which ship sank.

Patterns. The hider's whole choice is then a pattern: where the second 3 and the 2 sit relative to the first 3, counted in diagonals modulo 6 along the direction the hunter drew, and written "+1, +4" (the second 3 one diagonal over, the 2 four over); 36 choices, 12 up to swapping the 3s and reflecting.

Scoring. The pace is the plain plan's finishing time. Its law G(t), the chance it has finished by time t, is the same on every layout (section 3), so the yardstick does not depend on the hider. A strategy's race win rate is its chance to finish before an independent run of the plain plan: when the strategy finishes at T, the pace is still running with chance 1 − G(T), so

W = E[1 − G(T)].

Ties have probability zero. A strategy's guarantee is W on the hider's best choice.

What the model is not. It is not a race in which both players adapt: two optimal players race at 50% by symmetry, and that game is not solved. The finite boards' self-play mixtures race their own pace, so their "about 50%" measures something else.

2Why every third diagonal

A straight 3-cell ship covers three neighboring diagonals, x + y = s, s + 1 and s + 2, and exactly one of any three consecutive integers is divisible by 3. So the cells with x + y divisible by 3, every third diagonal and a third of the board, touch every 3 in exactly one cell.

Fewer is impossible. Each row splits into disjoint runs of three cells, and each run must hold a cell of the set, so a third of the board is the minimum (Dumitrescu and Tkadlec 2024 prove the general bound).

Only diagonals reach it (our short argument; Figure 2). Take a wrap-around board whose side n is a multiple of 3, and a set of cells with exactly a third of the board that touches every 3. Each row holds at least n/3 cells, one per run; with n²/3 in all, each holds exactly one per run, wherever the runs start. Runs starting at i and at i + 1 share two cells, so cell i is in the set exactly when cell i + 3 is: every third cell, at some phase. Columns likewise. A column holds a row's cell exactly when that row's phase matches the column, so for each column to get every third cell, the rows with each phase must themselves be every third row. So the phase is a function of the row number modulo 3 that takes each value once, and every such function is r ↦ r + k or r ↦ −r + k: two directions, three phases, six sets. On 10×10 the three phases of one direction hold 34, 33 and 33 cells.

A, B and C. The set's diagonals alternate checkerboard colors. A is its half on one color, picked at random as part of the offset; call that color dark. A 2-cell ship covers two neighboring diagonals, so it always has a dark square. A is a sixth of the board and does double duty: it hunts the 3s and is a third of the dark squares, which every 2 touches. B is the set's half on light squares (a sixth). C is the remaining dark squares (a third). The remaining light squares, a third of the board, never need a search shot.

Ship types. Under the random offset, a 3 has its one set cell in A or in B, with chance 1/2 each. A 2 touches A (1/3); is hooked to B, with one cell in B and one in C, and so has two chances (1/3); or lies only in C, next to a never-needed cell (1/3). A board type says where the 3s' set cells fall and which kind the 2 is. Figure 1 shows one ship of each kind.

3Three plans and their exact laws

The plain plan sweeps A and B together in random order (time 0 to 1/3), then C (1/3 to 2/3). Each 3 is found at a uniform moment of the first sweep, so by time t with chance 3·t. The 2 touches A or B with chance 2/3 and is then also uniform on the first sweep; otherwise it lies only in C and is uniform on the second. Multiplying, and adding up 1 − G for the average:

G(t) = 3t · 3t · (2/3)·3t = 18t³ up to t = 1/3, then G(t) = t + 1/3; E[T] = ∫(1 − G) = (1/3 − 1/18) + 1/18 = 1/3.

All three ships are found by the end of the first sweep with chance 2/3. No step depends on where the 3s lie, so the law is the same on every layout: the hider has nothing to aim at. 3-parity, which moves on to a second every-third set instead of C, has the same law, since a 2 that the first set misses has exactly one cell in each of the other two sets.

The adaptive plan sweeps A (0 to 1/6), then B while a 3 is afloat. At the second 3-sink it sweeps what is left of A together with C, in random order. Example, all three ships in A: the 3s are found at the later of two uniform moments of A's sweep, 1/9 of the board on average. One time in three the 2 is still afloat then; its cell is mixed into what is left of A and all of C and found halfway through on average, which adds 5/72. Together 13/72, against 1/8 if the plan finished A first; Table 1 gives every board type. By where the 3s lie, the averages are 19/72 (both in A), 1/3 (one in each half) and 19/54 (both in B), with chances 1/4, 1/2 and 1/4:

E[T] = ¼ · 19/72 + ½ · 1/3 + ¼ · 19/54 = 277/864 ≈ 0.321.

On its worst patterns (section 4), four of the 12, it still needs exactly 1/3. Adapting lowers the average, not the worst case (Figure 3).

Always jumping is the same plan, except that when the first 3 sinks during A it switches to B at once, and returns to A only if a 3 survives B. Against a random hider its average is exactly 1/3. Its worst is 41/108 ≈ 0.380 of the board, on a pattern where both 3s always share a half, so it often sweeps an empty B. In the worst case it is the slowest of the three.

4The race, and what a hider can aim at

Cases. The hunter's random offset turns a pattern into six equally likely cases, each of a board type in Table 1. The second 3's shift fixes how often the 3s split across A and B: never (shift 0), in 2 of 6 cases (shift 1 or 5), 4 of 6 (2 or 4), 6 of 6 (3); the 2's shift fixes which board types go together. That is all the hider controls: no ship's own chances change.

The adaptive plan's worst pattern, +1, +4: 47.56%. Its six cases come in three pairs. With the 3s split and the 2 only in C, the adaptive plan finishes A before B: 30.07%, against 39.96% for always jumping. With both 3s in A and the 2 hooked to B, its early switch to C gives up the 2's B route: 53.44% against 59.17%. With both 3s in B and the 2 touching A, the plans agree at 59.17%. So +1, +4 combines the adaptive plan's two weak spots: split 3s with the 2 in C, and 3s sharing a half with the 2 hooked to the other half. Splitting alone is not enough: the patterns that split the 3s in all six cases also bring easy ones and hold the plan only to 48.67%. The essay's short version, 3s in opposite halves, is the first weak spot.

Always jumping's worst pattern, 0, 0: 37.65%. The 3s share a half in every case. When both are in A and the 2 lies only in C, the jump sweeps an empty B: 21.56% against 53.44%.

Against a random hider the adaptive plan wins 50.95% and always jumping 49.04%; the adaptive plan's best pattern gives 55.88%. A hider who knows the rule turns the adaptive plan's one-point gain into a 2.44-point loss.

5Mixing, probing and the die roll

The blend. A secret coin, tossed before the game, picks always jumping with probability q and the adaptive plan otherwise (Figure 5, left). The best guarantee is 49.40% at q = 35.28%, where +1, +4 (which punishes never jumping) crosses 0, +3 (3s always in one half, which punishes always jumping). The hider's best reply plays +1, +4 in 39.5% of games and 0, +3 in 60.5%, so no blend of these two plans beats the plain plan. In shots the blend does break a third: at q = exactly 1/3 it needs at most 211/648 ≈ 0.3256 of the board, 2.3% below a third (Figure 5, right).

One plan beats 50%. The game's own randomness can stand in for the coin. Jumping only when the first 3 sinks in the first 19% of A's sweep already guarantees 49.93% (exact). A probe does better: after the first sink, sweep a set share b of B, then finish A, then B. B is fired in random order, so a 3 hidden there turns up in the probe with chance exactly b: the probe is a coin with odds b, tossed only when it matters. The second plan of Table 3, with two probes, guarantees 50.69% (exact; 50.73% with tuned probes). Beating the plain plan needs no die and no computed table.

The die roll, recomputed. The earlier study's mixed strategy rolls once before the game: two times in three the adaptive plan, one in four always jumping, one in twelve a responder, the best reply to a fixed mix of four hider patterns. Its table was not published, so we recomputed it by dynamic programming and scored it exactly (Table 2). With that responder the die roll guarantees 49.12% (exact, on +1, +4), not the reported 50.2%; the earlier study's own published decision maps give the same 49.1%. On +1, +4 the responder would need 64.1% and wins 50.7%. At a one-in-twelve share no responder can fix this: the die roll stays at most 50.38% (estimate, for responders that fire only A and B while a 3 is afloat). A three-rule responder (Table 3, right) gets 50.34% (exact).

The sweep-order rule. When the last sweep starts, a share w of B has been shot. Every cell of C has two neighbors in B. A surviving 2 hooked to B sits on an unshot B cell, so its C cell has two unshot B neighbors with chance 1 − w, one with chance w, never none. Sweep C in tiers: two unshot B neighbors first, then one, then none, holding (1 − w)², 2w(1 − w) and w² of C. Averaging over the hooked 2's tier, the share of C swept before it is hit falls to

(1 − w)³/2 + w(1 − w), against 1/2 in random order

(Figure 6). A 2 only in C is still found halfway through C on average, so the rule costs nothing. Sweeping B in long runs along its diagonals sharpens it: a C cell's two B neighbors sit side by side on one B diagonal and are shot together, so C splits into the cells next to unshot runs, which hold every surviving hooked 2, and the rest. The rule alone beats the plain plan by more than a point: the plain plan with this last sweep, the first plan of Table 3, wins exactly 37/72 ≈ 51.39% on every pattern, so the hider has nothing to aim at. The same last sweep also improves the essay's hand-played card on a real 10×10 board (the classic page). The die roll gains little from the rule: at most 50.81% (estimate). With probes, one plan does better: the third plan of Table 3 guarantees 51.77% (exact; 51.87% with probes of 24.3% and 79.5%).

6How much can cleverness buy?

Figure 7 collects the guarantees. Mixing four probe plans gives 50.86% with C swept at random and 51.95% with the sweep-order rule and B in runs (exact). Among plans that fire only A and B while a 3 is afloat and read which half each sunk ship was in, they are essentially the best. With C at random it is proven: the best such plan wins between 50.86297% and 50.86380% (Green's theorem, checked in interval arithmetic), so the probe plans are within 0.0003 points of it. With the rule, dynamic programming finds nothing better. The plain plan is not an equilibrium of the race.

Order within the parts. Ordering the shots inside the parts adds about 1.2 points, in four moves; against a random layout, where the gains add up: the sweep-order rule, +0.83 points; B in long runs, +0.15; C fired while the 3s are afloat, +0.13; slots, +0.05. For the third, cut the diagonals into long stretches and sweep B from one end of each, C from the other, which B never reaches: there one shot can find a 3 whose set cell is in B or a 2 that lies only in C. For the fourth, the two C diagonals beside each B diagonal take alternating slots there, so that a 3 lying across both is not hit twice.

A mixture of twelve computed plans that makes all four moves wins at least 52.14107% on every pattern, using only what the rules announce (certified), or 52.11614% reading only which half each sunk ship was in. Dynamic programming puts the best such play at about 52.143%. The best printable plan we found that orders shots this way guarantees 51.83% (exact), only 0.06 points above Table 3's third plan, with more rules.

The ceiling is proven. Against a random layout each cell lies in 6 of a 3's 2n placements and 4 of the 2's, so once a share t of the board is shot at most 3t of a 3's placements and 2t of the 2's have been hit, whatever the order and whatever the hunter knows. A time-change argument then makes each 3 no faster than an independent uniform time on [0, 1/3], and the 2 on [0, 1/2], so for every strategy

W ≤ E[1 − G(max(U₁/3, U₂/3, U₃/2))] = 2/9 + 5/36 + 1/6 = 19/36 ≈ 52.78%

with U independent uniforms. So the game's value lies between 52.14% and 52.78%. A sharper relaxation puts the ceiling near 52.5%, and near 52.18% for plans that fire only A, B and C before the last ship is found (estimates, not proofs).

One offset per block. The hider's leverage comes from the single offset. Draw a fresh one for each large block of the board (the blocks placed at random), and ships in different blocks get independent offsets whatever the hider does, so every plan wins its random-hider value on every such layout: 50.95% for the adaptive plan, 449/864 ≈ 51.97% with the sweep-order rule and B in runs, 52.15% with the order within the parts (exact, one plan). Ships within one block escape this.

7What finite boards show

The march toward a third (Figure 8). 3-parity's slowest layout falls from 0.79 of the board on 4×4 to at least 0.427 on 10×10, always 7 to 10 shots above a third. Our reading: that gap (finishing each ship, and the edges) grows only slowly with the board, so as a share of the board it shrinks roughly like the gap over the cell count: still at least 0.09 on 10×10.

The model's order on 10×10 (Figure 9). On the slowest layouts found, the four plans line up as in the model, always jumping less extreme. Averages line up less well: always jumping needs 40.2 shots and 3-parity 41.1, tied in the model.

The 10×10 self-play mixture opens like the plain plan: 88% of its opening search shots fall on one every-third set (picked after the fact; random shooting scores about half), with both halves mixed (the 10×10 page). Jumping in some games also peaks near a third of games there, against a different reference race (the 10×10 page).

Real cells, large board. On a 96 × 96 wrap-around board with real ships and finishing shots, one plan that orders shots within the parts (section 6) beats the plain plan by +2.00 ± 0.10 points on a random layout (+2.15 in the model), and every pattern lands within about 0.1 point of the model; finishing costs both players alike and changes nothing measurable.

8The classic fleet in the model

The yardstick moves. A 4 or a 5 crosses the every-third set once or twice, so the hunt for the 3s finds it along the way (the classic page sees the same on a real board). The plain plan still sets the pace, now against a random layout, but it is no longer the same on every layout: a hider who gives both big ships a second set cell on the boards where the 2 lies only in C holds it to 1259/2574 ≈ 48.91% (exact; Figure 10). So beating the plain plan means guaranteeing more than 48.91%; 50% is not special.

What the 4 and the 5 change. Table 3's third plan, playing the big ships as water, guarantees 49.85%. The sweep-order rule is worth more: ending the plain plan's sweep with it once the 3s, the 4 and the 5 are down adds 2.07 points, against 1.39 with three ships. New: after the first sink, alternate A and B to the end of A instead of probing B (51.42%), and read whether a sunk big ship touched both halves (0.16 points more; Table 4). The hider's lever moves to the big ships: against the tables below, 98.7% of its mix has exactly one big ship across both halves on every board.

How far it goes. Mixing cards from the menu we searched adds at most 0.1 point. A mixture of 28 computed tables, which start in A and bring in B as ships sink, guarantees 52.05% (exact; the best in its class is about 52.07%, extrapolated). No strategy guarantees more than 55.95% (proven, as in section 6). The value in between is open: section 6's order within the parts was not computed for this fleet, and one offset per block helps only if every ship has its own block (Sources).

9Proven, computed, open