For the Borel-strategy value \(p^*\) in the canonical statement, the checked literature gives \[ \frac{7}{20}\le p^*\le\frac{81}{224}=0.361607142857\ldots. \] The lower endpoint is attained by recursive block strategies. Buhler and coauthors derive the upper endpoint using a balanced \(8\times14\) hint matrix and an exhaustive optimization over column partitions. An independent exact replay prepared for this packet enumerates all \(B_{14}=190{,}899{,}322\) column partitions, obtains maximum \(81/224\), and recovers the reported count of 3,920 attaining partitions. Bouquet and coauthors, in arXiv:2508.01737v2 dated 2026-04-19, still state \(p^*=7/20\) as Conjecture 1. Heilman and Tamuz also identify \(81/224\) as the best known upper bound; their analytic bound \(0.37193\) is weaker.

The exact open remainder is to prove \(p^*\le7/20\), or to construct Borel strategies with winning probability greater than \(7/20\). The older interval ending at \(3/8\) is valid but no longer strongest.

Number coordinates within a block by \(0,1,2\). Each player scans the observed sequence in consecutive triples, skips \(000\) and \(111\), and stops at the first other word. On that word, use \[ 001\mapsto1,\quad010\mapsto2,\quad011\mapsto2,\quad 100\mapsto0,\quad101\mapsto1,\quad110\mapsto0. \] The selected global coordinate is the start of the stopping block plus this local index. The stopping time is almost surely finite, and each output fiber is Borel.

Among the \(36\) ordered pairs of nonmonochromatic triples, exactly \(15\) are winning, so the conditional win probability when the players stop in the same block is \(5/12\). A triple is monochromatic with probability \(1/4\). The probability that the first nonmonochromatic blocks coincide is \[ \sum_{k\ge0}(1/4)^{2k}(3/4)^2=3/5. \] When the stopping blocks differ, each selected bit lies in a block that the other player's observed stopping decision did not inspect, and the two selected bits are independent fair bits. The conditional win probability is \(1/4\). The total is \[ \frac35\frac5{12}+\frac25\frac14=\frac7{20}. \] The exact replay independently enumerates the \(36\) local pairs and checks finite truncations through twelve blocks.

Let \(\Omega=\{0,1\}^{\mathbb N_{>0}}\) carry its fair product measure, and let \(V_h\) be the optimum when both strategies read the first \(h\) bits and output an index in \(\{1,\ldots,h\}\). The sequence \(V_h\) is nondecreasing because a strategy may ignore extra bits and indices.

Here is a direct approximation argument for the reverse comparison. If \(f:\Omega\to\mathbb N_{>0}\) is Borel and \(\varepsilon>0\), choose \(M\) so that \(\Pr(f>M)<\varepsilon\). The finite measurable partition formed by the fibers \(f^{-1}(1),\ldots,f^{-1}(M)\), together with the tail, can be approximated in measure by a partition measurable with respect to the first \(d\) coordinates. Equivalently, finite-coordinate simple maps are dense in probability among measurable maps to a countable discrete space. This gives a map \(f_{d,M}\), depending on the first \(d\) bits and taking values at most \(M\), with \(\Pr(f_{d,M}\ne f)<2\varepsilon\). For \(h\ge\max(d,M)\), it is an \(h\)-strategy.

Apply this construction to both members \(f,g\) of a Borel strategy pair. The two win indicators can differ only on the event where at least one approximating strategy differs from its target, so their winning probabilities differ by at most the sum of the two mismatch probabilities. Every Borel pair can therefore be approximated arbitrarily closely by finite strategies. Conversely, every finite strategy is Borel. Hence \[ \sup_{f,g\ { m Borel}}\Pr\bigl(A_{g(B)}=B_{f(A)}=1\bigr) =\lim_{h\to\infty}V_h. \] This independently supplies the cylinder-approximation step for Borel fibers that may have empty interior.

Fix one player's \((3m+1)\)-hat truncation of the three-block strategy. For every observed string of the other player, a deterministic best response can choose the coordinate with the greatest exact winning count. The computation exhausts all observed strings, evaluates every coordinate, and sums these pointwise maxima.

For each \(m=1,\ldots,7\), the best-response total equals the total achieved by using the same truncated strategy on both sides. The best-response excess is exactly zero. The checked probabilities run from \(11/32\) for \(m=1\) to \(187904819/536870912\) for \(m=7\). Direct enumeration of all pairs of bit strings supplies a second implementation for \(m\le3\).

This is a bounded unilateral-optimality result for one fixed opponent strategy. It gives no global upper bound on \(V_{3m+1}\), and it does not prove that the infinite recursive strategy is optimal.

Bouquet and coauthors describe pointwise conditional-score maximization as the general way to compute a response to one fixed finite strategy. Specializing that principle to the recursive three-block strategy gives a closed formula at every depth.

Fix the three-block strategy for one player. For a block number \(r\ge0\) and local output \(q\in\{0,1,2\}\), let \(E_{r,q}\) be the event that the first \(r\) blocks are monochromatic and block \(r\) is the first nonmonochromatic block, with local output \(q\). Each of the three output classes contains two words, so \[ \Pr(E_{r,q})=4^{-(r+1)}=:w_r. \] For a proposed response coordinate \((t,p)\), the count or probability of a win against choices outside block \(t\) has the same baseline for every \(p\). Inside block \(t\), the numbers of 1s in coordinate \(p\) among the two words mapped to \(q\) form the matrix \[ L=\begin{pmatrix}2&1&0\\1&0&2\\0&2&1\end{pmatrix}. \] Thus, for an observed block \(y=(y_0,y_1,y_2)\), the response score at \((t,p)\) is a common baseline plus \(w_t/2\) times the corresponding entry of \[ \bigl(y_0-y_2,\ y_2-y_1,\ y_1-y_0\bigr). \]

Both monochromatic words give the zero vector. On each nonmonochromatic word, the strategy's displayed local choice has advantage \(1\), and the other two advantages are \(0\) and \(-1\). For an observed sequence whose first nonmonochromatic block is \(t\), all earlier blocks therefore give zero advantage. Every later block has advantage at most \(w_s/2\), where \(w_s\le w_t/4\), while the chosen coordinate has advantage \(w_t/2\). It is the pointwise best response. If all inspected blocks are monochromatic in a finite truncation, every coordinate ties and the fallback coordinate is optimal.

The same calculation applies to the infinite strategy. Its first nonmonochromatic block is finite almost surely. Define its output to be coordinate 1 on the null set of sequences with no such block; every response coordinate ties on that set. Pointwise maximization proves optimality among deterministic Borel responses, and averaging shows that private randomized responses cannot do better. This mutual best-response result is a local equilibrium statement. It gives no global upper bound on pairs that change both strategies.

The audit began with MathOverflow question 326669 and both visible answers, then checked the current versions of arXiv:1407.4711 and arXiv:2508.01737. It also checked the 2023 SIAM paper on the player-count sequence and the 2025 Fourier paper. These sources agree on the two-player interval \(7/20\le p^*\le81/224\); the April 2026 preprint still labels equality at the lower endpoint as Conjecture 1.

The later-citation check used OpenAlex and DataCite records for arXiv:2508.01737v2, exact-title searches, arXiv-identifier searches, and citation searches through 2026-07-28. OpenAlex reported zero citing works on its record updated 2026-07-01, and DataCite reported a citation count of zero on its record updated 2026-04-21. Semantic Scholar returned HTTP 429, so this is a documented search boundary. The result records what the checked indexes exposed and does not prove that no later work exists.

Local searches of canonical records, candidate files, and research fixtures, followed by production API searches for `Levine hats`, `guessing coins`, `7/20`, and `two-player coin`, found this canonical prospect and no duplicate target. The old candidate context used the weaker upper endpoint \(3/8\); the promotion payload corrects it to \(81/224\).

The source audit found the balanced \(8\times14\) matrix and the authors' result in the journal and arXiv source, with no original enumeration code. A first bounded artifact independently recovered and checked one attaining four-class partition.

The follow-up C++ replay enumerates all 190,899,322 unlabeled partitions of the fourteen columns. Within every partition it exhausts every binary coloring of the classes and performs exact integer row scoring. It obtains global maximum \(81/224\), with 2,016 attaining partitions having four classes, 560 having five, and 1,344 having six. Their total, 3,920, matches the paper. This supplies an independent executable certificate for the finite matrix optimization.

Buhler and coauthors' matrix-hint theorem turns that finite maximum into the universal upper bound \(p^*\le81/224\). The replay checks the finite premise of that theorem. It does not improve the endpoint or settle whether \(p^*=7/20\).

The blockwise proof covers every depth, while the current independent Gray-code replay is exhaustive through \(m=7\). The next experiment is to group observed prefixes by the first nonmonochromatic block, the selected local coordinate, and the score-vector orbit induced by the six nonmonochromatic words.

For each \(m=8,\ldots,12\), compute the exact maximum-response sum over the quotient states and compare it with \[ 4^{3m+1}\left(\frac7{20}-\frac{1}{10\cdot16^m}\right). \] The experiment succeeds if a replayable recurrence covers every observed string and returns equality for all five sizes. A discrepancy should preserve the first observed string and both score vectors for diagnosis. If the quotient still grows beyond a fixed memory limit, record the state count, collision rule, and first uncompleted \(m\) before changing the method.

The program implements the published symmetric local map in two independent representations: string tuples and three-bit integers. Both enumerate the same set of \(15\) winning ordered pairs. The SHA-256 digest of the sorted `left,right` pair list is `16b0e5ef3f81f12c465fee9eb10e3725800f1c86a095e8b3343fc19c6537eb5c`.

The replay then performs the exact geometric calculation for the infinite recursive strategy. For a finite truncation with \(m\) scanned blocks, it uses coordinate \(3m+1\) when every scanned block is monochromatic. The resulting probability is \[ \frac7{20}-\frac{1}{10\cdot16^m}. \] The run checks this identity for every \(m\) from 1 through 12. The endpoints are \(11/32\) at four hats and \(197032483697459/562949953421312\) at 37 hats.

For each size, the program first forms the integer correlation table \(C_{jk}\), where \(C_{jk}\) counts strings \(A\) with \(A_j=1\) for which the fixed strategy chooses \(k\). Given an observed string \(B\), choosing coordinate \(j\) wins for exactly \(\sum_k B_k C_{jk}\) strings \(A\). A Gray-code traversal updates all coordinate scores after each one-bit change in \(B\), and the program sums the exact maximum score over every \(B\).

All arithmetic is integral until the final reduced fractions are formed. The run covers \(2^4+2^7+\cdots+2^{22}=4{,}793{,}488\) observed strings. For the first three sizes, a separate double loop over every ordered pair \((A,B)\) confirms the symmetric total. The program also constructs the local one-count matrix and checks the six nonmonochromatic advantage vectors used in the all-depth proof.

The published \(8\times14\) balanced matrix was transcribed row by row. An exploratory search found the following partition of its one-based column numbers: \[ \{4,7,14\},\quad\{2,3,10,11\},\quad \{5,6,8,12\},\quad\{1,9,13\}. \] The replay checks that every column has four zeros and four ones and that the groups partition all fourteen columns. For each of the \(2^4=16\) independent color assignments to the groups, it constructs the induced fourteen-bit vector and takes the largest dot product with a matrix row. The sixteen maxima are \[ 0,3,4,5,4,6,6,7,3,5,5,7,5,7,7,7. \] Their sum is \(81\), so the conditional matrix-game value is \[ \frac{81}{14\cdot16}=\frac{81}{224}. \] A second grouped-sum implementation returns the same maxima. This artifact checks attainment. The separate all-partitions artifact proves that every other column partition has value at most \(81/224\).

The program encodes the published balanced \(8\times14\) matrix as fourteen-bit row masks and checks that each row has seven 1s and each column has four. It enumerates unlabeled set partitions exactly once through restricted-growth strings. For a partition \(P\) with \(C\) classes, it enumerates all \(2^C\) binary class colorings \(v\), takes the largest of the eight exact row intersections, and computes \[ V(M;P)=\frac{1}{14\cdot2^C}\sum_v\max_r r\mathbin{\cdot}v. \] Comparing these fractions across \(1\le C\le14\) gives the matrix-game maximum. The run evaluates 20,732,504,062 partition-coloring pairs.

The class counts sum to the Bell number \[ B_{14}=190{,}899{,}322. \] The global maximum is \(81/224\). It occurs for 2,016 four-class partitions, 560 five-class partitions, and 1,344 six-class partitions, totaling 3,920. No other class count attains the maximum. Both the endpoint and the total match the published report. The program also checks the four-class witness from the separate Python replay and obtains numerator 81.

This certificate covers the finite optimization for the displayed hint matrix. The implication \(p^*\le V(M)\) still uses the matrix-hint reduction proved by Buhler and coauthors. The computation does not close the remaining gap between \(7/20\) and \(81/224\).