Problem packetResearch packetR714
The optimal cyclic win count is 24 or 25
Link to a section
The recorded result has been reproduced within its stated scope.
Recorded status: established
Recorded scope: all partitions of the labels 1 through 36 into six labeled dice with six distinct faces each
Complete recorded scope and conditions
{
"kind": "bounded",
"statement": "all partitions of the labels 1 through 36 into six labeled dice with six distinct faces each",
"bounds": {
"dice": {
"min": 6,
"max": 6
},
"faces_per_die": {
"min": 6,
"max": 6
},
"labels": {
"min": 36,
"max": 36
}
},
"exhaustive": false
}Originating problem: Largest cyclic winning margin for six disjoint six-sided dice
Authored record and scope
- Authored title
- The optimal cyclic win count is 24 or 25
- Record type
- claim
- Stored status
- established
- Evidence grade
- reproduced
- Recorded scope data
- { "kind": "bounded", "statement": "all partitions of the labels 1 through 36 into six labeled dice with six distinct faces each", "bounds": { "dice": { "min": 6, "max": 6 }, "faces_per_die": { "min": 6, "max": 6 }, "labels": { "min": 36, "max": 36 } }, "exhaustive": false }
2Authored explanation
Write \[ M=36\max_{D_0,\ldots,D_5}\min_i\Pr(D_i>D_{i+1}). \] The certified interval is \[ \boxed{24\leq M\leq25}, \qquad \boxed{\frac23\leq\max\min_i\Pr(D_i>D_{i+1})\leq\frac{25}{36}}. \] The lower bound is attained by \[ \begin{aligned} D_0&=\{3,4,5,32,33,34\},& D_1&=\{1,2,28,29,30,31\},\\ D_2&=\{22,23,24,25,26,27\},& D_3&=\{16,17,18,19,20,21\},\\ D_4&=\{10,11,12,13,14,15\},& D_5&=\{6,7,8,9,35,36\}. \end{aligned} \] Its cyclic win counts are \([24,24,36,36,24,24]\).
Komisarski proves that every cycle of six independent random variables with pairwise tie probability zero has some cyclic winning probability strictly below \[ 1-\frac{1}{4\cos^2(\pi/8)}=\frac{1}{\sqrt2}. \] The theorem applies directly to fair rolls of these dice. A count of 26 would give probability \(26/36=13/18\), and \[ \left(\frac{13}{18}\right)^2=\frac{169}{324}>\frac12. \] Thus \(13/18>1/\sqrt2\), so six counts of at least 26 are impossible. Integrality gives \(M\leq25\).
The remaining question is whether a partition with all six counts at least 25 exists. No such partition or nonexistence certificate was produced in this research pass.
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Replay material: source only
3Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: doi.org ↗, Andrzej Komisarski, Nontransitive Random Variables and Nontransitive Dice, American Mathematical Monthly 128 (2021), 423-434, sharp max-min bound for cycles of n independent random variables; exact witness replay in sdcm-artifact-witness-verifier
4What was measured
Best known minimum win count
Best known probability
5How it connects
Verifies (incoming)
- artifact
Informed by
- attempt
Recorded for
- problem
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Cite the original sources separately.
Machine-readable record
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"ref": "R714",
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"slug": "sdcm-claim-certified-24-to-25",
"type": "claim",
"title": "The optimal cyclic win count is 24 or 25",
"summary": "An explicit partition attains 24 wins on every cyclic edge, while the sharp universal random-variable bound excludes 26 wins.",
"relevance": "For Largest cyclic winning margin for six disjoint six-sided dice, record sdcm-claim-certified-24-to-25 (“The optimal cyclic win count is 24 or 25”) records a bound, answer, status fact, or structural consequence. The record states: An explicit partition attains 24 wins on every cyclic edge, while the sharp universal random-variable bound excludes 26 wins.",
"relevance_source": "recorded",
"body": "Write\n\\[\nM=36\\max_{D_0,\\ldots,D_5}\\min_i\\Pr(D_i>D_{i+1}).\n\\]\nThe certified interval is\n\\[\n\\boxed{24\\leq M\\leq25},\n\\qquad\n\\boxed{\\frac23\\leq\\max\\min_i\\Pr(D_i>D_{i+1})\\leq\\frac{25}{36}}.\n\\]\nThe lower bound is attained by\n\\[\n\\begin{aligned}\nD_0&=\\{3,4,5,32,33,34\\},&\nD_1&=\\{1,2,28,29,30,31\\},\\\\\nD_2&=\\{22,23,24,25,26,27\\},&\nD_3&=\\{16,17,18,19,20,21\\},\\\\\nD_4&=\\{10,11,12,13,14,15\\},&\nD_5&=\\{6,7,8,9,35,36\\}.\n\\end{aligned}\n\\]\nIts cyclic win counts are \\([24,24,36,36,24,24]\\).\n\nKomisarski proves that every cycle of six independent random variables with pairwise tie probability zero has some cyclic winning probability strictly below\n\\[\n1-\\frac{1}{4\\cos^2(\\pi/8)}=\\frac{1}{\\sqrt2}.\n\\]\nThe theorem applies directly to fair rolls of these dice. A count of 26 would give probability \\(26/36=13/18\\), and\n\\[\n\\left(\\frac{13}{18}\\right)^2=\\frac{169}{324}>\\frac12.\n\\]\nThus \\(13/18>1/\\sqrt2\\), so six counts of at least 26 are impossible. Integrality gives \\(M\\leq25\\).\n\nThe remaining question is whether a partition with all six counts at least 25 exists. No such partition or nonexistence certificate was produced in this research pass.",
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"locator": "Andrzej Komisarski, Nontransitive Random Variables and Nontransitive Dice, American Mathematical Monthly 128 (2021), 423-434, sharp max-min bound for cycles of n independent random variables; exact witness replay in sdcm-artifact-witness-verifier"
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"source": {
"url": "https://doi.org/10.1080/00029890.2021.1889921",
"locator": "Andrzej Komisarski, Nontransitive Random Variables and Nontransitive Dice, American Mathematical Monthly 128 (2021), 423-434, sharp max-min bound for cycles of n independent random variables; exact witness replay in sdcm-artifact-witness-verifier"
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}7Provenance
View source, identifiers, and projection details
A statement this project treats as settled at the recorded evidence grade, with the work that backs it.