No lucky streaks, proved at last?
An AI preprint claims threshold repetition for entangled games
Covers result 277 in OpenAI’s catalog: its one paper. See its connections to other results and the work it cites.
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Chapters
The claim, in plain words
In a nonlocal game, a referee sends a question to each of two players, Alice and Bob, who cannot talk to each other, and checks their answers. In the CHSH game, players using shared randomness win 3/4 of the time; with entangled qubits they can win about 85%, and Tsirelson proved nothing does better.
Now play many rounds at once. The players may use one joint strategy across all rounds, so the rounds need not be independent: a classical strategy wins two rounds of CHSH with probability 5/8, more than (3/4)². How fast the chance of winning every round falls is the parallel repetition question. In August 2026, OpenAI claimed that for entangled players it falls exponentially for every finite two-player game they can’t always win, and Zhao Song improved the rate, with help from AI tools. A finer question is about thresholds: can players win clearly more than their fair share, a fraction above the game’s value, across many rounds? Before this paper, threshold theorems were known only for classical games and for special entangled games.
Result 277 is one OpenAI preprint, 22 pages, dated September 25, 2026. It claims that for every finite two-player game with entangled value v below one, and any margin δ up to 1 − v, the chance of winning at least a fraction v + δ of k rounds decays exponentially in k, for any finite-dimensional entangled strategy and any question distribution.
What it doesn’t claim
Infinite-dimensional or commuting-operator strategies, an explicit universal constant, more than two players, or games with quantum messages. The decay is very slow: its rate is δ⁵ divided by one plus the log of the number of answer pairs, times a universal constant the paper never computes, and this rate uses Song’s lemma. An explicit version, with a constant depending on the question distribution, falls like δ³. For CHSH with a margin of 0.1, that explicit bound needs about 50 billion rounds to drop below one in a million; independent play gets there in about 220 rounds.
Where it stands
- Posted as a preprint
- OpenAI, 22 pages, dated September 25, 2026
- Peer reviewed
- Not yet
- Formalized in Lean
- In part: a weaker rate, δ¹³, without Song’s lemma (docs)
Last checked October 19, 2026. The preprint has not been revised. The threshold theorem is formalized in Lean with a weaker rate, δ¹³, about 7,200 lines with no unproved steps, which we did not build; the Lean proof does not use Song’s lemma. We found no OSVerify review and no comment from specialists. On October 15, Rahul Jain posted a different proof with a better universal rate, citing this paper.
How the proof goes
- Condition on a few wins. The proof samples a few rounds at random and conditions on winning them. If winning many rounds were likely, the other rounds would look like a single game played unusually well, which Song’s lemma rules out. This turns a bound on conditional winning into a threshold bound; plugging in Song’s lemma gives the fifth-power rate.
The numbers in the video
- Ours: CHSH classical value 3/4 and entangled value (2+√2)/4 ≈ 0.854; a joint classical strategy wins two rounds with probability 5/8.
- Ours: for CHSH with margin 0.1, the paper’s explicit bound reaches one in a million after about 50 billion rounds; exact independent play after about 220.
Read the source
- Threshold Parallel Repetition for Finite-Dimensional Entangled Games (PDF), 22 pages.
- The Lean formalization notes.
- The collection’s catalog, where this is result 277.
The paper
OpenAI groups related papers into one result; this one has a single paper.
Transcript
Read the full script
0:00The question
In a nonlocal game, a referee sends a question to each of two players, Alice and Bob, who cannot talk to each other, and checks their answers. In the CHSH game, players using shared randomness win 3/4 of the time; with entangled qubits they can win about 85%, and Tsirelson proved nothing does better. Now play many rounds at once. The players may use one joint strategy across all rounds, so the rounds need not be independent: our computer finds a classical strategy that wins two rounds of CHSH with probability 5/8, more than 3/4 squared. How fast the chance of winning every round falls is the parallel repetition question. In August 2026, OpenAI claimed that for entangled players it falls exponentially for every finite two-player game they can’t always win, and Zhao Song improved the rate, with help from AI tools. A finer question is about thresholds: can players win clearly more than their fair share, a fraction above the game’s value, across many rounds? Before September, threshold theorems were known only for classical games and for special entangled games.
1:13The claims
Result 277 is one OpenAI preprint, 22 pages, dated September 25, 2026. It claims that for every finite two-player game with entangled value v below 1, and any margin δ up to 1 − v, the chance of winning at least a fraction v + δ of k rounds decays exponentially in k, for any finite-dimensional entangled strategy and any question distribution. The decay is very slow. Its rate is δ⁵, divided by one plus the log of the number of answer pairs, times a universal constant the paper never computes, and this rate uses Song’s lemma. An explicit version, with a constant depending on the question distribution, falls like δ³. For CHSH with a margin of 0.1, that explicit bound needs about 50 billion rounds to drop below one in a million; independent play gets there in about 220 rounds. Not claimed: infinite-dimensional or commuting-operator strategies, an explicit universal constant, more than two players, or games with quantum messages.
2:24How the proof goes
The proof samples a few rounds at random and conditions on winning them. If winning many rounds were likely, the other rounds would look like a single game played unusually well, which Song’s lemma rules out. This turns a bound on conditional winning into a threshold bound; plugging in Song’s lemma gives the fifth-power rate.
2:45Where it stands
The preprint is unrefereed and unchanged since release. The threshold theorem is formalized in Lean with a weaker rate, δ¹³, about 7,200 lines with no unproved steps, which we did not build; the Lean proof does not use Song’s lemma. We found no OSVerify review, and we found no comment from specialists. On October 15, Rahul Jain posted a different proof with a better universal rate, citing it. If it holds, finite-dimensional entangled players’ chance of beating their game’s value by any fixed margin shrinks exponentially, though astronomically slowly.
Spot a mistake?
This episode simplifies on purpose, but it shouldn’t be wrong. If something is, open an issue on GitHub or leave a comment on YouTube. Corrections are noted here with a date.