Can sharing an entangled pair help two players win a game with no communication?
Topics 11 to 15 asked who gains when several players choose. These nine go further: what changes when the moves come in turns (25), when players hold secrets (26), when they meet again (27), when they learn as they go (28), when a shared signal is allowed (29), when each chooses a route (30), when the question is how hard an equilibrium is to find (31), the one place where quantum physics changes a game's value (32), and then a market you design and attack yourself (33).
Here is a cooperative game where physics changes the answer. A referee gives Alice a secret bit x and Bob a secret bit y, each chosen at random. Alice answers with a bit a and Bob with a bit b, with no communication once the questions are dealt. They win if a XOR b = x AND y: if both questions are 1 they must give different answers, and otherwise the same.
They can plan beforehand. The best classical plan wins 3/4 of the time: answer 0 always, say, and you lose only when both questions are 1. No plan, however clever, and no shared random numbers, does better. That is the content of the CHSH inequality (Clauser, Horne, Shimony and Holt, 1969).
Now give them a pair of entangled qubits to share beforehand. Each measures their own qubit in a way that depends on their question and answers with the result. Without sending any message, they win about 85.36% of the time, more than any classical plan allows. Nothing is communicated: the questions stay private, and each person's answers on their own look like fair coin flips.
There is a ceiling: no quantum strategy exceeds cos2(π/8) = (2 + √2)/4 = 0.8536 (Tsirelson, 1980). Physics allows more than classical, and less than a game with no limit would allow.
Topics 11 and 12 were games of conflict. This one is a team game: Alice and Bob share one payoff and play against the referee's random questions. The widget on the quantum mechanics page lets you play it round by round; this topic computes the exact numbers.
The classical bound. A deterministic plan is a pair of rules a(x) and b(y): 2 × 2 × 2 × 2 = 16 plans. Listing them all: 8 win 1 of the 4 question pairs, 8 win 3 of the 4 question pairs. A random plan is a mixture of these, so it cannot beat the best of them, 3/4.
The entangled plan. Share the state (|00> + |11>)/√2. A measurement at angle θ in the x-z plane has outcome vectors (cos θ/2, sin θ/2) and (−sin θ/2, cos θ/2). For angles θA, θB the two answers agree with probability (1 + cos(θA − θB))/2 and differ with probability (1 − cos(θA − θB))/2.
Choose Alice's angles 0 and 90° and Bob's 45° and −45°. Three question pairs call for equal answers, each with angle difference 45°, so each is won with probability (1 + cos 45°)/2 = 0.8536. The fourth, (1, 1), calls for different answers and has difference 135°, won with probability (1 − cos 135°)/2 = 0.8536, the same. So the winning probability is
(1 + cos 45°)/2 = cos2(π/8) = (2 + √2)/4 ≈ 0.8536
The same number comes out of the full four-dimensional state-vector calculation on the page below, and of a calculation with the observables cos θ Z + sin θ X. Tsirelson's bound says no choice of state and measurements does better.
What this does and does not show. It shows that the correlations of quantum mechanics cannot be reproduced by any plan made in advance. It does not show a way to send messages faster than light (each side's answers, taken alone, are uniformly random), and it does not make quantum computers faster at games in general.
Move the four measurement angles. The page computes the winning probability exactly from a four-dimensional state and shows the three question pairs that call for equal answers and the one that calls for different ones. The best-angles button reaches the ceiling; the other button shows what happens if both measure the same way.
The experiments Aspect, Dalibard and Roger's 1982 experiment (Physical Review Letters 49, 1804) with photons, and the 2015 “loophole-free” tests with electron spins in diamond (Hensen et al., Nature 526, 682) and with photons (Giustina et al., Physical Review Letters 115, 250401; Shalm et al., 115, 250402), all find correlations that correspond to winning this game more than 3/4 of the time. The 2015 tests closed the detection and locality loopholes together, two of the main ways a classical explanation could survive.
In cryptography Ekert's 1991 key-distribution scheme (see the PQC page on QKD) uses a violation of this inequality to certify that nobody is listening.
In complexity theory Games of this kind, with two cooperating players who cannot talk, led to a famous result: Cleve, Høyer, Toner and Watrous (CCC 2004, 236) studied what such games can and cannot do with entanglement, and the line of work led to the 2020 result known as MIP* = RE (Communications of the ACM 64, 131 (2021)), about what a verifier can check from provers who share entanglement.
A contested idea: quantum games Eisert, Wilkens and Lewenstein (1999, doi:10.1103/PhysRevLett.83.3077) proposed a quantum version of the Prisoner's Dilemma in which players choose quantum operations, and claimed a new stable outcome. Benjamin and Hayden (Physical Review Letters 87, 069801 (2001)) replied that with a wider choice of quantum moves the claimed equilibrium disappears, and the status of such constructions is still argued. The CHSH game above is different: it is a game on a real physical test, and its numbers are not in dispute.
The sources Clauser, Horne, Shimony and Holt, Physical Review Letters 23, 880 (1969), doi:10.1103/PhysRevLett.23.880. Tsirelson (Cirel'son), Letters in Mathematical Physics 4, 93 (1980), doi:10.1007/BF00417500. Brunner et al., Reviews of Modern Physics 86, 419 (2014), doi:10.1103/RevModPhys.86.419.
No claim is made that quantum computers speed up game solving. The one place in this course where quantum physics changes a game's value is a game about physics itself.
Check yourself
In the CHSH game, which statement is correct?
All 16 deterministic plans win at most 3 of the 4 question pairs, and mixtures cannot beat the best. The entangled strategy wins cos²(π/8) ≈ 85.4%, and Tsirelson's bound says that is the most any quantum strategy achieves. Nothing is sent: each side's answers alone are random.