WikiJournal of Science/A card game for Bell's theorem and its loopholes/The car and the goats

Here we carefully prove Eq. (1), which calculates the penalty that will induce Alice and Bob to act like entangled particles by giving the same (even/odd) answer whenever they receive the same question suit (the &alpha;-strategy):


 * $$Q \ge \frac{4}{3}\left(\frac{1-P_S}{P_S}\right)\;,\;$$

Purpose and free-will represent an important distinction between humans and elementary particles in this simulation of a Bell's theorem experiment. Unlike entangled particles, partners in the card game are attempting to maximize a score. And, unlike entangled particles, humans have the freedom to give different answers to the same question. This adds mathematical complexity arising from the knowledge gained by a player upon seeing one of the question cards in phase 2. Consider, for example, the Monty Hall problem: A contestant in a game show is shown three doors and will win whatever is behind the door he or she selects. A new car is behind one door, while the other two doors hide (less desirable) goats. After the contestant selects a door, the host shows that a goat was behind a door not selected. The host then gives the guest the opportunity to change his or her selection by instead choosing the other unopened door. Should the contestant accept this offer? Under certain circumstances the answer is "yes".

Here we argue that in the partners' version of the card game, there is no advantage to modifying an answer choice after seeing the question. There is one caveat: We must assume that the referee selects randomly among the three card suits. Any discernible pattern by the referee to favor questions of one suit would give players an advantage that we shall not analyze. This equality among all three suits permits us to study just two cases: Either both players get the same question; or they get different questions. The probability that the referee asks different questions of both players is,


 * $$P_D = 1-P_S$$,

where $$P_S$$ is the probability that the referee selects two question cards of the same suit.

&alpha;-strategy
While typical Bell's test experiments do not use the three symmetrical polarization orientations shown in Fig. 1, this configuration greatly simplifies our analysis (see for other polarization measurement angles.)

To facilitate the changes of variables that allow symmetry arguments to establish to equivalent situations involving permutations of Fig. 8, it is helpful to refer to the "even" answer as the "majority" answer (since more answers are "even"), and "odd" as the "minority" answer. Likewise, "heart" is the "minority" question (since only heart has the answer that is different), while "spades" and "club" are the "majority" questions.

Suppose Alice announces that she will follow an &alpha;-strategy, for example by informing Bob that hearts will be "even" (e) while clubs and spades are "odd" (o), as shown in Fig. 8. Does this force Bob to also follow the &alpha;-strategy? Bob is clearly hoping that his question card will be "heart" because that guarantees a non-negative outcome if Bob also follows the same &alpha;-strategy. We now address the following question:


 * What should Bob do if his question is not the heart?

The answer depends on the penalty for the partners giving different answers to the same question, as well as the probability that they will be asked the same question. We will show that Bob's best strategy is to also play the same &alpha;-strategy ("even" to hearts and "odd" to spades and clubs.) Each case (favorable and unfavorable) must be considered separately, and expectation values for all possible strategies must be calculated:

Case 1: Bob sees the "unfavorable" card (spade or club). Since Bob does not know which question Alice will see, he must calculate the expected score for each three subcases (1a, 1b, 1c). And he must do this calculation for each (even/odd) option at his disposal. The 2&times;3 array the upper right corner shows all the outcomes associated with all subcases. Since Bob cannot know which subcase will occur, he must chose between the first or second column. In case 1, it is clear that the first column (e="even") is the better choice. Keep in mind that these subcases are not equally probable. The probability of 1c is $$P_D$$, and we refer to this as the "majority" subcase, since the referee has (unknowingly) selected the suit associated with the "majority" of Alice's answers (i.e., the "even" spades/clubs because Alice selected them to be the same.) The probability of either 1a or 1b occurring is $$P_S$$. Since we have already assumed that 1a and 1b are equally probable, Bob's expectation value for each choice is:


 * $$\begin{align}

E^{\alpha 1}_{maj}&=\; \frac{1-3}{2}P_D     &=&\; -\frac 1 2 P_D         & \,\text{  (majority:even)} \\ E^{\alpha 1}_{min}&= \;\frac{1-3}{2}P_D-QP_S &=&\; -\frac 1 2 P_D - QP_S & \text{   (minority:odd)} \end{align}$$

Here we have used the subscripts (maj/min) to denote the (majority/minority) card. It is clear that "even" is Bob's better choice whenever the penalty is positive $$(Q\ge 0)$$. This is the same answer Alice would give to a club or spade, and therefore we have concluded that in case 1, Bob should also follow the same &alpha;-strategy that Alice chose.

Case 2: Bob sees the (favorable) heart. The expectation values for both possible answers that Bob might give are easily shown to be:


 * $$\begin{align}

E^{\alpha 2}_{maj.}&=\;-3P_D - QP_S & \,\text{ (majority:even)} \\ E^{\alpha 2}_{min.}&=\quad P_D & \text{  (minority:odd)} \end{align}$$

Here it is clear that Bob's best choice is also to play the same (even/odd) answer to the heart that Alice would have played.

In conclusion, if Alice selects the &alpha;-strategy, Bob's optimal strategy is to also follow the same &alpha;-strategy. Combining the the best strategies of (3) and (4) to obtain the expectation value if the team uses the &alpha;-strategy, we have:


 * $$E_{\alpha} = -\frac 1 3 P_D$$

&beta;-strategy
Earlier we pointed out that another strategy is for Bob to give the opposite (even/odd) answer to each choice made by Alice. Here we also include the possibility that Bob attempts to override Alice's decision by giving the same answer as Alice. The expectation values for both of Bob's options are:


 * $$\begin{align}

E^{\beta}_{opposite}&=\; P_D - QP_S  & \text{  (opposite Alice)} \\ E^{\beta}_{same}&=\;-3P_D& \text{ (same as Alice)} \end{align}$$

The "neutral" scoring system associated with the equality in (1) is obtained by equating $$E_{\alpha}$$ to $$E^\beta_{opposite}$$ in (5) and (6). The other strategy associated with (6) is mathematically unsound, but psychologically feasible. Why would Bob opt for $$E^{\beta}_{same}$$ and select a strategy that is guaranteed to lose three points? Perhaps Alice is convinced that the referee will ask different questions and announces that all her answers will be "even". Bob disagrees and overrides Alice's decision because he is certain that the same question will be asked. From Bob's perspective, it is better to lose 3 points than incur the penalty of $$Q$$ points.

Completing the proof
There are eight ways Alice can select (even/odd) answers to each suit. Six of them are covered by Fig. 8, by interchanging the symbols (e,o) and/or (,&clubs;,&spades;). The other two fall under the &beta;-strategy. But what if Alice also changes her mind? This cannot improve the score because Bob has already optimized his response for each choice Alice might make; assigning probabilities to Alice's choices will not increase the expected score.