Please read Joy's short experimental paper http://arxiv.org/abs/0806.3078, or (just Section 4 of) the longer paper http://arxiv.org/abs/1211.0784.
Following Joy's explicit instructions, we will collect data from N runs. The data for each run consists of video films made by a battery of video cameras (or some other kinds of sensors) of two spinning hemispheres. We use them to compute the exact directions of the angular momentum s_k and minus s_k of each hemisphere in each run k; k=1, ..., N.
We now pick settings a and b. We *calculate* A_k(a) = sign(+s_k · a) and B_k(b) = sign (−s_k · b) for each run, and correlate them in the usual way: compute the average of the product. We call the result E(a,b).
Now we repeat this for other pairs of settings, same set of directions s_k if we like. Joy has said again and again explicitly that we are allowed to use the same set of runs.
I repeat: it was not my idea, it was Joy's idea, to use the same set of observed/computed spin directions s_k, k=1, ..., N, for each pair of setting directions a, b.
I quote from the other thread http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=31, the embedded quotes by Joy are taken verbatim from his experimental papers:
gill1109 wrote:Joy Christian wrote:These sensors will determine the exact direction of the spin angular momentum s_k (or −s_k) for each shell in a given explosion, without disturbing them otherwise so that their total angular momentum would remain zero, at a designated distance from the center.Joy Christian wrote:Once the actual directions of the angular momenta for a large ensemble of shells on both sides are fully recorded, the two computers are instructed to randomly choose a pair of reference directions, say a for one station and b for the other station.Joy Christian wrote:The correlation function for the bomb fragments can then be calculated as E(a, b) = lim_N 1/N sum_k {sign(+s_k · a)} {sign (−s_k · b)}
I am making a bet about the value of the correlation function at just four points: E(a, b), E(a, b'), E(a', b), E(a', b').
We are going to do one experiment in order to settle this bet, and it has one "N", we don't take a limit.
I had insisted on a CHSH style experiment where each run is assigned (by random choice) to just one of the four correlations. But I am happy to relinquish that demand if Joy really is happy to calculate the four correlations on the same set of N runs (N values of s_k).
I explained elsewhere why I am certain I will win the bet when we follow Joy's original specification (same set of N exploding ball data for each pair of directions) but only with rather large probability if we follow more closely a standard CHSH protocol (different sets of perhaps different numbers of exploding balls, for each pair of directions).
Joy is certain he will win the bet. Joy is certain his maths is right and mine is wrong. I am certain his maths is wrong and mine is right. So we are both entirely happy with his specification of the experiment and with our ensuing bet. The outcome of the bet is determined as follows: do we see a value of the usual CHSH criterium more than half way beyond local realism (+2) towards the best that QM can do (+2 sqrt 2), yes or no? If yes, Joy wins, if no, I win.
I believe that Joy and I already agreed on the main lines of the experiment. We have almost finished recruiting a team of highly respected, neutral, independent adjudicators, who will advise us in drawing up a definitive protocol and definitive agreement on the bet: who will win, under which circumstances. When that's all done we proceed to do crowd-funding to fund the experiment. Joy believes he has an experimenter who can do it for him. The experiment will be performed and the adjudicators will adjudicate.
If anyone has advice to either of us how to fill in the details they are welcome to give advice on the other thread (the thread on Joy's experiment, http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=31). Please be constructive, scientific. This could be the experiment of the century. I believe that it's good either for a Nobel prize or for an igNobel prize, depending on who wins.
In other words, we have a win-win situation.