Detection loophole in CHSH-inequality For the CHSH inequality, one can derive a bound on the detector efficiency required to avoid the fair sampling assumption, and thereby close the detection loophole. This bound is:
$$\eta > \frac{2}{1+\sqrt{2}}\approx 82.8\%$$
One slightly hand-wavy of getting this bound is to assign the value $+1$ to the non-detection of any photon, and like Bell (1971), assign the value $0$ to the detection of only one photon. Both photons are detected with probability $\eta^2$ and in the ideal case contribute with a correlation of $2\sqrt{2}$, the single photon events don't contribute at all, and the non-detection events happen with probability $(1-\eta)^2$ and contribute with $2$, the classical correlation bound.
All in all:
$$\eta^2 2\sqrt{2} + (1-\eta)^2 2 > 2
$$
and from this you get the correct bound on $\eta$.
Now, this is slightly dubious because the non-detection case should be included in the derivation of the CHSH inequality. But anyway, my question regards the choice of the value assigned to the non-detection event. Giving the value $0$ to the detection of only one photon makes intuitive sense to me, because this is the worst case scenario, where the photons are fully uncorrelated. However, assigning the value $+1$ to the non-detection event seems arbitrary to me, as this isn't the worst case scenario.
Choosing $0$ instead gives the bound $\eta > 2^{-1/4} \approx 84.09\%$ (same as you would get from Bell's derivation). Is it simply a matter of choosing the assigned values such that you can derive an inequality that allows you to minimize $\eta$? If so, is there a straight forward derivation of CHSH, akin to Bell (1971) that includes the non-detection events and arrives at this bound?
Can you justify the assignment by saying the photons are correlated in the sense that they both went undetected?
Thanks!
edit: maybe to clarify, if you choose to assign the value $+1$ to both the one-photon and the zero-photon detection events, then the detection efficiency bound is simply $1/\sqrt{2}$, and there are hidden variable models that can violate this bound, so it's wrong
 A: When only one photon is lost, one player doesn't know and still tries to use the quantum strategy. When two photons are lost, both players can use the best possible classical strategy.
In the CSCH game, a win probability of 75%  is attainable under local realism, corresponding to an $|S_{LR}|$ value of $2$. Using a strategy exploiting an entangled state results in a win probability of $P_{QM}= \frac12 + \frac1{2\sqrt2}$ and with $|S_{QM}|=2\sqrt2$.
With this game in mind, there are three outcomes with regards to unsuccessful measurements.


*

*Alice and Bob successfully measure — they win with $P = P_{QM}$ and corresponding $S=2\sqrt2$

*One of Alice or Bob successfully measures — they do no better than random guessing with $P(\mathrm{win}) = 50\%$

*Neither Alice or Bob successfully measure — they attain the classical win probability of $P_{LR}$ and $S = 2$
Why do they do no better than random chance if only one measures? If half a Bell state is lost, we know the resulting half is maximally mixed, resulting in a win rate no better than random guessing with a $50\%$ chance of winning. This means that only when two photons are lost, are the players able to use the optimal classical strategy and get $S=2$.
Using this logic gets you to the inequality you wrote down.
A: I think the reformulation of the CHSH inequality as the CHSH game makes this issue more clear: a referee gives inputs $x,y \in \{0,1\}$ to Alice and Bob, which give answers $a,b\in\{0,1\}$; the inputs are given with uniform probability, and Alice and Bob win the game if $a\oplus b = xy$. Now the answers are 0 and 1, but this is completely arbitrary: we can also say that $a,b\in\{-1,1\}$, and change the winning condition to $1-ab = 2xy$, and have exactly the same game. This should make it clear that there is nothing forcing Alice and Bob to assign a particular answer to the non-detection event; they assign whatever makes it easier to win the CHSH game.
Now one thing that they cannot do is assign the answer 0, as you suggested, because in that scenario $a,b\in\{-1,1\}$. To allow for a third answer is to change the nonlocal game, so it wouldn't be the CHSH game anymore.
It should be clear, though, that this is precisely what Alice and Bob should do: abandon the CHSH game, and go for a three-output game so that they can take into account all the information they have. Indeed this is the approach that Garg and Mermin took in 1987 to derive the bound $\eta > 2(\sqrt2-1)$ that you quote. They made a fatal mistake, though: since this three-output game is much harder to analyse, to simplify things they assumed that the best detection efficiency would be achieved with the maximally entangled state.
This assumption turned out to be false: actually the best detection efficiency is achieved by a state $|\psi\rangle = \cos\theta|00\rangle + \sin\theta|11\rangle$ in the limit $\theta \to 0$, that this, when the entanglement is arbitrarily small. It gives the bound on efficiency $\eta > 2/3$. It was found by Eberhard in 1993, and proven to be optimal by Branciard in 2010 (which illustrates the difficulty of analysing the three-outcome scenario).
The biggest irony is that the optimal efficiency is actually achievable without adding a third outcome, so none of this complication was actually needed. It is actually an elementary calculation to show that simply assigning the non-detection event to always 0 (or always 1) gets you $\eta > 2/3$, with the plain CHSH game. The key insight is to use the state with very little entanglement.
