How to set up an intercept-only mixed logistic regression in order to test for difference from 50% chance level? In my experiment, subjects repeatedly had to make a binary choice between A and B, and I want to test if subjects (as a group) differed from 50% chance in preferring A over B. Is there a way to test this using a mixed logistic regression, e.g. with glmer() in R?
The intercept only model, if I recall correctly, tests whether subjects as a whole differed from zero. But if I try to recode the dependent measure so that each score is a difference from chance (.5), and then test if that difference from chance differs from zero, then the model won't run because the value of the dependent measure has to be 0 or 1. 
Is there some other way to do this using glmer() or is there a more straightforward way that I'm overlooking? Does it depend on how many trials are making up the dependent measure? 
 A: I think you are confused about the role of intercept in logistic regression.
Logistic regression predicts the probability of some outcome, in your case e.g. the probability of the A choice. To do that, it forms a linear combination of predictors and passes it through a logistic function that "squeezes" real numbers from $-\infty$ to $+\infty$ into a $[0,1]$ interval. It looks like that (image from Wikipedia):
$\quad\quad\quad\quad$
So if the linear combination has value 0 then the output of logistic function is $p=0.5$. This means that a logistic regression without any predictors whatsoever and with zero intercept predicts 50% chance.
If you had 1 data point per subject, you could have used
glm(choice ~ 1, family='binomial')

and looked at the value and p-value for the intercept. In your case you have more than 1 data point per subject, so you can use
glmer(choice ~ 1 + (1|subject), family='binomial')

A: Suggest use of a Binomial test, with $p=0.5$. 
One common use of the binomial test is in the case where the null hypothesis is that two categories are equally likely to occur (such as a coin toss). Tables are widely available to give the significance observed numbers of observations in the categories for this case. However, the binomial test is not restricted to this case, and can be used for any probability.
Edit: The above assumes that each subject is "cut from the same cloth." That is, that the probability, whatever that is, of choosing A or B is the same for each participant, and that each choice of A or B is performed only once. As a counterexample, suppose that subjects 1 and 2 would have perfect discordance in choices between the A and B objects. In that latter hypothetical case, our ability to detect even a perfect bias for each subject would be diminished, e.g., Subject 1 chooses only B 100 times while subject 2 chooses only A 100 times for a $P=0.5$. Clearly, from that hypothetical example, a lack of bias for overall choices is not equivalent to a lack of subject preference.
There is not enough information provided about the experiment in the question above nor its participants to model this further. For example, the ability or inability to measure concordance is not presented and we have a perfect absence of context as to what is being adjudicated as A or B and how that occurs. For example, two judges pairwise adjudicating whether the A or B face of a dirty coin is showing is different than two judges adjudicating two separate piles of dirty coins with no overlapping opinions. We do not know if we are asking whether the judges are biased, if the "A vs B" coins are biased, and that neither collectively or separately. This problem is complicated enough, by itself, that it deserves separate treatment, so, I made it into a question of its own elsewhere.
