How to compare binary responses among six conditions where each respondent was in each condition? I have the following data:
There are three conditions, A, B and C (presented in random order). Each condition has two difficulties (i.e., six total). Each respondent is scored as 1 (if correct), or 0 (if incorrect). Each respondent did each condition once. With this we get the following:
Condition          Correct
A Easy             19/29
A Hard             16/29
B Easy             10/29
B Hard              0/29
C Easy             13/29
C Hard              6/29

I want to statistically prove that respondents are more correct in one condition compared with another. I stumbled upon the McNemar's test, but since B Hard is a constant, that will not work. Any other suggestions?
 A: Because your respondents are providing multiple scores, you need to account for the non-independence in your data.  The primary two ways to do that are to fit a generalized linear mixed effects model (GLMM), or use the generalized estimating equations (GEE).  To understand how these differ, and to decide which you might prefer, it may help you to read my answer here: Difference between generalized linear models & generalized linear mixed models in SPSS.  My guess is that you would prefer the GLMM.  
The catch is that GLMMs are finicky and sometimes just won't fit.  The 0 corrects in B Hard is especially likely to cause problems.  If you don't include an interaction term, you might be OK, though.  If you did try to fit an interaction term you will have separation (also known as the Hauck-Donner effect), and the GLMM fit is unlikely to converge.  Even if you don't include the interaction, and the fit does converge, you should probably fit nested models and use likelihood ratio tests instead of Wald tests.  
A: Have you considered Fisher's Exact Test?  https://en.wikipedia.org/wiki/Fisher%27s_exact_test
Using it I get p-values for Hard vs Easy of 0.1545 for A, 0.0004 for B, and 0.0340 for C.  Therefore, I think that you can probably reject the null hypothesis (no difference between easy and hard) for B and C, but not for A.
For A, I setup the table as follows:

choose(29,19)*choose(29,16)/choose(58,35) = 0.1545
Similar analysis (sum of easy and hard for each) shows all the types (A, B, and C) are different from one another (A and B p value is 1.4e-6, A and C p value is 0.0018, and B and C p value is 0.027).
