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Chi-squared test of proportion with zero cell counts?

I have a set of proportions:
for $(i,j) in {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,5),(6,6)}$, $r_{i,j}$.
The result of a chi-squared test for homogeneity of proportions with 3 df is:

How can I interpret the degree of freedom for each $i$ column?

Is it $1-sum_{i}^{‘}r_{i,j}^2$?
Is it $1-sum_{i}^{‘}sum_{j}^{‘}r_{i,j}^2$?
Is it $sum_{i}^{‘}r_{i,j}^2$ for $j in {1,2,3,4,5,6}$?

A:

If you don’t know that the counts are mutually exclusive, the degrees of freedom are 1 minus the variances. But, if the counts are mutually exclusive, then a more reasonable test is Fisher’s exact
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