Confused about the calculation of BLOSUM matrix
Ok I'll try to go through it myself and see what happens.

First, we get the pair counts: $$ c_{ij}=\begin{cases} \displaystyle\binom{n_i}{2} &amp; i=j\\ n_i n_j &amp; i\ne j \end{cases} \;\;\;\;\;\implies\;\;\;\;\; c_{11}=1, c_{12}=2, c_{22}=0 $$ Clearly, the normalizer is just $T=\sum_{i\geq j} c_{ij}=3$ and $c_{ij}\equiv c_{ij}^{(1)}$ since there's only one column. Then our pair distribution is: $$ q_{ij} = \frac{c_{ij}}{T} \;\;\;\;\;\implies\;\;\;\;\; q_{11}=\frac{1}{3},\;q_{12}=\frac{2}{3},\; q_{22}= 0  $$
Hence, the occurrence probability per residue: $$ p_i = q_{ii} + \sum_{i\ne j} \frac{q_{ij}}{2} \;\;\;\;\;\implies\;\;\;\;\; p_1 = \frac{2}{3},\;p_2=\frac{1}{3} $$
and the expected pair frequency:
$$ e_{ij} = \begin{cases} p_i^2 &amp; i=j\\ 2p_ip_j &amp; i\ne j \end{cases}\;\;\;\;\;\implies\;\;\;\;\; e_{11}=e_{12}=\frac{4}{9},\; e_{22}=\frac{1}{9} $$
Now we can finally compute the odds ratios:
$$ r_{ij} = \frac{q_{ij}}{e_{ij}} \;\;\;\;\;\implies\;\;\;\;\; r_{11}=\frac{3}{4},\; r_{12}=\frac{3}{2}, r_{22}=0 $$
Computing the log-odds $s_{ij} = \log_2( r_{ij} )$ gives the same values as you get.

So why is $r_{11} &gt; r_{12}$? Well, notice they have the same normalizer, i.e. $e_{11}=e_{12}$. So the difference comes from the fact that $ q_{11} &gt; q_{12} $, the empirical pair distribution. Essentially, because of the way we computed $q_{ij}$, we consider there to be only one pair QQ but two pairs of QS. 
I would also note that sometimes results are a little unintuitive for very small samples (data sparsity throws off the statistics essentially).