I am calculating two different $p$ values for the McNemar test and what I believe is the corresponding *exact* test, even for combined counts larger than 25.

Take, for example the contingency table

```
                   Posttest
Pretest      Outcome A   Outcome B
Outcome A       xx          50
Outcome B       65          xx

```

The chi-square statistic is 
$$\chi^2=\frac{(50-65)^2}{50+65}=1.96,$$
with 
$$p=0.1619. \tag{R}$$

According to Sheskin (2011, Test 20, VI.3, pg 844), the exact test for these situations is essentially a binomial sign test (for a single sample) with parameter $\pi=0.5$ and the two counts equal to the two the cells of interest in the contingency table. My understanding of this is that I'm calculating the probability that, out of $50+65=115$ trials with probability of success 0.5, at least 65 of them are successes. My attempt to calculate this for the particular situation above is:
$$
p
=
\sum\_{i=65}^{115}
\begin{pmatrix} 115 \\ i \end{pmatrix}
0.5^{i}\ 0.5^{115-i}
=
0.0957
\tag{Mathematica}
$$

Why do these two $p$-values differ so much? Am I making a simple mistake?