Using a,b,c,d convention of the 4-fold table, as here,
Y
1 0
-------
1 | a | b |
X -------
0 | c | d |
-------
a = number of cases on which both X and Y are 1
b = number of cases where X is 1 and Y is 0
c = number of cases where X is 0 and Y is 1
d = number of cases where X and Y are 0
a+b+c+d = n, the number of cases.
substitute and get
$1-\frac{2(b+c)}{n} = \frac{n-2b-2c}{n} = \frac{(a+d)-(b+c)}{a+b+c+d}$ = Hamann similarity coefficient. Meet it e.g. here. To cite:
Hamann similarity measure. This measure gives the probability that a
characteristic has the same state in both items (present in both or
absent from both) minus the probability that a characteristic has
different states in the two items (present in one and absent from the
other). HAMANN has a range of −1 to +1 and is monotonically related to
Simple Matching similarity (SM), Sokal & Sneath similarity 1 (SS1), and Rogers & Tanimoto similarity (RT).
You might want to compare the Hamann formula with that of phi correlation (that you mention) given in a,b,c,d terms. Both are "correlation" measures - ranging from -1 to 1. But look, Phi's numerator $ad-bc$ will approach 1 only when both a and d are large (or likewise -1, if both b and c are large): product, you know... In other words, Pearson correlation, and especially its dichotomous-data hypostasis, Phi, is sensitive to the symmetry of marginal distributions in the data. Hamann's numerator $(a+d)-(b+c)$, having sums in place of products, is not sensitive to it: either of two summands in a pair being large is enough for the coefficient to attain close to 1 (or -1). Thus, if you want a "correlation" (or quasi-correlation) measure defying marginal distributions shape - choose Hamann over Phi.
Illustration:
Crosstabulations:
Y
X 7 1
1 7
Phi = .75; Hamann = .75
Y
X 4 1
1 10
Phi = .71; Hamann = .75