# Is the percolation threshold strictly decreasing as dimension increases?

For sake of concreteness, we can consider bond percolation on an infinite $$\mathbb{Z}^d$$ lattice. Let $$p_b^{(d)}$$ be the percolation threshold in $$d$$ dimension. It is known that $$p^{(2)}_b = \frac{1}{2}$$ exactly and $$p_b^{(3)} \approx 0.245$$ from simulations. This seems to suggest that $$p^{(2)}_b > p^{(3)}_b$$.

My question is the following: is it rigorously known that when the dimension is increased, the bond percolation threshold must strictly decrease? In other words, given any pair $$d_2 > d_1$$, $$p^{(d_2)}_b < p^{(d_1)}_b$$ must hold?

There's a short argument that $$p_b^{(2d)} < p_b^{(d)}$$.
To prove this, look at the $$d$$-dimensional sublattice of $$\mathbb Z^{2d}$$ formed by the points where the last $$d$$ coordinates are $$0$$. Between two adjacent points $$(x_1, \dots, x_i, \dots, x_d, 0, \dots, 0) \qquad (x_1, \dots, x_i+1, \dots, x_d, 0, \dots, 0)$$ we can find two edge-disjoint paths: one is just the edge between them, and one is the path which takes a step in dimension $$d+i$$, then dimension $$i$$, then dimension $$d+i$$ again. By alternating the sign of the $$(d+i)$$-dimension step depending on the parity of $$x_i$$, we can ensure that none of these paths share any edges.
With probability-$$p$$ bond percolation in $$\mathbb Z^{2d}$$, the probability that both paths are closed is $$(1-p)(1 - p^3) = 1 - p - p^3 + p^4$$, so bond percolation in this $$d$$-dimensional sublattice is dominated by probability-$$(p+p^3-p^4)$$ bond percolation in $$\mathbb Z^d$$. We can always choose a value of $$p$$ such that $$p < p_b^{(d)} < p + p^3 - p^4$$. For such a value of $$p$$, bond percolation on $$\mathbb Z^{2d}$$ will have an infinite component with probability $$1$$, so $$p_b^{(2d)} < p$$.