In bond percolation on the square lattice, exactly one of the following occurs in the rectangle: an open left-to-right primal crossing, or a closed top-to-bottom crossing in the planar dual graph. These alternatives are disjoint and exhaustive by planar duality for rectangle crossings.

p = 1/2

, the closed dual edges have the same law as open primal edges. Rotating the dual rectangle through a right angle identifies its top-to-bottom crossing with the original left-to-right crossing; the slight difference between the side lengths

ℓ + 1

and

ℓ

is exactly the boundary shift introduced by dualization. Thus the event and its complement have equal probability, so

P_{1/2} (LR (ℓ)) = \frac{1}{2} .

(1)

Solved by gpt-5.6-sol high.

b

 0  0

Solution

 0  0

Write

π (r) = P_{1/2} (0 ⟷ \partial Λ_{r}) .

(1)

The Russo-Seymour-Welsh theorem and the Harris-FKG inequality give the standard one-arm extension estimate: there is

c > 0

such that

π (2 r) \geq c π (r)

(2)

uniformly in

r

. Indeed, on the one-arm event to scale

r

, a fixed finite collection of open rectangle crossings in the annulus

Λ_{2 r} ∖ Λ_{r}

, each having probability bounded below by RSW, joins that arm to

\partial Λ_{2 r}

; FKG multiplies the lower bounds. Iteration shows that

π (a r)

and

π (r)

are comparable for every fixed

a > 0

. This proves the estimate suggested in the hint.

Fix

x \in \partial Λ_{n}

and choose

r = ⌊ n /4 ⌋

. If

0 ⟷ x

, there is an open arm from

0

to distance

r

and another from

x

to distance

r

. These are independent events because they use disjoint edge sets, so the extension estimate gives

P_{1/2} (0 ⟷ x) \leq π (r)^{2} \leq C π (n)^{2} .

(3)

For the reverse inequality, take one-arm events from

0

and

x

at scale comparable with

n

, in disjoint boxes. A fixed collection of open crossings of rectangles of bounded aspect ratio joins the two arms. The Russo-Seymour-Welsh theorem bounds the probability of every added crossing below uniformly in