Let $R = [a,b] \times [c,d] $ be a rectangular box. Discretize $R$ by a hexagonal lattice of mesh size $\delta > 0$. Consider the usual critical percolation on it. Let $E_{wwbwwb}$ and $E_{wbwbwb}$ be the events that going from left to right in the box we can find six disjoint paths of colors white-white-black-white-black-black and white-black-white-black-white-black respectively. Show that $\mathbb{P}(E_{wwbwbb}) = \mathbb{P}(E_{wbwbwb}) $.
Hint: color flipping.
I thought to do that but I'm not sure: do you think that this work? It suffices to show that $ \mathbb{P}(E_{wwb})=\mathbb{P}(E_{wbw}) $ since by mirroring the paths and flipping the colors we get then $\mathbb{P}(E_{wwbwbb}) = \mathbb{P}(E_{wbwbwb}) $. Moreover we have again by flipping the color that $\mathbb{P}(E_{wbw})= \mathbb{P}(E_{bwb})$ and $\mathbb{P}(E_{wwb})=\mathbb{P}(E_{bbw})$. Now consider $ \mathbb{P}(E_{wwb})= \mathbb{P}(E_{w} \cap E_{wb}) $ and since $E_w$ is an increasing event and $E_{wb}$ a decreasing event, by FKG inequality we get that $$ \mathbb{P}(E_{wwb})= \mathbb{P}(E_{w} \cap E_{wb}) \leq \mathbb{P}(E_w) \mathbb{P}(E_{wb}) $$ again by flipping colors we get that $ \mathbb{P}(E_{w})=\mathbb{P}(E_{b})$ and then we deduce that $$\mathbb{P}(E_w) \mathbb{P}(E_{wb}) = \mathbb{P}(E_b) \mathbb{P}(E_{wb}) $$ now we get that $E_b$ and $E_{wb}$ are both decreasing events, then again by FKG inequality we get $$\mathbb{P}(E_{wwb}) \leq \mathbb{P}(E_b) \mathbb{P}(E_{wb}) \leq \mathbb{P}(E_{bwb})= \mathbb{P}(E_{wbw}) $$
For the other direction we have that $E_{b}$ and $E_{bw}$ are both decreasing events hence $$ \mathbb{P}(E_{wwb})= \mathbb{P}(E_{bbw})\geq \mathbb{P} (E_{b}) \mathbb{P}(E_{bw})$$ now similarly as above we have $$\mathbb{P} (E_{b}) \mathbb{P}(E_{bw}) = \mathbb{P}(E_w) \mathbb{P}(E_{bw})$$ and now $E_w$ is increasing and $E_{bw}$ is decreasing hence $$ \mathbb{P}(E_w) \mathbb{P}(E_{bw}) \geq \mathbb{P}(E_{wbw}) $$
hence the result.