TITLE: Prove there exists a 2-coloring of the points of the projective plane of order 11 such that every line has at least two points of each color
QUESTION [6 upvotes]: To clarify, the projective plane of order 11 has 133 lines and 133 points; any two lines intersect in one point, and any two points determine one line; there are exactly 12 points on each line, and each point lies on exactly 12 lines.
I've been thinking about the above problem for a while now, and I believe that I need to use pigeonhole principle. However, I'm not sure how to use pigeonhole principle to prove the existence of a special coloring rather than a guaranteed property of any coloring.
I've also looked at the smaller case of the projective plane of order 3**, and it seems that this plane has no 2-coloring such that each line has at least 1 point of each color. If that's true, I'm not sure how to prove it, although I believe that the strategy to prove this case could help me solve the original problem.
I would appreciate any help, especially a hint or approach that I can use to find the rest of the solution on my own!
EDIT*: I sketched a potential solution in the comments below; please feel free to critique it!
EDIT**: I meant order 2, with 7 points and 7 lines. Sorry about the confusion there!

REPLY [4 votes]: Reposted from the comments for more visibility:
Let $X$ be the discrete random variable that counts the number of lines that have at least 2 points of each color. Let $X_i$ be the indicator variable for line $i$.
Then $\mathbb{E}(X) = \mathbb{E}(\sum_{i=1}^{133} X_i) = \sum_{i=1}^{133} \mathbb{E}(X_i) = 133(1−\frac{26}{2^{12}}) = 133 − \frac{1729}{2048} > 132.$ There exists a 1-coloring where $X=0$, but $\mathbb{E}(X) > 132$, so there must exist a 2-coloring where $X=133$. Therefore, there exists a 2-coloring such that all 133 lines each have at least 2 points of each color.