This is a repost of a question I saw here that was deleted.
Suppose that we have some number $m\geq 1$ of frogs. These frogs form $2^{n-1}-1$ groups, each consisting of $n>1$ frogs. Each frog can be part of multiple groups.
Suppose that each frog is colored red or blue. I want to show that, no matter how the frogs choose their groups, I can always assign blue or red to each frog such that each group contains at least one blue and one red frog.
My attempt: I thought about using induction. Here are my thinking steps for small $n$:
If $n=2$ then there is only one group with two frogs. So I can assign "blue" to one frog and "red" to the other.
If $n=3$ then there are three groups, each with three frogs. Now I already don't know what to do. I could easily brute force every possibility of the frogs to form groups, however that does not help my understanding of a potential induction step.
If the statement is true for $n$ then it is true for $n+1$: I don't know how to prove this.
A random coloring will have the desired property with positive probability.
More precisely: color each frog red or blue with probability $1/2$ each. The probability that any given group of $n$ frogs is monochromatic (all red or all blue) is $1/2^{n-1}$. By linearity of expectation, the expected number of monochromatic groups is $(2^{n-1}-1)\cdot1/2^{n-1} = 1-1/2^{n-1}<1$. In particular, there must be a positive probability that there are $0$ monochromatic groups.