[shard]

Does this formulation work?

You have 6 wooden blocks, each can be 1 of 6 colors. There will always be either (A) 3 blocks of the same color, or (B) 3 blocks of different colors.

Iterating through all possibilities:

6 of 1 color - case A

5 of 1 color, 1 of another - case A

4 of 1 color, 2 of another - case A

4 of 1 color, 1 of another, 1 of yet another - case A and B

3 of 1 color, 3 of another - case A

3 of 1 color, 2 of another, 1 of yet another - case A and B

3 of 1 color, 1 of another, 1 of yet another, 1 of another another - case A and B

2 of 1 color, 2 of another, 2 of yet another - case B

2 of 1 color, 2 of another, 1 of yet another, 1 of another another - case B

2 of 1 color, 1 of another, 1 of yet another, 1 of another another, 1 of another another another - case B

all colors different - case B

[cookmeplox]

Your formulation is not equivalent. The actual Ramsey theory formulation is something more like...you need to color, either red or blue, all of the edges of a fully connected graph of 6 elements (there are 15 such edges). No matter which coloring you choose (there are 2^15 of these), there will always be a "triangle" between three nodes where the entire triangle is either fully red, or fully blue. If you were to instead restrict yourself to a graph with 5 elements instead of 6, it's possible[1] to color the edges so there's no triangle where all the elements are the same.

As an exercise, try repeating your same argument for 5 colors/blocks, and note that it still works, when it shouldn't.

[1] - https://commons.wikimedia.org/wiki/File:RamseyTheory_K5_no_m...