did you use a probabilistic method in your proof?

Official solution: The number of 4-colorings of the set $M$ is equal to $4^{1987}.$ Let $A$ be the number of arithmetic progressions in $M$ with $10$ terms. The number of colorings containing a monochromatic arithmetic progression with $10$ terms is less than $4A \cdot 4^{1977}.$ So, if $A < 4^9$, then there exist 4-colorings with the required property. Now we estimate the value of $A$. If the first term of a 10-term progression is $k$ and the difference is $d$, then $1 \leq k \leq 1978$ and $d \leq [\frac{1987-k}{9}].$ Hence \[A=\sum_{k=1}^{1978} \bigg[\frac{1987-k}{9}\bigg] < \frac{1986+1985+\cdots + 9}{9}=\frac{1995 \cdot 1978}{18} < 4^9. \]

Is it also valid to say the probability of an arithmetic sequence being monochromatic is $\frac{1}{4^9}$ so the probability $P$ of having a monochromatic sequence in a random coloring is less than or equal to $\frac{A}{4^9}$ where $A$ is the total number of ten term arithmetic sequences. Then if we can prove that $\frac{A}{4^9}\le 1$, then there is a positive probability that a random coloring does not have any monochromatic sequences so such a coloring is possible?

Yes, that works. Equivalently, this $\frac{A}{4^9}$ is the expected number of monochromatic arithmetic progressions; if the expectation of a $\mathbb{Z}_{\geq 0}$-valued random variable is less than 1, then it takes the value 0 with positive probability.

Randomly color all the points with equal probabilities. Let $A_{ij}$ denote the event that the $10$ term AP starting from $i$ with common difference $j$ is monochromatic, where $1 \le i, j \le 1987.$ Note that we must have $i+9j \le 1987$. Now we claim that $\mathbf{P}[A_{ij}] = 4^{-9}$. Indeed, any color for $i$ works, and then there is a probability of $1/4$ for each of the other $9$ integers to be of the same color. Hence, $$\mathbf{P}\left [ \bigcup_{i,j} A_{ij} \right] \le \sum_{i,j} \mathbf{P}\left [ A_{ij} \right]=4^{-9} \sum_{i=1}^{1987} \left \lfloor \frac{1987-i}{9} \right \rfloor =4^{-9} \left( 220 \times 7+ 9(219+218+\cdots +0) \right)=\frac{218350}{262144}<1$$Hence there is a non-zero probability that a desired coloring exists, and so we are done.

https://artofproblemsolving.com/community/c6h1567712p9607276 Phillipines 2018 ManuelKahayon wrote: Each of the numbers in the set \(A = \{1,2, \cdots, 2017\}\) is colored either red or white. Prove that for \(n \geq 18\), there exists a coloring of the numbers in \(A\) such that any of its n-term arithmetic sequences contains both colors.