, 2021

It will be easier to compute the probability that no monochromatic triangles exist. Suppose one of the vertices, say $A$, has $3$ segments of the same color connecting it to $3$ other vertices, say $B$, $C$, and $D$. If one of the edges of $\triangle BCD$ has the same color as edges $\overline{AB}$, $\overline{AC}$, and $\overline{AD}$, then a monochromatic triangle exists. Otherwise, $\triangle BCD$ forms a monochromatic triangle of the other color. Therefore in order for there to be no monochromatic triangles, each vertex must be incident to exactly $2$ red and $2$ blue segments. This is possible only if the coloring creates a loop of $5$ segments all colored red and a loop of $5$ segments all colored blue.

There are $\frac{4!}{2} = 12$ choices for the red loop because the loop can always be viewed as starting at a particular vertex and can go in two different directions. There are $2^{10}$ different colorings of the $10$ segments. Therefore the requested probability is
$$1 - \frac{12}{2^{10}} = 1 - \frac{3}{256} = \frac{253}{256}.$$