I was asked the question
The edges of the graph $K_n$ are coloured randomly with three colours (chosen with equal probability). What is the probability of finding a monochromatic $r$-clique in it?
What I did was I picked $r$ vertices out of $n$ in $\binom nr$ ways and with that came $\binom r2$ edges. This is an $r$-clique iff all the $\binom r2$ edges are monochromatic. But, the probability that $\binom r2$ edges are monochromatic when randomly coloured with three colours is $$\binom nr \times 3\times \frac 1{3^{\binom r2}}$$ But, in this formula, if we put $r=3$ and $n=6$, we get $$\binom 63\times 3\times \frac 1{3^{\binom 32}}=20\times 3\times \frac 1{27}>1$$ This is no surprise since in our counting argument, we have counted a lot of edges more than once. So, the expression we got is sort of an upper bound for the required probability.
But, I couldn't find any ways to sharpen this upper bound to reach the exact answer. Any help will be appreciated.
Also, here's a result pointed out by Manuel Lafond in the comments that says that if $n$ is large enough compared to $r$, then the required probability is $1$.