If $K_6$ has its edges coloured using $2$ coulours, it contains two monochromatic triangles NOTE: I am looking for a probabilistic solution only. For a non-probabilistic solution, see here.

Show that if the edges of $K_6$ are coloured using two colours, then it contains two monochromatic triangles.

We first define the random variable $X$. Given any colouring of $K_6$, $$X=\text{number of monochromatic triangles for that colouring}$$
Now, we number the vertices of the  graph with numbers from $1$ through $6$. We define $$X_{i,j,k} = \begin{cases}
  1, & \text{if i, j, k form a monochromatic triangle}  \\
  0, & \text{otherwise}.
\end{cases}$$
It is easy to see that $X=\sum X_{i,j,k}$. Now, by linearity of expectation, $$\mathbb E[X]=\sum \mathbb E[X_{i,j,k}]$$
Since there are $\binom63=20$ triangles, the sum on the right side has exactly $20$ terms. By symmetry, we can say that $\mathbb E[X_{i,j,k}]=\mathbb E[X_{a,b,c}]$. Hence, $$\mathbb E[X]=20 \mathbb E[X_{1,2,3}]$$
Now note that the triangle formed using $1,2,3$ can form $2$ monochromatic triangles (one of them blue, the other red), and the remaining $15-3=12$ edges can be coloured in $2^{12}$ ways. Hence there are $2^{13}$ edge-colourings where $1,2,3$ form a monochromatic triangle. The total number of edge-colourings is $2^{15}$, and so $\mathbb E[X_{1,2,3}]=\dfrac{2^{13}}{2^{15}}=\dfrac14$. Thus $$\mathbb E[X]=20 \mathbb E[X_{1,2,3}]=20\cdot \dfrac14=5$$
But this is clearly incorrect. There can't always be $5$ monochromatic triangles. So what am I doing wrong?
Thanks in advance.
 A: I don't think there is a very simple solution using the probabilistic method since the "common" bounds given by the probabilistic method's bound for $R(k,k)$ are not sharp at all, and a direct consequence of what we want to prove is that $R(3,3)\leq 6$. While most direct probabilistic method bounds for $R(3,3)$ are much larger.
Here is what I found:
Suppose we have a $K_6$ colored with black and white, wlog there is a black $K_3$ called $T$, Let $a,b,c$ be the other $3$ vertices (the ones not in $T$). Then each of them has at least two white edges toward $T$. This means if you pick two vertices $u,v$ from the set $\{a,b,c\}$ there is a vertex $x$ in the $T$ with white edges towards $u$ and $v$. Therefore none of the $3$ edges between $a,b,c$ can be white (because if edge $uv$ is white then $u,v,x$ forms a white triangle). Therefore the $3$ edges must be black, but this also gives us an extra triangle.
Note the bound is sharp when you color a $K_{3,3}$ in black and the rest in white (you get two white $K_3$'s and no black $K_3$).
