World Cup: what group stage result (vector) is most likely?
First let's list the possible vectors and their probablities in terms of $p$ and $q=1-p$. As the number of draws determines the sum of the points, we can only get the same vector if we have the same number of draws.
\begin{array}{c|c|c} \text{#draws}&\text{vector}&\text{probability}\\\hline 6&[3,3,3,3]&p^6\\\hline 5&[5,3,3,2]&6p^5q\\\hline 4&[7,3,2,2]&3p^4q^2\\ 4&[5,5,3,1]&3p^4q^2\\ 4&[5,5,2,2]&3p^4q^2\\ 4&[5,4,3,2]&6p^4q^2\\\hline 3&[9,2,2,2]&\frac12p^3q^3\\ 3&[6,5,2,2]&\frac32p^3q^3\\ 3&[5,5,3,2]&\frac32p^3q^3\\ 3&[5,5,5,0]&\frac12p^3q^3\\ 3&[7,5,2,1]&3p^3q^3\\ 3&[7,4,2,2]&3p^3q^3\\ 3&[5,5,4,1]&3p^3q^3\\ 3&[7,4,3,1]&3p^3q^3\\ 3&[5,4,4,2]&3p^3q^3\\ 3&[4,4,4,3]&p^3q^3\\ \hline 2&[9, 4, 2, 1] & \frac32p^2q^4\\ 2&[7, 5, 4, 0] & \frac32p^2q^4\\ 2&[7, 4, 4, 1] & \frac94p^2q^4\\ 2&[7, 6, 2, 1] & \frac32p^2q^4\\ 2&[4, 4, 4, 4] & \frac38p^2q^4\\ 2&[5, 4, 4, 3] & \frac32p^2q^4\\ 2&[7, 7, 1, 1] & \frac38p^2q^4\\ 2&[7, 4, 3, 2] & \frac32p^2q^4\\ 2&[7, 5, 3, 1] & \frac32p^2q^4\\ 2&[6, 5, 4, 1] & \frac32p^2q^4\\ 2&[6, 4, 4, 2] & \frac32p^2q^4\\ \hline 1&[9, 4, 4, 0] & \frac38pq^5\\ 1&[7, 6, 4, 0] & \frac34pq^5\\ 1&[9, 4, 3, 1] & \frac34pq^5\\ 1&[9, 6, 1, 1] & \frac38pq^5\\ 1&[6, 6, 4, 1] & \frac34pq^5\\ 1&[6, 4, 4, 3] & \frac98pq^5\\ 1&[7, 6, 3, 1] & \frac34pq^5\\ 1&[7, 4, 3, 3] & \frac34pq^5\\ 1&[7, 7, 3, 0] & \frac38pq^5\\ \hline 0&[9,6,3,0]&\frac38q^6\\ 0&[9,3,3,3]&\frac18q^6\\ 0&[6,6,6,0]&\frac18q^6\\ 0&[6,6,3,3]&\frac38q^6 \end{array}
(I worked these out by hand down to $3$ draws, then I was missing one with $3$ draws and couldn't be bothered to figure it out and coded it up after all. :-)
Only the vectors with the highest coefficients for each monomial can ever be the most likely, and it's straightforward to work out at which values of $p$ the crossovers among the monomials occur. The probability $3p^3q^3$ for the case of $3$ draws is the only one that never dominates:
\begin{array}{c|c|c} \text{#draws}&\text{vector}&\text{probability}&\text{domain}\\\hline 6&[3,3,3,3]&p^6&p\in[\frac67,1]\\\hline 5&[5,3,3,2]&6p^5q&p\in[\frac12,\frac67]\\\hline 4&[5,4,3,2]&6p^4q^2&p\in[\frac{2\sqrt6-3}5,\frac12]\approx[0.38,\frac12]\\\hline 3&[7,5,2,1]&3p^3q^3\\ 3&[7,4,2,2]&3p^3q^3\\ 3&[5,5,4,1]&3p^3q^3\\ 3&[7,4,3,1]&3p^3q^3\\ 3&[5,4,4,2]&3p^3q^3\\ \hline 2&[7, 4, 4, 1] & \frac94p^2q^4&p\in[\frac13,\frac{2\sqrt6-3}5]\approx[\frac13,0.38]\\ \hline 1&[6, 4, 4, 3] & \frac98pq^5&p\in[\frac14,\frac13]\\ \hline 0&[9,6,3,0]&\frac38q^6&p\in[0,\frac14]\\ 0&[6,6,3,3]&\frac38q^6&p\in[0,\frac14] \end{array}
I wouldn't know how to derive this result just based on symmetry arguments :-)
There are $3^6=729$ ways to put between $0$ and $6$ arrows on the edges of a labeled $K_4$. Let $q:={1-p\over2}$. An assignment containing $r\in [0..6]$ arrows then has probability $q^rp^{6-r}$. We now have to go through the $729$ cases and collect the probabilities of the various possible scoring vectors. I don't think it pays to set up a Polya counting scheme in order to exploit the occurring symmetries. The resulting probabilities all are polynomials of degree $\leq6$ in $p$, and some search will be necessary to identify the most probable scoring vector in terms of $p$.
I went through the cases, and it turned out that there are $40$ possible scoring vectors, numbered $1$ to $40$ in my setup. I let Mathematica compute the resulting polynomial $s_j(p)$ for each of them, and then determined for the values $p_k={k\over400}$ $(0\leq k\leq400)$ which scoring vector had the highest probability. The result is in the following figure:
Inspection of this figure allows to determine a posteriori the $p$-values where the jumps take place. They are $${1\over4},\quad{1\over3},\quad{2\sqrt{6}-3\over5}=0.3798,\quad{1\over2},\quad{6\over7}\ ,$$ as in Joriki's answer. Note that some of the probabilities $s_j(p)$ have respectable values, as can be seen from the following figure which shows a plot of all $40$ functions $p\mapsto s_j(p)$:
