Are the rules of this tournament fair? My daughter just took part to a volleyball tournament and she wonders whether the rules of the tournament were fair or not.

There are 10 teams, gathered into 3 groups: Group 1 with 4 teams and Groups 2 and 3 with 3 teams. Each team plays a match against each other team in the same group. The result of each match being either 2-0 or 2-1, 1-2 or 0-2 (there is no tie in volleyball), it gives each team either 2, 1, -1 or -2 points for the tournament ranking.

After this first series of matches, the teams are redistributed into three new groups as follows: Group A (4 teams: the three winners of each group and the best among the seconds of the three groups), Group B (3 teams: the remaining seconds of each group and the best among the thirds of groups 1, 2 and 3), Group C (the remaining 3 teams).

The key rule explains how to decide which team is best second (respectively best third). The rule consists to add the points obtained by each team and to divide it by the number of matches. For instance, if a team of Group 1 won the first match by 2-1, won the second one by 2-0 and lost the third one by 1-2, this team would get a score of $(1 + 2 - 1)/3 = 2/3$. In case of tie, the total number of points scored in each set can be used.

Question. My daughter had the feeling that the teams in Group 1 (the one with 4 teams) had a slight advantage to end up being in either Group A or Group B. Is this feeling justified?

In a more mathematical setting, assuming that the results of the matches are randomly distributed with equal probability 1/4 for each score $(-2, -1, 1, 2)$, what is the probability for a team of Group 1 (respectively 2 or 3) to end up in Group A (respectively B and C)?

Although my daugther is mainly interested in the case of 10 teams divided into 3 groups, a mathematical argument for the case of $n$ teams divided into $r$ groups would be appreciated.
tl;dr: Group 1 teams each have about a $2$ percent higher probability of advancing to Group A than their Group 2 and 3 counterparts; they each have about a $2$ percent lower probability of moving on to Group B than their Group 2 and 3 counterparts.

If I've understood the situation correctly, Group 1 teams do indeed have an advantage, at least as far as Group A is concerned.  Intuitively, this is easy to understand: Keep in mind that all games are zero-sum.  If one team gains a point, its opponent loses a point.  If one team gains two points, its opponent loses two points.  At the end of round-robin play, the points of the teams in each numbered groups sum up to zero.

This is the crux (the key, if you will).  By symmetry, the second-place teams in the three-team Group 2 and Group 3 will have zero points on average.  By contrast, the second-place team in Group 1 is on the plus side, by as much as the third-place team in Group 1 is on the minus side.  Thus, we should expect that Group 1 will contribute the best second-place team more often that Group 2 or Group 3, even when their extra game is taken into account.

This was borne out by numerical computation.  There are $12$ games in all, with $4$ equally likely results for each.  Accordingly, I simulated round-robin play in all $4^{12} = 16777216$ scenarios.  For each scenario, I kept track of both the group with the best second-best (third-best, resp.) team, as well as the average points-per-game for that team.  In case of a two-way or three-way tie, I considered it equally likely for each Group to contribute the team to Group A (Group B, resp.).

The results were quite interesting.  The average second-best team in Group 1 had about $0.33984$ points per game, as compared with $0$ points per game in Groups 2 and 3.  Accordingly, Group 1 contributed its second-best team to Group A a whopping $0.64924$ of the time—almost two-thirds—as compared with Groups 2 and 3, which contributed their second-best team only $0.17538$ of the time.  (All results have been rounded to five significant digits.)

The results for the best third-place team were similar.  Here, we are comparing the second-worst team in Group 1 with the very worst teams in Groups 2 and 3.  The third-place team in Group 1 had an average points-per-game of $-0.33984$ (the flip side of the second-place team, as expected), whereas the third-place teams in Groups 2 and 3 had an average points-per-game of $-1.2188$.  This led to an almost landslide contribution from Group 1 to Group B of $0.79220$—nearly four-fifths of the time—as compared to Groups 2 and 3 with their contribution of $0.10390$.

Interestingly, the average first-place teams in Groups 2 and 3 had higher points per game than their counterparts in Group 1: $1.2188$ (again, the flip side of the third-place team) as opposed to $1.1055$.  But points-per-game have no bearing on a first-place team.

Counteracting this seeming dominance is the fact that Group 1 has more teams than Groups 2 and 3; that's why the rule was instituted in the first place.  Accounting for that, the average team from Group 1 has a $0.41231$ chance of advancing to Group A, respectively; this compares with $0.39179$ for Groups 2 and 3.  I think the figures for Group B are $0.28574$ for Group 1, and $0.30951$ for Groups 2 and 3, but I haven't confirmed that.
Well, out of the four teams in Group 1, one team that gets the highest score is placed in Group A, but that winning score doesn't matter in relation to the team scores from other groups. For example, the winning score from Group 1 can be 1, and the winning score from Group 2 can be 2, but both teams still go to Group A. In this first case where the winning teams are sent, the teams from Group 1 are at a slight disadvantage, since a team has a 1/4 chance of winning rather than the 1/3 chance teams from other groups face (does this make sense? Group 1 teams have to beat out 3 other teams rather beating out two like the teams from Group 2/3 need to). 

Now, we must consider the fourth team that progresses to Group A. As you previously mentioned, the KEY RULE counteracts any advantage teams may gain by playing more/less games by dividing by the number of games played, thus making the probability of the fourth team in group A being from Group 1 1/3 (equal chance). Hence, if I were a team in Group 1, I would have a probability of getting into Group A of (1/4)+(1/3)=(7/12), whereas if I were from Group 2/3, I would have a probability of (1/3)+(1/3)=(2/3)=(8/12). This means that a team originally from Group 2/3 would have a slightly higher chance of going to Group A. 

This can be intuitively understood through simple casework: probability of winning in your group and probability of being the "best second". Due to the KEY RULE, the latter is equal for all groups. However, in order to win in your group, it's harder in Group 1 since you have to beat 3 teams, as opposed to 2 other teams in Group 2/3.
 This method can be generalized to n teams r groups through simple substitution of n and r into the probabilities.