Multi sports tournament $10$ teams $6$ sports simultaneous I'm hosting a bar sports tournament with $10$ teams and $6$ different sports (pool, darts, table tennis, foosball, beer pong and cornhole). Trying to get the fixtures as fair as possible so that each team plays each sport twice and playing against the same team multiple times is minimised. Are there any formulas to follow? There is potential for $11$ or $12$ teams to end up competing. I feel like having $12$ or $13$ would make this task a lot easier!
Edit: The plan is to have 12 rounds with each sport being played in each round. There is only one pool table, dart board etc available so the same sport cannot be played in the same round. Are there simpler solutions for say 12 teams?
Edit 2: I haven't done any mathematics beyond high school 10 years ago. I've just tried to figure it out as best I can drawing up tables and assigning teams against each other in different slots but ending up with teams playing the same sport or against each other again as there are limited playing slots. Ideally I'd like to know if there is a formula that can be applied to x amount of teams/sports/rounds for multi sports tournaments as all the fixture generators I've tried using only account for one sport being played.

Label the teams $0$ to $9$ and the sports $0$ to $5$. Here $k\,\%\,6$ denotes the remainder of $k$ modulo $6$.
First let each team  $i$ play each other team $j$ in the sport $i+j\,\%\,6$:
$$
\matrix{&1&2&3&4&5&0&1&2&3}\\
\matrix{1&&3&4&5&0&1&2&3&4}\\
\matrix{2&3&&5&0&1&2&3&4&5}\\
\matrix{3&4&5&&1&2&3&4&5&0}\\
\matrix{4&5&0&1&&3&4&5&0&1}\\
\matrix{5&0&1&2&3&&5&0&1&2}\\
\matrix{0&1&2&3&4&5&&1&2&3}\\
\matrix{1&2&3&4&5&0&1&&3&4}\\
\matrix{2&3&4&5&0&1&2&3&&5}\\
\matrix{3&4&5&0&1&2&3&4&5&}\\
$$
The three sports that each team is missing are the ones that would have been on the diagonal and in the two columns you'd get if you'd extend the table to the right. With the diagonal in the first column and the two extension columns in the second and third column, this is:
$$
\matrix{
0&4&5\\
2&5&0\\
4&0&1\\
0&1&2\\
2&2&3\\
4&3&4\\
0&4&5\\
2&5&0\\
4&0&1\\
0&1&2
}
$$
Now let each team $i$ for $0\le i\le8$ play team $i+1$ in sport $i+5\,\%\,6$. That takes care of the second and third columns, except for the game in sport $4$ for team $0$ and the game in sport $2$ for team $9$. These two games together with the ten missing games from the diagonal in the first column are four games each in the sports $0$, $2$ and $4$, and you can form two pairs for each of these sports in whatever way you like because none of the possible pairs have played a second game yet.
