How do I create a 4 on 4 tournament where every player plays with every other player an equal number of times? I would like to create a 4 on 4 tournament with 8 players (4 players on a team where two teams play against each other each game), where every player plays with every other player an equal number of times.  A simple example of this would be if you had a 2 on 2 tournament with 4 players then:
12 v 34
13 v 24
14 v 23
If it were 6 players doing 3 v 3 then you could have 10 games covering the 20 combinations possible. (i.e. 123 v 456 and so forth).
With 4 v 4 using 8 players it is difficult or at least impossible and impractical to cover all combinations with 8 Choose 4 being 105.  I would like to determine a 'very close' practical solution that would require no more than 10 games, so really an ideal would be 7 games where each player plays with every other player 3 times total on their team.  I haven't been able to figure out a good algorithmic way to approach this aside from doing it by hand and adjusting as I go to ensure player 1 plays with all others 3 times, then player 2 plays with all others (3-8) 3 times, then player 3 plays with all others (4-8) 3 times, making changes that preserve the previous counts.  Any suggestions or solutions?
Second Update: 
I have solved the problem by hand below, where each player has each other player as a teammate for exactly 3 matches:
1235 4678
1458 2367
1347 2568
1278 3456
1368 2457
1246 3578
1567 2348
I performed this by hand by looking for imbalances and attempting a rebalance that preserved the partner match count for player 1. For instance if there was one match with 46 paired but 5 for 48 paired then I looked to change a 48 pairing into a 46 and then preserve the balance of the matchups for player 1 by changing yet another 48 into a 46. Then, recheck to verify all pairings up to the "4s" were still balanced and continue. I feel like it was dumb luck paired with a generally sound higher probability approach that enabled me to reach this perfect solution. 
 A: You are looking for a Balanced Incomplete Block Design, or BIBD. A $(v,k,\lambda)$-design puts $v$ players into groups of $k$ at a time, and any two players will play in exactly $\lambda$ groups.
In your case, you want an $(8,4,\lambda)$-design, and it happens that an $(8,4,3)$-design exists. This example comes from the Wikipedia article on Block Designs:
  0123  0124  0156  0257  0345  0367  0467  1267  1346  1357  1457  2347  2356  2456

Every group will play against its complement, so the first game is $0123$ vs. $4567$, and so on. There are $14$ games in total.
A: I started thinking of it this way: 
We have $28$ pairs of players that need to be satisfied three times. Each game can accommodate six pairs. If we can find an optimal configuration of pairings for a certain number of games such that each pair is satisfied, we can just play that configuration three times.
Say we have eight players $P = \{A, B, ..., H\}$. The pairs that need to be satisfied are:
$$\begin{array}{ccccccc}
(AB) & (AF) & (BD) & (BH) & (CG) & (DG) & (EH) \\
(AC) & (AG) & (BE) & (CD) & (CH) & (DH) & (FG) \\
(AD) & (AH) & (BF) & (CE) & (DE) & (EF) & (FH) \\
(AE) & (BC) & (BG) & (CF) & (DF) & (EG) & (GH) \\
\end{array}$$
Where each pair represents one player playing with another player. For the first game, it doesn't matter who we pick—the same number of pairs will be satisfied regardless (12 pairs). For example, say our first game is team $\{A, B, C, D\}$ vs. team $\{E, F, G, H\}$: 
$$(AB)(BC) (CD) \times (EF)(FG)(GH)$$
The pairs $(AB), (BC), (CD), (EF), (FH),$ and $(GH)$ are satisfied, but so are $(AC), (AD), (BC), (BD)$ and $(EH), (EG), (FH), (FG)$. Keep in mind that, if you changed the teams (for the first choice), the particular pairs satisfied would be different, but $12$ of them would always be satisfied. (Think of just changing the labels, for example. What if $P = \{H, G, F, ... , A\}$?)
That leaves $16$ pairs. Now we have to switch teams up. We want three new pairs on each team. Let's say...
$$(AE)(AH)(BE) \times (CF)(CG)(DF)$$
This is just team $\{A,B,E,H\}$ vs. $\{C,D,F,G\}$. This satisfies pairs $(AE), (AH), (BE), (CF), (CG),$ and $(DF)$, but also $(AE)$ and $(CF)$. Again, [I don't think] you can get more efficient than this. Trying to remove any of the redundancies results in a team that's too big. (For example: In the first team, the redundancies are $(AE) + (AH)$ and $(AE) + (BE)$ (since $E$ and $H$ were on the same team already, $A$ and $B$ were on the same team already, and $A$ and $A$... well, are always on the same team.) If you try to switch out one of these pairs for another, less redundant one, you get a team that's too big---or you increase the redundancies on the other team.
For the third game, we have team $\{A, B, F, G\}$ vs. $\{C, D, E, H\}$:
$$(AF)(BF)(BG) \times (DH)(CH)(CE)$$
... which satisfies another $8$ pairs. This leaves us with only two pairs, $(BH)$ and $(DG)$.
So the most optimal teams to get everyone to play with one another once is: 


*

*$\{A, B, C, D\} \times \{E, F, G, H\}$

*$\{A, B, E, H\} \times \{C, D, F, G\}$

*$\{A, B, F, G\} \times \{C, D, E, H\}$

*$\{B, H, x, x\} \times \{D, H, x, x\}$


With $x \in P$.

I believe this is the most efficient way to shift teams around. Disclaimer my argument isn't very mathematical (yet), so I haven't actually proven any of this, but I believe it can be made into one—the primary argument just that "Switching teams around gives you either the same result, or a worse one" ... which is really just another way of saying "it's the most efficient."
It might follow that the most efficient way to pair everyone three times is by following the above configuration three times (resulting in $12$ games). I'm more skeptical of this claim, though, because the open spaces in $(4)$ lead me to believe there's some space with which to work and make things more efficient. My gut instinct is that there isn't, but $\text{gut instinct} \neq \text{math}$.
