Golf Question when you have uneven players teams how can you make it simple and fair We have various sizes groups sometimes they are even sometimes not.
Most of the time we have foursomes equal groups sometimes we might have 3 foursomes one threesomes.
We play golf we come into the club house and we then PICK from a deck of cards to find out whose team we are on.
(The reason is when everyone is on the same team everyone gives putts and therefore everyone’s scores are skewed by those who are not following the game properly)
So to prevent this we just pick cards.
So if there were for example 15 guys playing we would have 15 cards
4 aces
4 kings
4 queens
3 jacks
Everyone picks a card and that is the team you are on.
We add up all the scores and the team with the lowest total score wins……
The question I would like for you to analyze for me is the following!!
Is there any mathematical reason for the team with the three players to have a mathematical disadvantage when they have only three players?
It is my contention that since the game is total RAMDOM (Picking cards to determine your team) That to make the figuring simple to calculate….. IT makes no difference from a mathematical sense…..  I am not a mathematical genius but it just seems to me as simple logic if it is all random it should not make any difference
If the groups have all four players and one team has three players
Or
If all the teams are made up of 4 teams of three and one team of four!!
I am praying that you will give to my friends your mathematical analysis.
We just want everything to be simple, quick and fair
Thanks for considering my question.
 A: First of all: before the cards are drawn nobody is in a disadvantage with regard to another player, because the cards are drawn randomly the conditions are equal.
Is there a mathematical reason why the team with three players is at a disadvantage?
We have to analyze how the golf scoring works. According to about about.golf scoring in golf works by seeing how much better you did than the par. So if you took $70$ shots and the par was $80$ your score is $10$. If you took $90$ shots your score is $-10$.
So after doing this to find everyones score we add the scores, however scores highest wins.
So who is at a disadvantage? It depends at how good you are. If players are terrible and get negative scores then having three players will be better. Because you have one less player to bring the score down. If on the other hand players are good then having three players will be bad.
I suggest that you look at the average instead of the sum. So the average for a team of $3$ is the sum of the scores divided by $3$, while the average for a team of $4$ is the sum divided by $4$. I believe this will reduce the differences between being in a team of $3$ and a team of $4$ greatly.
A: The randomness only makes this solution fair in the sense that any given person has an equal chance of ending up on any team. But the team with the fewest players is certainly at an advantage once the game actually starts. 
We'll use the example you provided with three teams of four players and one team of three players. Let's say that everyone plays exactly the same game and every individual golfer ends up with a score of $60$. If the scoring were really fair, then the teams ought to tie. But this isn't what happens. Instead, following your scoring method, the teams with four players end up with scores of $$60+60+60+60=240,$$ while the team with three players ends up with a score of $$60+60+60=180.$$ Then that team ends up winning, even though no player has done better than any other.
A more reasonable scoring method would be to assign each team's final score as the average of the scores of the players on that team. This accounts for the differences in size. Using the above example, the teams of four would now score $$\frac{60+60+60+60}{4}=60$$ while the team of three would also score $$\frac{60+60+60}{3}=60.$$ That is, since everyone has done equally well, now the teams actually tie. (There will tend to be more statistical variability on teams with fewer players, but this is something that is much harder to control for. As far as simple solutions go, averaging is about as fair as you're going to get, and it should be sufficiently fair for your purposes.)
Edit: It seems I missed a subtlety in scoring that dREaM picked up on: to get the final score, the number of strokes is traditionally shifted by the par for the course. Luckily, this bias is also eliminated through averaging. We can see why mathematically using our previous example. Let's call the par for the course $p$. Then to get each score, we subtract this value from the number of strokes. Then the averages for the teams of four are $$\frac{(60-p)+(60-p)+(60-p)+(60-p)}{4}=\frac{60+60+60+60}{4}-\frac{4p}{4}=60-p$$ and the average for the team of three is likewise $$\frac{(60-p)+(60-p)+(60-p)}{3}=\frac{60+60+60}{3}-\frac{3p}{3}=60-p.$$ Now it doesn't matter how good or bad each golfer is: whether subtracting the par results in negative or positive scores, each team's result is shifted by the same amount.
A: The average is a good method.
But it has this property: the team of three is more likely to be the winner - and also more likely to be the worst.
Extra players on the team drag the score towards the middle.
A formula to avoid this might go:
1. Find the average score of everyone.  Call it M.
3. The team of 3 gets $$\left(\frac{a+b+c}3-M\right)*1.732$$
4. Teams of 4 get $$\left(\frac{a+b+c+d}4-M\right)*2$$
The key is that, after subtracting $M$ from the average, you multiply by the square-root of the number of players in your team.  It's the Central Limit Theorem.
