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I'm still following this. But here's a new question, which several hundred high school volleyball coaches are discussing today as we're considering making beach volleyball a formal CIF high school sport in SoCal.
What's the best scoring system for 5 pairs, so that a school is rewarded more for winning the 1's pair than winning the 2's pair, more for winning the 2's pair than the 3's pair, and so on? We're considering the strongest pair to be pair #1 and the weakest to be pair #5. One example would be to give 7 for winning the 1's pair, 5 for winning the 2's pair, 3 for winning the 3's, 2 points for winning the 4's and 1 point for winning the 5's. But those are numbers I've just made up without a lot of math thought. I'm thinking mathmaticians may come up with a more reasoned, logical approach.
Currently, NCAA does NOT stagger, so winning the 5's pair counts the same as winning the 1's pair. Most of the coaches think that the other coaches do not follow the rule that requires that the 1's pair be stronger than the 2's pair, etc, and then justify it by saying of the higher ranked team "They beat them in practice", even though the coaches know that was a fluke and the pair they're ranking lower is actually the stronger pair. Even though the now higher ranked pair may have beaten the other pair in practice, the now higher ranked pair does not win the majority of the time against the other pair, so they should be ranked lower, but the coaches will use a single victory as a justification. It's common but it cause the public not to understand the sport. So, staggering makes sense. The question is, what is the best way mathematically to stagger? It should probably require winning 3 of the 5, and winning either the 1's or the 2's, so you can't win by winning 3s, 4s & 5s.
Let me know if I need to do a better job of explaining the problem.
I'm following it closely. There are a lot of coaches who will be very interested in these numbers. Beach volleyball is the fastest growing NCAA sport in history. It is also growing in the So Cal high schools and in other states also, and as a club sport.
I'm learning from watching this process!
First of all, THANK YOU for all this great input and thinking!
So, is this the consensus:
I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?
ANSWER: (2*n)!/(n!*2^n). In our case n is the number of pairs of athletes. The answer is the product of the first n odd numbers: 1*3*5*7*9 = 945.
Here's a further question. We actually have 13 athletes on the roster. 10 of them can play in a match (5 doubles pairs). How many possible combinations of 5 pairs can I make from 13 athletes?
ANSWER: (2n+3)!/(3!*n!*2^n) - For our problem, n is the number of doubles pairs (5), from which (2*5+3)!/(3!*5!*2^5) = 270270.
I coach beach volleyball. We field 5 pairs of athletes. How many possible combinations of 5 pairs I can make from 10 athletes?
Here's more detail:
There are 10 athletes.
We make 5 pairs to play in matches.
Any of the 10 athletes can be paired with any of the other 9 to make a pair.
There are always 5 pair.
So far, the answers posted are 45, 250 and 945.
Having made these pairs for 10 seasons, I've realized there are a lot of possible pairings. I was thinking it would be 5,000 to 6,500 or around 25,000. I'm a Harvard grad but not a math whiz, so I appreciate your input greatly.
If you'll share your formula so all who are watching can help, that would be great.
Thank you to all of you who have helped so far.
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