Sorry, but I can't answer your ranking question. Maybe someone else here can.
I googled around a bit and came across quite a range of coverage about this topic on the net. Here are some search suggestions:
"Sports teams' ranking"
"Sports rating system methods"
"Sports power ranking"
Hopefully a ready-made system exists for your model!
]]>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.
]]>From the point of view of EM, the problem is solved.
I'm quite sure of that too, so here's the full list of 945 possible combinations of 5 pairs from 10 players, condensed down to 4 image files:
After getting more details i found that the total no. Of possibilities are 10C2 or C(10,2) =10!/(2!*8!)=45
bobbym and I both got 945 possibilities. The full list is a bit large to post, but in the hidebox below is a random sample of 50 from the list. The entries are in strict numerical order.
In post #14, my image in the 'level 5' hidebox shows the first 60 and the last 60 possibilities from the 945 (entries are in numerical order). 4 from the list of 50 random samples are in there too, so that makes 166 possibilities that I've posted.
They all seem to be valid to me, and if you check them I think you'll agree that there are many more possibilities than your answer of 45.
So I suppose there's an error in your formula, but I'm sorry, I don't know enough about this kind of maths to be able to show where it is.
]]>On the front page go into Help Me and look for the Post New Topic button. Hit that and you are ready.
]]>Understanding the problem, finding the logic required to solve it, applying that logic correctly with EM, using that logic for a mathematical approach that confirms the EM results, having two people from opposite sides of world who weren't looking over each other's shoulder arrive at identical results...I think we've done that.
I just hope that one of us didn't misunderstand the problem and influence the other ESP-ly...we are cousins, after all!
]]>I have some confidence in the result now.
I'm convinced!
]]>Then I did 2 pair out of 4,5,6,7,8,9,...players etc.
]]>Well, that gets 270270 alright, but I have no idea what that code means.
I fiddled around with it a bit to see if I could get to understand it, and got the first 4 levels {10,105,1260,17325) that we'd got before.
So, including your code, this is the pattern:
(1/3840)(m-9)(m-8)(m-7)(m-6)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->13 = 270270
(10/3840)(m-7)(m-6)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->11 = 17325
(80/3840)(m-5)(m-4)(m-3)(m-2)(m-1)m/.m->9 = 1260
(480/3840)(m-3)(m-2)(m-1)m/.m->7 = 105
(1920/3840)(m-1)m/.m->5 = 10
It looks interesting - like it should mean something - but I don't really know what it was that I did!
]]>Thanks, my machine is the same as yours and runs out of memory too.
Coming at the problem in another way, I get for 5 pairs out of 13 players:
(1/3840) (m - 9) (m - 8) (m - 7) (m - 6) (-5 + m) (-4 + m) (-3 + m) (-2 + m) (-1 + m) m /. m -> 13
I have some confidence in the result now. I would like to have a solution in generating functions too but so far this has not been possible.
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Could you please run this when you get a chance and post a screenshot of the notebook that pops up.
Needs["Benchmarking`"]
BenchmarkReport[]
Does your code solve for 5 pairs out of 13 athletes?
My computer can't handle that number of permutations. M stops working straightaway, giving this message:
"General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation."
I noticed that when I ran the M code for 11 athletes, it used up all available RAM (20GB or so) and then began to use virtual memory from the hard drive...which turned out to be enough to get a result. The extra amount of memory needed to solve for 13 athletes would be absolutely humongous, so I let that one slide and just took OEIS's word for it that 270270 was the next element after 17325 in the sequence.
Their formula, (2n+3)!/(3!*n!*2^n), and their M code, Table[(2n+5)!!/3-(2n+3)!!,{n,1,5}]/2, confirm that. Of course, I don't understand the how/what/why etc of those formulas!
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