BTW I just realised that p1*(1-a1)=p2*(1-a2) just means that the total number of times team 1 loses their serve is equal to the total number of times team 2 loses theirs. That makes sense as obviously you can't lose your serve twice without the other team losing theirs in between. In the limit of lots of serves, the two tend to be equal (because the +/-1 serve loss is irrelevant)
Volleyball Math Problem Need help from Markov Mavens
#21
Posted 2015-February-14, 00:36
... and I can prove it with my usual, flawless logic.
George Carlin
George Carlin
#22
Posted 2015-February-15, 02:59
mgoetze, on 2015-February-12, 19:30, said:
OK so, uh, what would be the first thing you try?
You start at 38%-38% for r1=r2.
Then you change for 36%-X (you will have to run hundreds of matches to balance X untill you find a correct data around 40% that matches the 38% Total).
Now you calculate r1 and r2 for 36%-X (total points teams scored)
Repeat for 34%-X
In the end you have a table of r1/r2 (average number of points for team 1 vs team 2) and their corresponding win % when serving. And a table of corresponding % for receiving. You try to extrapolate functions from that table.
It will look something like this (I just invented all data)
serv r1/r2 rec --------------------------- 38% 1.000 62% 37% 0.974 63.84% 36% 0.954 65.75% 35% 0.922 67.44% 39% 1.064 60.17% ..... .....
#23
Posted 2015-February-15, 04:32
You don't need to extrapolate when there is an analytic function available.
I was trying to look at the idea of fixing (1-a1)/a2=(1-A)/A but then (of course) (1-a2)/a1 cannot simultaneously be equal to (1-A)/A.
I tried the other idea where you have
a1=A*r1/(r1+r2)
a2=A*r2/(r1+r2)
where A=0.76 is the loss of rating due to serving
This immediately gives us the ratio of the won points, but it will not be r1/r2:
p1/p2=(r1+r2(1-A))/(r2+r1(1-A))
This means that r1/r2 only gives the ratio of points if we eliminated the disadvantage of serving (A=1), for example if we had the teams serve an equal number of times.
How far away does this deviate from the restriction that exactly 38% points are won by the serving team?
At, say, r2=r1*0.5, we have that 41.26% of the serves are won (a1=50.8%, a2=25.2%, p1/p2=1.68). So it does not deviate that much from the restriction at mildly unbalanced teams, so maybe this is a better model. I don't know.
Is anyone still in this thread btw?
I was trying to look at the idea of fixing (1-a1)/a2=(1-A)/A but then (of course) (1-a2)/a1 cannot simultaneously be equal to (1-A)/A.
I tried the other idea where you have
a1=A*r1/(r1+r2)
a2=A*r2/(r1+r2)
where A=0.76 is the loss of rating due to serving
This immediately gives us the ratio of the won points, but it will not be r1/r2:
p1/p2=(r1+r2(1-A))/(r2+r1(1-A))
This means that r1/r2 only gives the ratio of points if we eliminated the disadvantage of serving (A=1), for example if we had the teams serve an equal number of times.
How far away does this deviate from the restriction that exactly 38% points are won by the serving team?
At, say, r2=r1*0.5, we have that 41.26% of the serves are won (a1=50.8%, a2=25.2%, p1/p2=1.68). So it does not deviate that much from the restriction at mildly unbalanced teams, so maybe this is a better model. I don't know.
Is anyone still in this thread btw?
... and I can prove it with my usual, flawless logic.
George Carlin
George Carlin
#24
Posted 2015-February-15, 13:05
I promise this is the last model I propose! It occurred to me that the formula I wrote before had the drawback that the serve success was at most 0.76, no matter what the skill difference. It makes more sense to reduce the rating by a constant factor, not the % of points won on serve. Then we would have
a1=B*r1/(B*r1+r2)
where B=0.38/(1-0.38)=0.613 to satisfy a1=0.38 at r1=r2.
Then, we have
p1/p2=(1-(B*r2/(r1+B*r2))/(1-B*r1/(B*r1+r2)).
At r2=r1*0.5, we have:
a1=55.2%; a2=23.3%; weighted average=43.5%. For smaller difference in skills, the weighted average will be closer to 38%.
a1=B*r1/(B*r1+r2)
where B=0.38/(1-0.38)=0.613 to satisfy a1=0.38 at r1=r2.
Then, we have
p1/p2=(1-(B*r2/(r1+B*r2))/(1-B*r1/(B*r1+r2)).
At r2=r1*0.5, we have:
a1=55.2%; a2=23.3%; weighted average=43.5%. For smaller difference in skills, the weighted average will be closer to 38%.
... and I can prove it with my usual, flawless logic.
George Carlin
George Carlin