Open scotthue opened 10 years ago
I don't understand why the weighting was changed from the previous version to the update suggestion by @rdunlop.
Minor Falls | Major Falls | |
---|---|---|
Old Weighting | 0.25 | 0.5 |
New Weighting | 0.125 | 02.5 |
My suggestion is just to correct the mistake in the formula and get the following:
Mistake Score = 0.5*[#major] + 0.25[#minor]
Final Dismount Score = 10*(1- [Mistake Score]/[#riders])
@scotthue , the weighting was changed because it felt like giving a score of 0 for 2-major falls was a little harsh. But, I think I'm starting to agree with the corrected formula. (though I would put "max(0, score)" into the calculation instead of only in the text).
If I run my examples again:
Example: Individual: with 2 major, 1 minor:
Mistake Score = 0.5* major +0.25_minor Mistake Score = 0.5_2 + 0.251 = 1 + .25 = 1.25 Final Dismount Score = 10·(1− mistake score/number of riders) Final Dismount Score = 10(1 - 1.25/1) Final Dismount Score = 10*(-0.25) Final Dismount Score = -2.5 Final Dismount Score = 0 (cannot be below 0)
Group with: 5 major, 4 minor, 10 riders:
Mistake Score = 0.5_5 + 0.254 = 3.5 Final Dismount Score = 10(1 - 3.5/10) Final Dismount Score = 10_(1 - 0.35) = 6.5
Group with 0 major, 1 minor, 10 riders:
Mistake Score: 0.50 + 0.25 = 0.25 Final Dismount Score = 10(1 - 0.25/10) = 9.75
Group with 8 major, 0 minor, 4 riders:
Mistake Score: 0.58 + 0 = 4 Final Dismount Score = 10(1 - 4/4) = 10*0 = 0
There was a formula in the last rulebook committee discussion. I thought this was going into the rulebook, too. Where is the last rulebook committee software running?
@gossi, the rulebook committee website doesn't work anymore because it was a subdomain of iufinc not unicycling.org. Perhaps you could fix that.
Ah, we transferred that domain the last night. @olarf is now in charge of that. I hope to move the IUF to its new server by the end of the week. So we should look into this after the migration.
From Robin:
Example: Individual: with 2 major, 1 minor:
Group with: 5 major, 4 minor, 10 riders:
(this appears to work....but)
Group with 0 major, 1 minor, 10 riders:
Why would they have such a low score if they had only a single minor?
Group with 8 major, 0 minor, 4 riders:
This is why this calculation is broken.
I propose: "max(0, 10 (1 - ((# major falls * .25) + (# minor falls \ .125)))/num_members)"
With this, a single competitor would get 10 if they have 0 major falls, 0 if they have 4 major falls. If there are pairs, they get 8 major falls before they get to 0. If there are 3 competitors, it goes to 12 major falls....etc etc.
I look forward to hearing what we actually want to use for the calculation.