Fix freestyle fall count calculation

[scotthue]

From Robin:

Example: Individual: with 2 major, 1 minor:

• Mistake Score = 10 - 0.5* major -0.25*minor
• Mistake Score = 10 - 0.5_2 - 0.25_1 = 10 - 1 - .25 = 8.75
• Final Dismount Score = 10·(1− mistake score/number of riders)
• Final Dismount Score = 10*(1 - 8.75/1)
• Final Dismount Score = 10*(-7.75)
• Final Dismount Score = -77.5

Group with: 5 major, 4 minor, 10 riders:

• Mistake Score = 10 - 0.5_5 - 0.25_4 = 6.5
• Final Dismount Score = 10*(1 - 6.5/10)
• Final Dismount Score = 10*(1 - 0.65) = 3.5

(this appears to work....but)

Group with 0 major, 1 minor, 10 riders:

• Mistake Score: 10 - 0.5*0 - .25 = 9.75
• Final Dismount Score = 10*(1 - 9.75/10) = .25

Why would they have such a low score if they had only a single minor?

Group with 8 major, 0 minor, 4 riders:

• Mistake Score: 10 - 0.5*8 - 0 = 6
• Final Dismount Score = 10(1 - 6/4) = 10-.5 = -5

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.

[scotthue]

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])

[rdunlop]

@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

[gossi]

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?

[scotthue]

@gossi, the rulebook committee website doesn't work anymore because it was a subdomain of iufinc not unicycling.org. Perhaps you could fix that.

[gossi]

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.