Message Detail
Board: ARCHIVE - SpC2k5 to SC2k6Topic: SC2k7 Oracle Challenge Discussion Topic
From: King Morgoth
Posted: 8/27/2007 8:31:57 PM
Second theory time. And we'll shake everything up.
Nothing totally incompatible with we've been using. In fact, it doesn't change the results we're usually getting in 2-way matches.
Right now we're usually something we could call "sticks", i.e. the longer sticks basically have an advantage over EVERYONE else for the "section" of the vote they have "over" them. Say if Cloud (.5) faces CATS (.2) and L-Block (.1), as long as he "rolls" over .2, he's getting that vote. And that's what we're going to change...
Starting from a 1-on-1 match.
Let's say we get Cloud 60% - Mega Man 40%
We'd give Cloud a range of [0-1.0] and Mega Man a range of [0-0.8]
Instead let's say this:
Cloud has a propability of likeability p of 100%
Mega Man's p is 80%
The idea is this. Instead of being a "max", the values we have is the probability you'll "like" the character. If you like more than one, you pick one of them randomly.
******
Cloud gets P[A¬B] + P[AB]/2 = P[A]-P[AB] + P[AB]/2 = 100%-80% + 80%/2 = 60%
Mega Man gets P[B¬A] + P[AB]/2 = P[B]-P[AB] + P[AB]/2 = 80%-80% + 100%*80%/2 = 40%
It's magic! Except it's not! We didn,t think it that way but the x-stats follow the same idea.
******
Let's throw in a 3rd character in there
Cloud's p of 100%
Mega Man's p is 80%
Ryu's p is 60%
Cloud gets P[A¬B¬C] + P[AB¬C]/2 + P[A¬BC]/2 + P[ABC]/3
=1*.2*.4 + 1*.8*.4/2 + 1*.2*.6/2 + 1*.8*.6/3 = 46.00%
Mega Man gets P[¬AB¬C] + P[AB¬C]/2 + P[¬ABC]/2 + P[ABC]/3
=0*.8*.4 + 1*.8*.4/2 + 0*.8*.6 + 1*.8*.6/3 = 32.00%
Ryu gets P[¬A¬BC] + P[¬ABC]/2 + P[A¬BC]/2 + P[ABC]/3
=0*.2*.6 + 0*.8*.6/2 + 1*.2*.6/2 + 1*.8*.6/3 = 22.00%
Direct proportionality gives : 41.67%|33.33%|25.00%
Old x-stats theory says : 52.50%|32.50%|15.00%
For some reason, I think the first one looks better!
******
This is where it hurts....4 characters
Cloud's p of 100%
Mega Man's p is 80%
Ryu's p is 60%
Terra's p is 40%
Cloud gets P[A¬B¬C¬D] + P[AB¬C¬D]/2 + P[A¬BC¬D]/2 + P[A¬B¬CD]/2 + P[A¬BCD]/3 + P[AB¬CD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 1*.2*.4*.6 + 1*.8*.6*.4/2 + 1*.2*.6*.6/2 + 1*.2*.4*.4/2 + 1*.2*.6*.4/3 + 1*.8*.4*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 4.8% + 9.6% + 3.6% + 1.6% + 1.6% + 4.267% + 9.6% + 4.8% = 39.87%
Mega Man gets P[¬AB¬C¬D] + P[AB¬C¬D]/2 + P[¬ABC¬D]/2 + P[¬AB¬CD]/2 + P[¬ABCD]/3 + P[AB¬CD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 0 + 1*.8*.4*.6/2 + 0 + 0 + 0 + 1*.8*.4*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 9.6% + 4.267% + 9.6% + 4.8% = 28.27%
Ryu gets P[¬A¬BC¬D] + P[A¬BC¬D]/2 + P[¬ABC¬D]/2 + P[¬A¬BCD]/2 + P[¬ABCD]/3 + P[A¬BCD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 0 + 1*.2*.6*.6/2 + 0 + 0 + 0 + 1*.2*.6*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 3.6% + 1.6% + 9.6% + 4.8% = 19.60%
Terra gets P[¬A¬B¬CD] + P[¬A¬BCD]/2 + P[¬AB¬CD]/2 + P[A¬B¬CD]/2 + P[¬ABCD]/3 + P[A¬BCD]/3 + P[AB¬CD]/3 + P[ABCD]/4
= 0 + 0 + 0 + 1*.2*.4*.4/2 + 0 + 1*.2*.6*.4/3 + 1*.8*.4*.4/3 + 1*.8*.6*.4/4
= 1.6% + 1.6% + 4.267% + 4.8% = 12.27%
End result : 39.87%|28.27%|19.60%|12.27%
Direct prop.:35.71%|28.57%|21.43%|14.29%
Old x-stats: 51.39%|31.39%|13.89%|3.33% (cute)
************
Conclusion : I don't know if people are going to agree with me on this one....but i like it!
Better yet, it's very classic and basic probabilities theory. And we can actually throw SFF in there if we want!
---
Kaxon is a figurehead. Morgoth has all the real power. - Janus5000
SpC2k4 Triple Crown WINNER (Oracle, Betting, Spread Betting)
Nothing totally incompatible with we've been using. In fact, it doesn't change the results we're usually getting in 2-way matches.
Right now we're usually something we could call "sticks", i.e. the longer sticks basically have an advantage over EVERYONE else for the "section" of the vote they have "over" them. Say if Cloud (.5) faces CATS (.2) and L-Block (.1), as long as he "rolls" over .2, he's getting that vote. And that's what we're going to change...
Starting from a 1-on-1 match.
Let's say we get Cloud 60% - Mega Man 40%
We'd give Cloud a range of [0-1.0] and Mega Man a range of [0-0.8]
Instead let's say this:
Cloud has a propability of likeability p of 100%
Mega Man's p is 80%
The idea is this. Instead of being a "max", the values we have is the probability you'll "like" the character. If you like more than one, you pick one of them randomly.
******
Cloud gets P[A¬B] + P[AB]/2 = P[A]-P[AB] + P[AB]/2 = 100%-80% + 80%/2 = 60%
Mega Man gets P[B¬A] + P[AB]/2 = P[B]-P[AB] + P[AB]/2 = 80%-80% + 100%*80%/2 = 40%
It's magic! Except it's not! We didn,t think it that way but the x-stats follow the same idea.
******
Let's throw in a 3rd character in there
Cloud's p of 100%
Mega Man's p is 80%
Ryu's p is 60%
Cloud gets P[A¬B¬C] + P[AB¬C]/2 + P[A¬BC]/2 + P[ABC]/3
=1*.2*.4 + 1*.8*.4/2 + 1*.2*.6/2 + 1*.8*.6/3 = 46.00%
Mega Man gets P[¬AB¬C] + P[AB¬C]/2 + P[¬ABC]/2 + P[ABC]/3
=0*.8*.4 + 1*.8*.4/2 + 0*.8*.6 + 1*.8*.6/3 = 32.00%
Ryu gets P[¬A¬BC] + P[¬ABC]/2 + P[A¬BC]/2 + P[ABC]/3
=0*.2*.6 + 0*.8*.6/2 + 1*.2*.6/2 + 1*.8*.6/3 = 22.00%
Direct proportionality gives : 41.67%|33.33%|25.00%
Old x-stats theory says : 52.50%|32.50%|15.00%
For some reason, I think the first one looks better!
******
This is where it hurts....4 characters
Cloud's p of 100%
Mega Man's p is 80%
Ryu's p is 60%
Terra's p is 40%
Cloud gets P[A¬B¬C¬D] + P[AB¬C¬D]/2 + P[A¬BC¬D]/2 + P[A¬B¬CD]/2 + P[A¬BCD]/3 + P[AB¬CD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 1*.2*.4*.6 + 1*.8*.6*.4/2 + 1*.2*.6*.6/2 + 1*.2*.4*.4/2 + 1*.2*.6*.4/3 + 1*.8*.4*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 4.8% + 9.6% + 3.6% + 1.6% + 1.6% + 4.267% + 9.6% + 4.8% = 39.87%
Mega Man gets P[¬AB¬C¬D] + P[AB¬C¬D]/2 + P[¬ABC¬D]/2 + P[¬AB¬CD]/2 + P[¬ABCD]/3 + P[AB¬CD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 0 + 1*.8*.4*.6/2 + 0 + 0 + 0 + 1*.8*.4*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 9.6% + 4.267% + 9.6% + 4.8% = 28.27%
Ryu gets P[¬A¬BC¬D] + P[A¬BC¬D]/2 + P[¬ABC¬D]/2 + P[¬A¬BCD]/2 + P[¬ABCD]/3 + P[A¬BCD]/3 + P[ABC¬D]/3 + P[ABCD]/4
= 0 + 1*.2*.6*.6/2 + 0 + 0 + 0 + 1*.2*.6*.4/3 + 1*.8*.6*.6/3 + 1*.8*.6*.4/4
= 3.6% + 1.6% + 9.6% + 4.8% = 19.60%
Terra gets P[¬A¬B¬CD] + P[¬A¬BCD]/2 + P[¬AB¬CD]/2 + P[A¬B¬CD]/2 + P[¬ABCD]/3 + P[A¬BCD]/3 + P[AB¬CD]/3 + P[ABCD]/4
= 0 + 0 + 0 + 1*.2*.4*.4/2 + 0 + 1*.2*.6*.4/3 + 1*.8*.4*.4/3 + 1*.8*.6*.4/4
= 1.6% + 1.6% + 4.267% + 4.8% = 12.27%
End result : 39.87%|28.27%|19.60%|12.27%
Direct prop.:35.71%|28.57%|21.43%|14.29%
Old x-stats: 51.39%|31.39%|13.89%|3.33% (cute)
************
Conclusion : I don't know if people are going to agree with me on this one....but i like it!
Better yet, it's very classic and basic probabilities theory. And we can actually throw SFF in there if we want!
---
Kaxon is a figurehead. Morgoth has all the real power. - Janus5000
SpC2k4 Triple Crown WINNER (Oracle, Betting, Spread Betting)
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