[Attempt at] New PBEM rating system - continued

[blackice]

so the leading player ends up with INFINITY and the losing player with either - INFINITY or 0 ?

That would be the same for any system so the obvious answer would be yes.

Why make it all so complex, with all that capped stuff? Can we not simply go with merely 10% original stuff?

Actually this is very simple, the original 10% was designed for a game ending ranking. So doing this every 9th turn showed up flaws.

The players in the individual games would either take turns or one person would do the math and hopefully if all goes well...enter it to the automated rankings. They would then just have to post a zip file if asked to comfirm the scores. The scores for the individual games would be posted in thier respective threads.

Only the final score for the over-all rankings would be shown in the main rankings thread.

This takes out the work as someone just has to check from time to time to make sure it is working ok.

Easy huh?

We could use programs such as DaDaBIK to allow people to update thier own game players scores to an automated PHP form. Or a score keeper with the password to update them only. That would be all the score keeper would have to do, update and delete, no math no lawyers.
Simple enough?

[Mathemagician]

Originally posted by blackice

so the leading player ends up with INFINITY and the losing player with either - INFINITY or 0 ?

That would be the same for any system so the obvious answer would be yes.

now here you got something badly wrong !
neither of the systems i proposed during the course of this thread or its predecessor was meant to have the cumulative effect that i heard (at least one of) the original system(s) had.

the actual score gained from a game would be replaced every time a scoring turn (9' or whatever) comes around.
so if you got 1.54 points at turn 9 and then 0.31 points at turn 19, you'd effectively get +1.54 at turn 9, then -1.23 at turn 19 to make the effective score +0.31...

[Mathemagician]

AND ... it could be done automated too, using Excel or whatever...
all the players do is enter the current ranking at each scoring turn, which is the least work necessary for ANY system that wants to be updated before the end of the game.

[blackice]

the actual score gained from a game would be replaced every time a scoring turn (9' or whatever) comes around.
so if you got 1.54 points at turn 9 and then 0.31 points at turn 19, you'd effectively get +1.54 at turn 9, then -1.23 at turn 19 to make the effective score +0.31...

So if you won each and every round 9 turn until the end of the game? Which was your query so lets talk apples to apples.

So if you got 1.54 on turn 9 and won with 1.54 in turn 19 you would get 1.54 until the end of the game if you won all rounds?

That being the case then each game would be capped at a maximum loss and win amount, correct? And we would end up with the "Klair" factor.

So I guess we conclude all systems if you won or lost each and every turn would end up with status quo points. While yours does not cumulate the score remains the same from start to finish if you win or lose all rounds.

Capped in other words, explain the advantage of this capped method?

[Mathemagician]

Originally posted by blackice

the actual score gained from a game would be replaced every time a scoring turn (9' or whatever) comes around.
so if you got 1.54 points at turn 9 and then 0.31 points at turn 19, you'd effectively get +1.54 at turn 9, then -1.23 at turn 19 to make the effective score +0.31...

So if you won each and every round 9 turn until the end of the game? Which was your query so lets talk apples to apples.

my query ? as far as i understood it, you stated that each system would have the problem (i see it as a problem) that the leading player would gain massive points and thus the length of the game would let him lead the rankings more than the number of his games.

i simply stated that this wasnt so.

So if you got 1.54 on turn 9 and won with 1.54 in turn 19 you would get 1.54 until the end of the game if you won all rounds?

you would GET 1.54 once (the first time you make place 1) and would KEEP it for the rest of the game (as long as you retain place 1).

That being the case then each game would be capped at a maximum loss and win amount, correct?

there would be a maximum number of points you could gain or lose, yes.

And we would end up with the "Klair" factor.

sorry, i dont know what that is.

So I guess we conclude all systems if you won or lost each and every turn would end up with status quo points. While yours does not cumulate the score remains the same from start to finish if you win or lose all rounds.

yes. as long as you 'hold' your position, be it positive or negative, your score wouldnt change.

Capped in other words, explain the advantage of this capped method?

the advantage of the points not being cumulatively collected, if you meant that, is that

1.) your actual score in the ranking reflects, and only reflects, the number of games you play and the positions you have in these games.

in some variants your rating would also depend on the strengths of your respective opponents in one or the other way, like my attempt at an multiplayer ELO system, or Solver/quinns's proposal of calculating a MP game as several SP games using an ELO or quadratic formula with the player scores as input.

2.) to make sure to lead the ranking table, you need to play more games, ergo you have more risk to not win them all, ergo someone else with a similar number of games and a higher winning percentage would overtake you.

also, depending on the specific system and the player numbers of your games, several 2nd places would sum up to be worth one 1st place, several 3rd's would be worth a 2nd and so on.

[Mathemagician]

how about this suggestion...

everyone who likes to can propose ONE rating/ranking system.
each of those people starts a ranking table according to his own system. the to-be-ranked games start.

once a a certain amount of results are in that meet certain criteria (for example: 10 total ranked games, 5 of which have reached turn 30, 2 of which have reached turn 60) the table of each ranking system is presented and all participating players vote for their most liked candidate.

this might have the effect that one votes for the system most benefitting himself, but the differences in that shouldnt be too severe as all systems would use the same game results...

[blackice]

Ok so you too would cap win/loss in a game.

Well toss out your formula the 10% is there take two minutes to create a three game scenario and post it.

I will do the same.

[Mathemagician]

i'll leave it to Solver to do that using quinns' quadratic formula and counting a MP game as several SP games.

i believe i have finally found a formula specific for MP games. im still experimenting, but it seems to work.

[Mathemagician]

yip !
i gave the MP ELO system a second chance, and i succeeded !!

my formula (which is remotely based on the exponential ELO 2 player system) now is:


Drat = K1 * ( r(i) - s(i) )


or more precisely:

Drat = K1 * ( r(R(i)) - ( P(i) - ( Sum{ P(j); j != i } / (n - 1) ) ) / Pmax - Pmin + K2 )


where:

i = the pure index of the current player (=you)
Drat = your rating change due to this game
n = number of players in the game
K1 = constant that determines how many points can be won or lost in that game
K2 = constant that determines how important the rating (=point) difference of the participating players will be for the outcome
R(i) = the rank in the power graph that player i had on the last rated turn (9*)
r(R(i)) = the modifier associated with player i's rank in the game (-1 to +1)
P(i) = the current rating of player i, counting only his base score and his finished games
Pmax = the score of the best player playing in this game
Pmin = the score of the worst player playing in this game

thus:

s(i) = ( P(i) - ( Sum{ P(j); j != i } / n - 1) ) / Pmax - Pmin + K2 = everything right of r(R(i)), your modifier for your expected result, based on your rating, ranging from -1 to +1, though a little bit more leaning towards 0.

Sum{ P(j); j != i } / n - 1 = the average score of all players in the game except you

( P(i) - ( Sum{ P(j); j != i } / n - 1) ) / Pmax - Pmin + K2
the difference between your score to the average player score in this game, weighed towards the total score difference of the game, plus a little surplus, so that the total result varies roughly from -1 to +1

i assumed both K1 and K2 to have value 10.

so here is the explained, hopefully more comprehensive, version:


your rating change =
10 * (rank modifier - rating modifier)


each modifier ranges from -1 to +1, so the total score variation might possibly be between -20 and +20

assuming a base rating of 1000.00 for new players, this should be appropriate.
constants K1 and K2 can easily be changed to adjust the characteristics of the system.

here is an example:

in a 3 player game, A(dam), B(rian) and C(hris) are playing.
the ratings before the game:
A : 1050
B : 1020
C : 980

Adam, as the best player, will gain only +4 points if he wins (the winner always comes out positive). if Adam places second, though, he loses (-)6 points, as his expected result, being the best player, was a win.
if he totally scrubs out and loses (3rd place) he would lose (-)16 points.

Brian, on the other hand, is expected to finish 2nd (microscopic tendency towards 1st place, due to averaging).
if he wins, he gains +9 points. second place means -1 (due to the slightly positive expectation tendency mentioned above), if he places 3rd he will make -11 points.

Chris, as the total outsider, will always win points, unless he loses the game.
1st place: +17
2nd place: +7
3rd place: -3

(points are rounded for whole values)

so, for several finishing orders:

ABC : +4 / -1 / -3 // SUM = 0
ACB : +4 / -11 / +7 // SUM = 0
BAC : +9 / -6 / -3 // SUM = 0
BCA : +9 / +7 / -16 // SUM = 0
CAB : +17 / -6 / -11 // SUM = 0
CBA : +17 / - 16 / -1 // SUM = 0

[Grandpa Troll]

Originally posted by Mathemagician
yip !
i gave the MP ELO system a second chance, and i succeeded !!

my formula (which is remotely based on the exponential ELO 2 player system) now is:


Drat = K1 * ( r(i) - s(i) )


or more precisely:

Drat = K1 * ( r(R(i)) - ( P(i) - ( Sum{ P(j); j != i } / n - 1) ) / Pmax - Pmin + K2 )


where:

n = number of players in the game
Drat = your rating change due to this game
K1 = constant that determines how many points can be won or lost in that game
K2 = constant that determines how important the rating (=point) difference of the participating players will be for the outcome
i = the pure index of the current player (=you)
R(i) = the rank in the power graph that player had on the last rated turn (9*)
P(i) = the current rating of player i, counting only his base score and his finished games
Pmax = the score of the best player playing in this game
Pmin = the score of the worst player playing in this game

thus:

r(R(i)) = your score modifier due to your rank in the game, ranging from -1 to +1

s(i) = ( P(i) - ( Sum{ P(j); j != i } / n - 1) ) / Pmax - Pmin + K2 = everything right of r(R(i)), your modifier for your expected result, based on your rating, ranging from -1 to +1, though a little bit more leaning towards 0.

Sum{ P(j); j != i } / n - 1 = the average score of all players in the game except you

( P(i) - ( Sum{ P(j); j != i } / n - 1) ) / Pmax - Pmin + K2
the difference between your score to the average player score in this game, weighed towards the total score difference of the game, plus a little surplus, so that the total result varies roughly from -1 to +1

i assumed both K1 and K2 to have value 10.

so here is the explained, hopefully more comprehensive, version:


your rating change =
10 * (rank modifier - rating modifier)


each modifier ranges from -1 to +1, so the total score variation might possibly be between -20 and +20

assuming a base rating of 1000.00 for new players, this should be appropriate.
constants K1 and K2 can easily be changed to adjust the characteristics of the system.

here is an example:

in a 3 player game, A(dam), B(rian) and C(hris) are playing.
the ratings before the game:
A : 1050
B : 1020
C : 980

Adam, as the best player, will gain only +4 points if he wins (the winner always comes out positive). if Adam places second, though, he loses (-)6 points, as his expected result, being the best player, was a win.
if he totally scrubs out and loses (3rd place) he would lose (-)16 points.

Brian, on the other hand, is expected to finish 2nd (microscopic tendency towards 1st place, due to averaging).
if he wins, he gains +9 points. second place means -1 (due to the slightly positive expectation tendency mentioned above), if he places 3rd he will make -11 points.

Chris, as the total outsider, will always win points, unless he loses the game.
1st place: +17
2nd place: +7
3rd place: -3

(points are rounded for whole values)

so, for several finishing orders:

ABC : +4 / -1 / -3 // SUM = 0
ACB : +4 / -11 / +7 // SUM = 0
BAC : +9 / -6 / -3 // SUM = 0
BCA : +9 / +7 / -16 // SUM = 0
CAB : +17 / -6 / -11 // SUM = 0
CBA : +17 / - 16 / -1 // SUM = 0

Ya know.. I deal with Theoretical Values on a dailey basis and specific gravities to deal with yield to produce 27 cubic feet for each cubic yard of concrete.

I now see that not only do I need to learn a lot more about this game I love, but on TOP OF THAT I gotta be a Calculus/Algebra/and MEGA MATH MAJOR!!


I gotta tell ya, Ill play you tell me if I get any points or if I lose any..because I sure as heck dont how in the heck anyone thinks THIS SYSTEM meets criteria known as "SIMPLE"




I am for real..you guys are A#-1 in mathematics but for real cant we just have a simple method..this is crazy to try to figure out..

BUT..I know i know..I dont HAVE to play..if i dont like it..

Peace

Grandpa Troll

a.k.a. The Dumb@$$-who-aint no-math-major!

[Mathemagician]

i was just posting this for the people interested.
basically i stick to my suggestion made above:

we start out with several systems and a few weeks into the season the players vote for one system to continue with.

no math knowledge involved !

[Grandpa Troll]

Originally posted by Mathemagician
i was just posting this for the people interested.
basically i stick to my suggestion made above:

we start out with several systems and a few weeks into the season the players vote for one system to continue with.

no math knowledge involved !

Yeah..snicker snicker snicker..

I used to gamble...

LOVED playing cards

we had a saying Math..

"Bring your money..we'll play with your pockets"

or another

"Earn-as-u-learn"

I am just saying..although you bright individuals may understand CLEARLY..this old country boy, knowing a good bit..am-dumb-as-dirt when it comes to above formulas..

THATS math and thats not simple..

Might I suggest...

K-keep

I-it

$-Simple

$-Stupid

Peace

Grandpa Troll

By The Way

Pete was in the games lobby ifin your interested

[blackice]

You see Math this is exactly what people want, simple.

A formula may or may not work but does the system?

How many rules are going to be needed on top of the already hard to figure out formula?

Now from what I see you have recreated the old system.
One loss to a newbie like paul and you are never going to see the light of day.

One good player like paul in one game and he is top dog until he retires. Yes I know you have added the multiple game factor. So he starts another and another until he gets the best starting position and bingo!

Paul just using you as an example no reality applied here.

You have seen how simple the 10% system is, you want to cap it as yours is so be it. At least it takes into account land and all the other unique aspects of this game.

A game of skill I.E. chess my have a sweat factor but it does not have even 1 tenth the variables (unknown) as this game does.

That is the major problem with using a solely based on skill ranking... It sucks, is complicated and does not this game justice...

[ricketyclik]

I like the idea of a change in scores only with a change in rankings between x9 turns. It gives trailing players an incentive to stay the distance.

I also think that a penalty should be applied for dropping a game. How about: the retiring player immediately drops to last place in the game, scores are recalculated and assigned, the replacement player gets the rank reflected by the PG and only gains or loses points by a change in rank from thereon?

[Mathemagician]

@blackice
so could you please tell me understandably where exactly you see a flaw in this system. your prior statements are just insinuations and dont really help me.

@ricketyclik
retiring players would be handled in that the player takes up the last slot (but ahead of any previously retired players) for the rest of the game.
i admit, i didnt think about replaced players yet, but im sure this can be incorporated, like everything can if one wants it.