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Edit:
After a bit of pondering, I have come to the conclusion, that my system of on-numbers does not reflect the phenomenon commonly referred to as "ownage" well enough.
Frag ratio is not enough to decide what is "ownage". This was brought to my attention by the example of these two games:
50:10
and
5:1
somewhere in the comments. 50:10 is ownage, but 5:1 isn't, but both have the same frag ratio 5. Thank you.
My initial problem with frag-difference remains. I still think that
50:40 is pretty close and 10:0 is not that close (both games have the same frag-difference 10).
I believe it's common sense, that the higher the total frag count, the less the influence of frag-difference should be.
This could be expressed with a new formula for a new ranking number "RN":
RN = 1/(1 + FT) * FD,
where FT = number of total fragsin the game
and FD = frag-difference.
It's a rational number between 0 and 1. The closer it is to 1, the more "ownage".
A very straight-forward formula and closely related to
FD/FT obviously, but with the advantage that FT = 0 is not a problem anymore. But it's somewhat more abstract, since it's not a percentage of frag-counts per total frags for example.
A few number examples:
15:0 => RN = 0,94
30:0 => RN = 0,97
50:10 => RN = 0,66
5:1 => RN = 0,57
10:10 => RN = 0
11:10 => RN = 0,05
50:40 => RN = 0,11
10:0 => RN = 0,91
I think, this is a little better than my initial "on".
The advantages are:
* Number of frags are taken into account
* If both players are matched up for a long time and they make a lot of frags, the final frag difference is not weighed as much as this difference would be represented by an x:0 game,
e.g. a 30:25-game gets only a 0,09, but a 5:0-game gets a 0,83 - the winner likely was either able to defend an early lead, or he could clearly win a shorter nerve-batthe in the end of the game.
* The RN of games like x:0 increase with increasing x.
------------------------------
Original (old) article:
Need a handy way to determine how much somebody owned in a Q3-game? Seems easy.
Heuristic point: If we knew the number of frags of two 1on1- games beforehand, wouldn't it be reasonable to say, that the game where the winner got the bigger percentage of total frags had a higher "ownage-number"" (i.e. was a better victory)?
I think, that absolute frag difference doesn't mean anything. If the result is 100:90, it's obviously not as much pwnage as 11:1. But both games have the same frag difference 10, since 100 - 90 = 11 - 1 = 10.
We have to take into account how many frags have been made in total. Let's define the "ownage number" (on) for a given player as
on(player) = player-score / #(total frags in the game),
where # = "number of" and
suicides count as negative frags in the number of total frags in the game
Let's look at the ESWC finals and the on's (approximate values given):
hub: 41:22 for czm, on(czm) = 41 / 63 = 0,65
DM6: 16:12 for cooller, on(cooller) = 16 / 28 = 0,57
DM13: 13:3 for cooller, on(cooller) = 13 / 16 = 0,81
Conclusion: cooller owned czm harder on DM13 than czm owned cooller on hub, because his on is higher! Since he won the other game, cooller played better. QED
P.S. Side remarks: If at least one player has a non-negative number of frags, A) and B) are true:
A) If player1 and player2 are competing, it is clear that
on(player1) + on(player2) = 1
B) A game is won by "player" if and only if
on(player) > 0,5
C) Collect as many games as possible from two arbitrary players, calculate the on for each game and each player. And take the arithmetic mean to determine the ultimate ownage number "uon" to determine who is the best player.
D) Of course, an alternative version of on can be expressed via the ratio of the scores of the players. "Frag difference" should be replaced with "Frag ratio" or the on.
E) Let n be an integer. If x1 and x2 are two integers with x1 > n and x2 > n. the following is true:
x1 > x2
<=>
on for the x1:n-game > on for the x2:n-game
<=>
frag difference of x1:n-game > frag difference of x2:n-game
"E)" yields the following:
If comparing two games with the same frag count for the loser (or winner), it makes no difference if the games are compared by frag difference or on. But in general, the on-system results in a different ranking of games than the frag-difference-system.: 10:5 >> 8:4 by frag difference, but 10:5 ~ 8:4 by on,
where ">>" means "ranked higher" and "~" means "has equal ranking".
After a bit of pondering, I have come to the conclusion, that my system of on-numbers does not reflect the phenomenon commonly referred to as "ownage" well enough.
Frag ratio is not enough to decide what is "ownage". This was brought to my attention by the example of these two games:
50:10
and
5:1
somewhere in the comments. 50:10 is ownage, but 5:1 isn't, but both have the same frag ratio 5. Thank you.
My initial problem with frag-difference remains. I still think that
50:40 is pretty close and 10:0 is not that close (both games have the same frag-difference 10).
I believe it's common sense, that the higher the total frag count, the less the influence of frag-difference should be.
This could be expressed with a new formula for a new ranking number "RN":
RN = 1/(1 + FT) * FD,
where FT = number of total fragsin the game
and FD = frag-difference.
It's a rational number between 0 and 1. The closer it is to 1, the more "ownage".
A very straight-forward formula and closely related to
FD/FT obviously, but with the advantage that FT = 0 is not a problem anymore. But it's somewhat more abstract, since it's not a percentage of frag-counts per total frags for example.
A few number examples:
15:0 => RN = 0,94
30:0 => RN = 0,97
50:10 => RN = 0,66
5:1 => RN = 0,57
10:10 => RN = 0
11:10 => RN = 0,05
50:40 => RN = 0,11
10:0 => RN = 0,91
I think, this is a little better than my initial "on".
The advantages are:
* Number of frags are taken into account
* If both players are matched up for a long time and they make a lot of frags, the final frag difference is not weighed as much as this difference would be represented by an x:0 game,
e.g. a 30:25-game gets only a 0,09, but a 5:0-game gets a 0,83 - the winner likely was either able to defend an early lead, or he could clearly win a shorter nerve-batthe in the end of the game.
* The RN of games like x:0 increase with increasing x.
------------------------------
Original (old) article:
Need a handy way to determine how much somebody owned in a Q3-game? Seems easy.
Heuristic point: If we knew the number of frags of two 1on1- games beforehand, wouldn't it be reasonable to say, that the game where the winner got the bigger percentage of total frags had a higher "ownage-number"" (i.e. was a better victory)?
I think, that absolute frag difference doesn't mean anything. If the result is 100:90, it's obviously not as much pwnage as 11:1. But both games have the same frag difference 10, since 100 - 90 = 11 - 1 = 10.
We have to take into account how many frags have been made in total. Let's define the "ownage number" (on) for a given player as
on(player) = player-score / #(total frags in the game),
where # = "number of" and
suicides count as negative frags in the number of total frags in the game
Let's look at the ESWC finals and the on's (approximate values given):
hub: 41:22 for czm, on(czm) = 41 / 63 = 0,65
DM6: 16:12 for cooller, on(cooller) = 16 / 28 = 0,57
DM13: 13:3 for cooller, on(cooller) = 13 / 16 = 0,81
Conclusion: cooller owned czm harder on DM13 than czm owned cooller on hub, because his on is higher! Since he won the other game, cooller played better. QED
P.S. Side remarks: If at least one player has a non-negative number of frags, A) and B) are true:
A) If player1 and player2 are competing, it is clear that
on(player1) + on(player2) = 1
B) A game is won by "player" if and only if
on(player) > 0,5
C) Collect as many games as possible from two arbitrary players, calculate the on for each game and each player. And take the arithmetic mean to determine the ultimate ownage number "uon" to determine who is the best player.
D) Of course, an alternative version of on can be expressed via the ratio of the scores of the players. "Frag difference" should be replaced with "Frag ratio" or the on.
E) Let n be an integer. If x1 and x2 are two integers with x1 > n and x2 > n. the following is true:
x1 > x2
<=>
on for the x1:n-game > on for the x2:n-game
<=>
frag difference of x1:n-game > frag difference of x2:n-game
"E)" yields the following:
If comparing two games with the same frag count for the loser (or winner), it makes no difference if the games are compared by frag difference or on. But in general, the on-system results in a different ranking of games than the frag-difference-system.: 10:5 >> 8:4 by frag difference, but 10:5 ~ 8:4 by on,
where ">>" means "ranked higher" and "~" means "has equal ranking".
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Edited by agardenchair at 21:35 GMT, 14th Jul 2005 - 44197 Hits
i'd suggest though that the on isn't linear but weighed according to the relation of frag difference to total amount of frags in order to compare two matches. thereby cooller would've owned czm twice as hard on dm13 as czm did own cooller on hub.