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 Run around the Pond

Discuss about this new multiplayer game or comment current runs. (includes all versions of the game)

Game link..... Ponds
Ratings link..... Regular Pond Ratings -and- Dark Pond Ratings -and- Run in the Rain Ratings
Winners link..... All Winners - (Regular Ponds Only) - (Dark Ponds Only) - (Run in the Rain Only)


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6. 二月 2005, 08:25:18
Walter Montego 
<20 player pond Ed's proposal
Z = 20
1st gets 40, 2nd gets 20, 3rd gets 10 , 4th gets 5, 5 to 8 get 0, 9th loses 24, 10th loses 27, 11th loses 30, 12th loses 33, 13th loses 36
14th loses 39, 15th loses 42, 16th loses 45
17th loses 48, 18th loses 51, 19th loses 54
20th loses 57
I suppose when two or more people drop on the same round the points would work the same. Or you could add up the total and divide equally rounding where necessary. Seems like it'd work. Start everyone at zero. Nothing wrong with a negative rating in my book. Or start at whatever is the starting point here and limit the lowest to 100. Tripling the score for the early droppers seems kind of harsh to me. Why not just double it or leave it single based on when you drop? Perhaps a scale would work to smooth out the edges too. A power function with 2 as the base? Or "e" the base of the natural logrithms?
Here's a formula that'd work as a smooth curve and wouldn't be too harsh to people that drop early while only awarding the most points to the very last players in. At the half way point the award for winning starts. Before then deduction takes place, less as you last longer.
Z = The starting amount of people
e = the base of the natural logrithms
P = final position of player
C = constant (this can be changed to increase or decrease award given. In my examples that follow I use a value of C = 1)
R = rating points awarded or deducted
Top half finishers award:
R = (Z/2) times [(e raised to the reciprocal of P)-C]
Bottom half finishers deduction:
R = (Z/2 - P) + .6 then rounded
In the 20 player game example the values would be
1st gets 17
2nd gets 6
3rd gets 4
4th gets 3
5th gets 2
6th gets 2
7th gets 2
8th, 9th, and 10th get 1
11th stays even
12th loses 1
13th loses 2
14th loses 3
15th loses 4
16th loses 5
17th loses 6
18th loses 7
19th loses 8
20th loses 9
You could double or triple the deductions if you wanted to.
In a 100 player pond the awards would be
1st 86, 2nd 32, 3rd 20, 4th 14, 5th 11
6th 9, 7th 8, 8th 7, 9th 6, 10th 5
11th 5, 12th 4, 13th 4, 14th 4, 15th 3
16th through 20th-- 3
21st through 33rd-- 2
34th through 50th-- 1
51st stays even
52nd loses 1 and so on

It's kind of hard to write mathematical formulars on this keyboard, but those that know this stuff can see what I'm angling at. A curve with a gentle slope until the very end when the reward for winning or coming in close increase a lot. If you make it half way, you'll not lose rating points. The deduction can be varied also, but keeping it a straight linear type of penality is easy to understand and most people go along with that. Lowering the constant C will increase the award. I hope the numbers line up after I enter them to post.
20 player pond
-------> C = 1 C =.5 C = 0 C = -.5
1st place 17 22 27 32
2nd place 6 11 16 21
3rd place 4 9 14 19
10th place 1 6 11 16
100 player pond
1st place 86 111 136 161
2nd place 32 57 82 107
3rd place 20 45 70 95
10th place 5 30 55 80
25th place 2 27 52 77
50th place 1 26 51 76
The divisor of Z can be varied to, but first this whole idea has to be figured with if it's acceptable for using as a rating system. Also using C = 1 you could multiply the whole award by some number to put the range of awards to whatever value seems appropriate for the size pond in question.


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