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Algorithm for puntuation
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Semagon wrote
at 12:42 PM, Tuesday February 13, 2007 EST |
0 people think this is a good idea
Replies 1 - 3 of 3
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Xifis wrote
at 12:45 PM, Tuesday February 13, 2007 EST I propose this algorithm:
Y Puntuation of X player = --- x 700 - 350 Z where: Y = Number of dices losts in that game for player X. Z = Number of dices losts in the game (in the moment of the die) for all the players. Thank you. |
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Xifis wrote
at 12:46 PM, Tuesday February 13, 2007 EST y
- x 700 - 350 = puntuation z |
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Semagon wrote
at 7:38 AM, Wednesday February 14, 2007 EST This has something to do with Fermat's Last Theorem, doesn't it?
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