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more payouts in the ABT - pointwise

Posted By: Tom Keith
Date: Saturday, 28 February 2015, at 1:10 a.m.

In Response To: more payouts in the ABT - pointwise (Bob Koca)

Do you have a specific proposal for how to give out ABT points including getting points for each win?

Here is one idea. This could be used stand-alone or on top of the existing formula.

Single elimination tournaments

Let N be the total number of matches played by all players in the tournament. (For example, in a field of 64, N = 63.) Let n be the number of matches won by a particular player.

The number of points awarded to a player is 100 × n / N. (First-round losers get 0 points; second-round losers get 100 × 1 / 63 = 1.587 points; the winner of the tournament gets 100 × 6 / 63 = 9.524 points.)

Multiple elimination tournaments

Tournaments in which you can lose and still not be out need one more wrinkle. To calculate n for a particular player, take their number of wins and subtract their number of non-eliminating losses*. (For example, in a tournament where players are out after three losses, n for a player with 3 wins would be 3 − 2 = 1. [3 wins minus 2 non-eliminating losses equals 1.])

Some players will have a negative n. Treat those as zero. (We don't want to subtract ABT points from players who do poorly.)

Big N is now the sum of all the little n's. The number of points awarded to a given player is: 100 × n / N.
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* The loss that puts you into the Consolation round counts as a "non-eliminating loss" and a rebuy counts as a "non-eliminating loss."

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