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Playoff with three finalists

Posted By: Nack Ballard
Date: Monday, 11 January 2010, at 12:47 a.m.

In Response To: Playoff with three finalists (Sam Pottle)

Here's an equivalent formulation of Neil's/Nack's method:

Randomly draw for a first round bye. The winner of the semifinal match gets 1/4 of the prize money; the loser gets 1/12. The winner of the finals gets 1/2 of the prize money; the loser gets 1/6.

The point is that Bill's suggestion is not as dissimilar to this method as it may appear.

I agree. With both the N/N and Bill's methods, the players all still have 1/3 prize pool equity after the bye is randomly drawn. The difference is the prize distribution.

With the N/N method, the distribution ends up [6:5:1] if the bye guy wins the finals, or [9:2:1] if the non-bye guy wins the final. (If the finalists hedge all -- to simplify the wording say they split without playing -- it's [7:4:1] in favor of the finalist who won an extra match.)

As I understand Bill's method, the distribution is [8:4:0] either way (or [6:6:0] if the finalists hedge all).

I'm not going to brand one method better or worse, but in a nutshell:

With N/N, a higher prize goes to the person who wins the tournament by playing two matches instead of one, and third place gets something. With Bill's method, the guy who wins the extra match to get to the finals and the guy who doesn't are on equal footing, they have an extra 12th of the pot to split, and third place gets nothing.

Nack

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