Tournament tiebreaks

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kevinj
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Tournament tiebreaks

Post by kevinj »

Every so often we get the situation where 2 or more people in a tournament finish with the same score. This can result in players not knowing how the final result was adjudicated and organisers having to decide at the time how to resolve it, neither of which is satisfactory for any of the parties concerned. I think it would be a good idea to have a standard way to resolve tiebreaks so that everyone knows what to do when it happens and how the eventual winner was determined. The following (in no particular order) are all methods that have been used or suggested:

1) Result of any head-to-head game between the players concerned.
2) Number of Army Breaks inflicted.
3) More granular scoring (using decimal values).
4) Results against common opponents.
5) Relative strength of opposition faced (based on their total scores/positions).

Which option (or sequence of options) do you think produces the fairest outcome?

I'm going to copy this thread on the Fog R forum as it happens there too.
Scruff
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Re: Tournament tiebreaks

Post by Scruff »

What about checking to see how many points they bled to the opposition?

ps not too clear on how the existing scoring works lol
philqw78
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Re: Tournament tiebreaks

Post by philqw78 »

IMO

1) Result of any head-to-head game between the players concerned.
2) Relative strength of opposition faced (based on their total scores/positions).
3) More granular scoring (using decimal values).
4) Number of Army Breaks inflicted.
5) Results against common opponents.

Head to head must be the major one, just because it makes sense
Then having better results against the higher scoring opponents, means in the swiss scoring that they had a harder route to their tie.(not sure how you work that out though)
phil
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spike
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Re: Tournament tiebreaks

Post by spike »

Tie's and Tiebreakers

Is not that this life and death..... how about "an honorable draw", being exactly that.
A tie is a tie that's it, no count back and no recrimination on organiser's for it.

S
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hazelbark
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Re: Tournament tiebreaks

Post by hazelbark »

philqw78 wrote: Then having better results against the higher scoring opponents, means in the swiss scoring that they had a harder route to their tie.(not sure how you work that out though)
Countbacks are pretty easy to calculate in a spread sheet
dave_r
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Re: Tournament tiebreaks

Post by dave_r »

I normally work this out on the spreadsheets.

It goes something like this:

(Sum all of the points all of your opponents scored) - (Sum all of the points your opponents scored against you)

So in the Challenge my countback would have been 246 and I think Grahams was 311.
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prb4
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Re: Tournament tiebreaks

Post by prb4 »

Surely only "More granular scoring (using decimal values)" is required.

If the scores are calculated with a spreadsheet automatically rather than relying on antiquated ready reckoners (which all have errors) then there would be almost no chance of a draw.

Any number of decimal places can be taken into account to determine who actually has the higher score.

In the event that the scores really are exactly the same to any number of decimal places then there really is a draw.
philqw78
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Re: Tournament tiebreaks

Post by philqw78 »

How granular did the scoring need to be in 2010 for me to agree with you?
phil
putting the arg into argumentative, except for the lists I check where there is no argument!
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