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Withdrawal -- What happens to previous results?

#1 User is offline   Vampyr 

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Posted 2015-September-10, 12:14

A recent thread got me thinking about something that happened a few years ago. OK, it was in backgammon and not bridge, but the principle is the same.

Suppose you have a league, which is all-play-all. (In the actual case it was all-play-all x2, which does seem to add a layer of complication but maybe not). Anyway a team withdraws before the end of the competition. Now what do we do with the results obtained against this team, and how do we score the teams that didn't get a chance to play them? If you have a default VP formula is it fair if the actual results were significantly different to what the formula would give? But in any case, suppose the method of scoring was W/L?
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#2 User is offline   FrancesHinden 

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Posted 2015-September-10, 13:13

That's why you need regulations to cover this possibility.
There are two standard options, either

1. Remove all the scores obtained against this team
2. Give the remaining teams a calculated score (a typical formula used is the highest of 12, the team's average over other matches, and 20 (max VPs) minus the withdrawn team's average over all the matches they played.

I think the strictly fairest approach is to remove all scores achieved against the withdrawn team. However in practice people don't like having time they invested playing made 'worthless' (particularly if each match in the league was a whole evening), so which of these I would do (i.e. which I would write in my regulation) depends on how many of their scheduled matches the withdrawn team has played: fewer than half and I expunge results, more than half and use the formula.

The EBU Premier League has the following rules (after saying that you aren't allowed to withdraw)

If a contestant has completed fewer than half the scheduled matches, the results of all such matches will be expunged.
If such a contestant has played at least half the matches, all results will stand. The unplayed matches will be scored using the formula in Appendix C

[appendix C says for matches on a 20 VP scale]
The non-offending side receive the higher of 13 VPs or the result of this formula.
15 * (x / (20-x)) / (x / (20 - x)) + (y / (20 - y)) + 5
where
x = Non-offending side’s average result in VPs
y = Offending side’s average result in VPs

Appendix C doesn't say so, but if x = 20 I believe they get 20
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#3 User is offline   Vampyr 

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Posted 2015-September-10, 13:17

Thanks Frances. What do you do if the scoring is W/L?
I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones -- Albert Einstein
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#4 User is offline   blackshoe 

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Posted 2015-September-10, 19:51

I must be reading it wrong. It looks to me like if x=20, the formula blows up because you're dividing by zero.
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#5 User is offline   Cascade 

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Posted 2015-September-10, 20:21

View PostFrancesHinden, on 2015-September-10, 13:13, said:


[appendix C says for matches on a 20 VP scale]
The non-offending side receive the higher of 13 VPs or the result of this formula.
15 * (x / (20-x)) / (x / (20 - x)) + (y / (20 - y)) + 5
where
x = Non-offending side’s average result in VPs
y = Offending side’s average result in VPs

Appendix C doesn't say so, but if x = 20 I believe they get 20


I think you have some parentheses misplaced because (x / (20-x)) / (x / (20 - x)) = 1 and so the formula as written always gives you more than 20 VPs at least for the likely non-negative y.

The formula also has a problem if y=0. Which reminds me I have seen a worse case in which a team on a negative score withdrew. They had lost every match 25-0 and had incurred a penalty for arriving late to one match.
Wayne Burrows

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#6 User is offline   gordontd 

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Posted 2015-September-11, 09:16

View PostCascade, on 2015-September-10, 20:21, said:

I think you have some parentheses misplaced because (x / (20-x)) / (x / (20 - x)) = 1 and so the formula as written always gives you more than 20 VPs at least for the likely non-negative y.

The formula also has a problem if y=0. Which reminds me I have seen a worse case in which a team on a negative score withdrew. They had lost every match 25-0 and had incurred a penalty for arriving late to one match.

Yes, it should be 15 * (x / (20-x)) / ((x / (20 - x)) + (y / (20 - y))) + 5

y = 0 doesn't give a problem: 0/20 = 0 and the NOS get 20.

x=20 (and y=20) do theoretically give a problem, but it's easily resolved in practical terms in the unlikely event of either of them occurring, by noting that the result of the calculation tends towards 20 as x gets bigger, it tends towards 5 as y gets bigger, and it's 12.5 whenever x=y. Note also that whenever it's below 13 the NOS get 13.
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#7 User is offline   Vampyr 

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Posted 2015-September-11, 18:01

Does anyone have any ideas about what to do if it is W/L? And the special case where it is a double all-play-all so teams have played the withdrawn team 0, 1 or 2 times?
I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones -- Albert Einstein
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