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arrowswitch advice

#1 User is offline   shevek 

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Posted 2014-August-28, 02:31

I'm running a 2-session cumulative pairs event.
Say it's13 tables, want all play all.

Session 1 = 13 x 2 Mitchell
Session 2 = 13T interwoven Howell, 13 x 2
So they play all opponents with 1 repeat.

Question: Do I need to arrow-switch the Mitchell?
Or does the Howell take care of that?

TIA
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#2 User is offline   pran 

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Posted 2014-August-28, 03:31

View Postshevek, on 2014-August-28, 02:31, said:

I'm running a 2-session cumulative pairs event.
Say it's13 tables, want all play all.

Session 1 = 13 x 2 Mitchell
Session 2 = 13T interwoven Howell, 13 x 2
So they play all opponents with 1 repeat.

Question: Do I need to arrow-switch the Mitchell?
Or does the Howell take care of that?

TIA

Your "situation" was thoroughly discussed in a guide for Norwegian tournament Directors written back in 1973.

It is possible to achieve acceptable Balance (not favouring any contestant over another) if the number of contestants divided by 3 gives a remainder of 1, i.e if the number of tables is 8, 11, 14, 17 or so on. The basic principle is to play 3 sessions, each with one Howell and two Mitchell Groups. I shall not bother you with the detailed description which may seem a bit complicated, but only remind you that any other schedule (except simple Howell and round robin baromether) will be a compromise between balance and ease of operation.

So my answers to your questions are:
1: I don't think it really makes much difference (contrary to popular belief).
2: No.
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#3 User is offline   campboy 

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Posted 2014-August-28, 03:38

View Postshevek, on 2014-August-28, 02:31, said:

Question: Do I need to arrow-switch the Mitchell?
Or does the Howell take care of that?

Yes, you do. The Howell actually makes it worse -- unswitched, you are entirely competing against your own line in the Mitchell, and mostly competing against your own line in the Howell as well -- so actually you should switch about 1/4 of the rounds (instead of 1/8 for a 1-session Mitchell).
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#4 User is offline   gordontd 

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Posted 2014-August-28, 05:20

View Postshevek, on 2014-August-28, 02:31, said:

So they play all opponents with 1 repeat.

It's usual to avoid the repeat by not playing the first round of the Mitchell.
Gordon Rainsford
London UK
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#5 User is offline   Vampyr 

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Posted 2014-August-28, 05:33

View Postpran, on 2014-August-28, 03:31, said:

It is possible to achieve acceptable Balance (not favouring any contestant over another) if the number of contestants divided by 3 gives a remainder of 1, i.e if the number of tables is 8, 11, 14, 17 or so on.


Ummm... like none of these numbers? If the remainder really is one, say you have 16 pairs, you put one pair at the stationary Howells position for all three sessions and divide the remaining 15 into three lines. They will play two Mitchells and a Howell.
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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#6 User is offline   shevek 

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Posted 2014-August-28, 05:54

View Postgordontd, on 2014-August-28, 05:20, said:

It's usual to avoid the repeat by not playing the first round of the Mitchell.


Prefer all players play all boards in play.
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#7 User is offline   bixby 

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Posted 2014-August-28, 08:00

This is how the 2-session final of the North American Pairs qualifier is run in my area. There is no arrow switch. But one significant detail: we do the Howell movement in the first session, and the Mitchell in the second. Players are tired in the second session and there would be many mistakes if that was the Howell session.
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#8 User is offline   gordontd 

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Posted 2014-August-28, 08:25

View Postshevek, on 2014-August-28, 05:54, said:

Prefer all players play all boards in play.

I'm sure those who play the strongest pair twice won't agree!
Gordon Rainsford
London UK
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#9 User is offline   gordontd 

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Posted 2014-August-28, 08:26

View PostVampyr, on 2014-August-28, 05:33, said:

Ummm... like none of these numbers? If the remainder really is one, say you have 16 pairs, you put one pair at the stationary Howells position for all three sessions and divide the remaining 15 into three lines. They will play two Mitchells and a Howell.

These are numbers of tables. Multiply them by two to get the number of pairs and then Sven's rule does work.
Gordon Rainsford
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#10 User is offline   pran 

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Posted 2014-August-28, 09:28

View PostVampyr, on 2014-August-28, 05:33, said:

Ummm... like none of these numbers? If the remainder really is one, say you have 16 pairs, you put one pair at the stationary Howells position for all three sessions and divide the remaining 15 into three lines. They will play two Mitchells and a Howell.

Correct. I wrote tables - not pairs, so 16 pairs is the same as 8 tables!
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#11 User is offline   shevek 

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Posted 2014-August-28, 19:24

View Postgordontd, on 2014-August-28, 08:25, said:

I'm sure those who play the strongest pair twice won't agree!


Okay but they like a full session. Also, 2 x 26 bds allows adding percentages for outright, rather than matchpoints.
More meaningful to the players.
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#12 User is offline   Vampyr 

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Posted 2014-August-29, 04:15

View Postpran, on 2014-August-28, 09:28, said:

Correct. I wrote tables - not pairs, so 16 pairs is the same as 8 tables!


Oops -- sorry.
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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#13 User is offline   Vampyr 

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Posted 2014-August-29, 07:08

View Postshevek, on 2014-August-28, 19:24, said:

Okay but they like a full session. Also, 2 x 26 bds allows adding percentages for outright, rather than matchpoints.
More meaningful to the players.


The repeat is a big problem; I guess it depends on how serious you want the event to be. The simplest solution would be to have another table, but perhaps Sven or Gordon can advise whether this can be made to work.
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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#14 User is offline   pran 

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Posted 2014-August-29, 08:01

View PostVampyr, on 2014-August-29, 07:08, said:

The repeat is a big problem; I guess it depends on how serious you want the event to be. The simplest solution would be to have another table, but perhaps Sven or Gordon can advise whether this can be made to work.

I am sure it can be made to work somehow, but I fear that it will require very careful nursing to avoid errors.

For more than 25 years the tradition in Norway has been competitions for pairs to be organized as baromether events with pre-duplicated boards dealt from computer programs, and I wouldn't spend any time at all considering how to organize OP's event in any other fashion.

Baromether is so simple and straight forward that it far outweights its alternatives when it comes to possible (managing) irregularities during the event.
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#15 User is offline   Vampyr 

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Posted 2014-August-29, 09:54

View Postpran, on 2014-August-29, 08:01, said:

Baromether is so simple and straight forward that it far outweights its alternatives when it comes to possible (managing) irregularities during the event.


The OP may not have enough board sets available to do a barometer event; in any case, avoiding meeting pairs twice is, IMO, much more important.
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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