[shevek]

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

[pran]

shevek, 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.

[campboy]

shevek, 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).

[gordontd]

shevek, 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.

[Vampyr]

pran, 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.

[shevek]

gordontd, 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.

[bixby]

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.

[gordontd]

shevek, 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!

[gordontd]

Vampyr, 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.

[pran]

Vampyr, 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!

[shevek]

gordontd, 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.

[Vampyr]

pran, on 2014-August-28, 09:28, said:

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

Oops -- sorry.

[Vampyr]

shevek, 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.

[pran]

Vampyr, 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.

[Vampyr]

pran, 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.