Why only 1 in 8?
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Why only 1 in 8?

Arrow-switches

by John Probst, London, England UK

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This article sets out to show, both empirically and also with some fancy math, that arrow switching about 1/8 of the boards in a Mitchell allows one to declare an overall winner with a great degree of fairness, and that arrow switching more than this is incorrect.

I must acknowledge my debt to David Martin who ran a brilliant seminar on this topic in 1997 and got me really interested in the subject. I should also acknowledge the contributions of the late John Manning to the subject and his book of movements that every TD should carry.

As a start point we will consider the 8-round share-and-relay movement because it illustrates the effects of the switch very clearly. For simplicity we will have "1-board rounds". The matchpointing will be by the English method where 2 points are awarded for beating another pair and one for a tie. ACBLers should just divide by two.

We will designate that pair 1NS are world experts who always make 1NT and that at all other tables the hand is passed out. This is just to keep the numbers small and easy to follow. EW have added 10. The travellers are scored on first available line with Pair 1 above Pair 8 (the sharers) and Boards 1 & 8 are below. The heading MPs shows the scores without a switch and the Switch column shows the score where the last round is switched (denoted by *s. For Board 1 we can see that 13 vs 2 get 6 and 8 respectively).

Board 1 Board 8
PairsScore MpsSwitch PairsScore MpsSwitch
NSEWNSEWNS EWNSEWNSEWNS EW
11190140140 71706886
81806868 61506886
71606868 51306886
61406868 41806886
51206868 31606886
41706868 21406886
31506868 1*1290140014
2*1306868 8*1106886

Setting out the scores for Boards 1-8 on a recap sheet with no switch we get:

Boards
Pair123456 78Total %age
1141414141414 1414112100.00
2-866666666 4842.86
11088888 885650.00
12888888 805650.00
etc Check Total: 112x8 = 896

This is as we expect in that Pair 1 takes 7.14% from Pairs 2-8 for their extra 50%

Now let us score it with the arrrow switch (denoted by the *), and something strange happens. Not only do pairs 2-8 get an extra 2 mps from the fact they get the benefit of the EW line when they switch, they also gain 2 mps from the board when Pair 1 switches and they don't. They gain 4 matchpoints.total.

Here is the recap sheet for the one round arrow switch:

Boards
Pair123456 78Total %age
1141414141414 1414112100.00
2866666 685246.43
3686666 685246.43
4668666 685246.43
5666686 685246.43
6666668 685246.43
7666666 885246.43
8666666 664842.86
1108888 8885650.00
12888888 805650.00
13688888 065246.43
14-185x8 + 2x6 + 1x0 5246.43
Check Total: 1x112 + 2x56 + 12x52 + 1x48 = 896

Some Sponsoring Organisations tend to switch half the rounds. Let us see what happens if we arrow switch the last 4 rounds

Boards
Pair123456 78Total %age
11414141414 141414112100.00
2866686 665246.43
3886688 665650.00
4888688 866053.57
5688868 886053.57
6668866 885650.00
7666866 685246.45
8666666 664842.86
1106866 8684842.86
12886886 805246.45
13688668 064842.86
14868880 685246.45
15686806 864842.86
16868088 685246.45
17680668 864842.86
18806886 885246.45
Check Total 112 + 60x2 + 56x2 + 52x6 + 48x5 = 896

Consider the one round arrow switch:
It isn't quite perfect but it's darned close. The losers are those who share with the "experts" and the gainers those who switch the same board as the experts but play in the opposite direction. (Board 8 is switched twice as it's shared, and board 4 sits out the last round) A double- weave movement would work even better but it requires player co- operation. NB If it worked perfectly all pairs (except Pair 1) would score an impossible fractional 52 plus a bit mps.

And the 4 round switch:
Depending on our seat there's a difference to us of 11% with a pair of "world experts" in the game. This is still 2% if the experts run a 60% game. It is self evident that the one round arrow switch works better. What we have done is to show the degree of competition between each other pair and a particular pair (Pair 1) with different switching strategies

Let's look at this more closely. We'll use the 8 Table Mitchell, one round switched.
[Set arm-waving on].
We can see that arrow switching one round causes a whole lot of matchpoints to change lines, compared with no switches. If I arrow-switch one round then so do all the other pairs in my line - and so compared with each other we have arrow switched one quarter of the boards.. So 3/4 of the competition comes from our line and 1/4 from the other line. Which means that nett, 3/4-1/4=1/2 of the competition is from our line, and the rest (the other half) comes from the other line. Which means we have balanced the competition between the two lines. It now doesn't matter if the strong or weak pair sits in our line or the other line, on average we get the same competition from them regardless.
[Set arm-waving off]

Probst's Rule for Seating Strategy with imperfect arrow switching:
If more than one eighth of the rounds are switched, sit half the number of tables away from the strongest pair in the room in the same line as them. It's worth up to 2%

Failing which, sit as close as possible to the weakest pair, making sure you switch the same boards as they do. That's worth up to 2%

Combine both if you can. Enjoy!

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I promised you some fancy maths and here it comes:

While you are playing a board there are three other sorts of pair:

  1. Your Direct opponents (A) You can get R-1 mps from them compared to average
  2. Your Indirect Opponents (B) playing the board the same way round. +1 mp
  3. Your Allies (C) playing the board the other way round -1 mp

If you get a top then; A loses R-1 mps, all of B loses 1 mp and all of C gains 1 mp So the degree of competition (compared with average) between any two pairs (i,j) in one line but not near each other is expressed as (R- 1).Aij + Bij - Cij matchpoints.

For R rounds with Q rounds arrow switched the competition against each other pair in the same line is 0 + (R-2Q) -2Q matchpoints. There are three terms reflecting: direct opponents (A); the pairs in our line (B); the pairs in the other line (C). The 2Q happens because both we and they switch in our line and on different boards

Since one never plays any of the people in one's own line we can ignore (R-1).A, hence the zero term (A). The R-2Q refers to those playing this board in my line (B); and the -2Q refers to the opposite line's players that I'm not currently playing (C) So we can say that the competition with each other pair in our line is R-4Q

The total amount of competition is expressed as R(T-1) [the check total], so the average competition available is R(T-1)/(2T-1) mps where 2T-1 is the number of other pairs. In other words we want to fight for the same number of matchpoints with every other pair to a grand total equal to the check total.

So for fairness in a movement we can say R-4Q must be equal R(T-1)/(2T- 1). This expression says "The competition between any two pairs in our line" must be equal to "the average competition between any two pairs regardless of line"

We rewrite this as
R-4Q must equal R(T-1)/(2(T-1) + 1)
This is unwieldy but if we drop the +1 term, then the RHS gets a bit bigger
And R-4Q must be slightly less than R(T-1)/(2(T-1),
whence R-4Q must be slightly less than R/2
and so 4Q must be slightly more than R/2
and so Q, the number of arrow switches must be slightly more than R/8,

We must arrow switch slightly more than 1/8 of the rounds for fairness. Anything else is WRONG!!!

In another article I'll be looking at multi-session, and multi-section events

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Gopher Editor's note:

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