*Editor's note:*

*This article arose out of an email correspondence involving Max Bavin, EBU Chief Tournament Director, Hermand De Wael of Belgium, David Martin, John Probst and David Stevenson of England*

Note that there is another article also following from this correspondence.**Max Bavin wrote**- If you arrow-switch 1/4 of the boards, you will be comparing with all the other pairs an equal number of times (actually, you have to make it more than 1/4, because some of the switches will cancel each other out: hence Herman's figure of 1/3).

If this were an all-play-all competition, this would be ideal and we'd all be in total agreement. However, it isn't an all-play all competition.

**The point is that we don't actually want equal comparisons (as distinct from equal balance) when it's not an all-play all event.**

Some pairs don't play against each other. As compensation for this, we want them to have to compare scores more frequently with one another.

So, in a Mitchell movement, we don't want to switch 1/4 (or 1/3) of the boards - it needs to be something less than this in order that the pairs who dont play each other (i.e. the ones sitting in the same line) compare scores quite frequently. **Herman De Wael wrote**- I understand this, but I would like to receive some indication that this is the huge difference between 1/3 and 1/8. Please David ?

The above is the reason for the difference between 1/3 and 1/8.

The 1/3 is really an approximation of a more complicated formula used to balance **comparisons only** when head to head competition is being ignored, eg. when playing an all-play-all event (or any other event if you are Swedish!). You may be familiar with the nCr statistical formula. This states that the number of different combinations of r balls that can be taken out of a sack containing n balls is given by n!/r!(n-r)!. An example. If I have a sack of 9 balls (n) then there are 36 [9!/2!(9-2)! = 9*8/2 = 36] different combinations of 2 balls (r) that can be picked from the bag. In general, if we want to pick 2 objects from a set of n then there will be n(n-1)/2 ways of doing it, ie. 9*8/2 = 36 in the earlier example.

If I have a 9 table Mitchell with no arrow-switch then there will be 36 comparisons available between different NS pairs on every board (and the same number for EW amongst their line). If I wish to balance comparisons only then I need to reduce the 36 down to 18 by the use of arrow-switches. Consider only stationary pairs who are usually NS in non-arrow-switched rounds and EW during arrow-switches.

If I arrow-switch each board once then there will now be only be 28 comparisons available between the different stationary pairs on every board, ie. 8*7/2 = 28.

If I arrow-switch 2 rounds then I will have 22 comparisons available between the different stationary pairs on every board, ie. 7*6/2 = 21 amongst the stationary pairs who are NS **plus** 1 comparison between the 2 stationary pairs who both arrow-switched the board.
If I arrow-switch 3 rounds then I will have exactly the desired 18 comparisons available between the different stationary pairs on every board, ie. 6*5/2 = 15 amongst the stationary pairs who are NS **plus** 3*2/2 = 3 comparisons amongst the 3 stationary pairs who arrow-switched the board.

Thus, three arrow-switches has perfectly balanced the number of comparisons in this case. However, it is pure coincidence that we switched **exactly** 1/3 of the rounds in this example. We would need to arrow-switch more than 1/3 if we had more tables. For example, with a single section of 20 tables, we would need to arrow-switch 8 rounds, with a single section of 30 tables we would need to arrow-switch 12 rounds, and with 2 sections, each of 30 tables, we would need to arrow-switch 13 rounds. The limiting case for an infinite number of sections or for infinitely large sections or both is that we would need to arrow-switch exactly half of the rounds.

If we wish to balance competition, ie. where we also take into account the head-to-head competition as well as the indirect competition and alliance then we can show mathematically that we now need to switch slightly greater than 1/8 of the number of rounds. The proof is as follows.

Consider a complete Mitchell movement of T tables playing N boards and arrow-switching exactly q boards. Further, consider the competition between two NS pairs who are sufficiently separated that they have no boards which they both arrow-switch in common. The indirect competition between them is N-2q. The alliance between them is 2q. There is no direct, ie. head-to-head, competition. Thus, the total amount of competition between them is N-2q-2q = N-4q.

Now, since the movement is complete, every board is played at every table and hence there is T-1 units of competition available in total to each of the two NS pairs when they play a board. Since the movement is complete, they will also play all N boards and, hence, the total amount of competition available to each NS pair is N(T-1). They each have 2T-1 opponents if we are to produce a single ranking list and, hence, we would like N(T-1)/(2T-1) units of competition with each opponent in the field. Thus, we can set N-4q = N(T-1)/(2T-1) and solve for q to find the amount of arrow-switching that will give exactly the correct amount of competition between the NS pairs that we are considering. If I subtract 1 from the denominator of the RHS of the equation, I make it bigger and get N(T-1)/((2T-1)-1) which is equivalent to N(T-1)/2(T-1) which is equivalent to N/2. Thus, I get the inequality N-4q < N/2. Rearranging, I get 4q > N-N/2, ie. 4q > N/2. Hence, q > N/8.

Finally, your wish to have a completely fair movement is eminently desirable but completely impossible in practice. Consider a simple four table Howell which is perfect, ie. I play each pair exactly once, they sit in my direction in exactly three rounds and they sit in the opposite direction in exactly three rounds. If one pair of world-wide experts (who get a top on every board they play) take part and everyone else has exactly equal ability then my ranking list will be:

Pos | Names | Score | % |

1. | World-Wide Experts | 42 | 100.00 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

2= | Normal | 18 | 42.86 |

So far, so good. Exactly what we would expect. Now replace one normal pair with a second set of world-wide experts.

Pos | Names | Score | % |

1= | World-Wide Experts | 36 | 85.71 |

1= | World-Wide Experts | 36 | 85.71 |

3= | Normal | 17 | 40.48 |

3= | Normal | 17 | 40.48 |

5= | Normal | 16 | 38.10 |

5= | Normal | 16 | 38.10 |

7= | Normal | 15 | 35.71 |

7= | Normal | 15 | 35.71 |

Now hang on! This movement was supposed to be perfect. What has happened? Well, the two lucky pairs who came 3= found that, on both occasions when a pair of experts was beating them up, the other experts were their allies and, hence, they got a joint bottom rather than an absolute bottom. The two unlucky pairs who came 7= found that, on both occasions when a pair of experts was beating them up, the other experts were their indirect opponents and, hence, they got an absolute bottom rather than a joint bottom. The two pairs that came 5= had mixed fortunes in that once the other experts were their allies and once they were their indirect opponents hence they had one joint and one absolute bottom.

Thus, even though the movement is perfect or fair, there is still luck as to the timing of who is at what position at each other table in each round. This is the effect that I refer to as *'movement dynamics'*. A completely fair movement would have to have an astronomical number of rounds to ensure that every pair played one board with all possible combinations of opponents at the other tables. Even then, some boards would be swingy whilst others are flat.

*Editor's note:*

- If you want to comment on this article, why not write direct to David Martin?
- He will be pleased to hear from you!

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