I am often asked about arrow-switching, why there should not be more (or less), so here is an attempt to explain the ideas without (much) recourse to algebra.
In a pairs event the total amount of competition can be defined as the total number of match-points awarded to NS on all the travellers. Or EW - the total is the same.
This total is the product of four factors:
Factor (i) is determined by the number of players who turn up.
Factors (ii) and (iii) depend on the time allocated - in most clubs the product of Factors (ii) and (iii) is likely to be in the range 24-28.
Only factor (iv) is at the disposal of the Director, who should maximise it by choosing a complete movement, where all the players play all the boards, and half a top is one less than the number of tables in play. An incomplete movement simply cuts down the amount of competition, and makes the result more of a lottery, less of a test of skill.
The total amount of competition ought to be as large as possible - otherwise we might as well play rubber bridge.
For example, in a 7-table Mitchell, (and suppose for simplicity that we are playing 1-board rounds), the total amount of competition is 7 x 7 x 1 x 6 = 294.
This figure of 294 can be split up amongst the various pairs of pairs:
Pair 1 competes against each of the other 13 pairs, Pair 2 competes against Pairs 3 to 14, and so on, 91 pairs of pairs in all. (If there are P pairs competing, the number of pairs of pairs is P x (P-1)/2 ).
A movement is balanced if each pair competes fairly and equally against each of the other participating pairs.
If you want a single-winner 7-table Mitchell movement each pair of pairs ought to compete by 294/91 = 3.23
This cannot be arranged exactly, because all measures of competition are whole numbers, but the movement would be as balanced as possible if all the amounts of competition were 3 or 4. By judicious arrow-switching we can approach the ideal. This is how to get a precise measure of the amount of competition between two pairs, say Pair 1 & Pair 2:
Consider each board (or board set) separately and allocate a competition score thus:
|If Pair 1 and Pair 2
play the board
|against each other||score half a top|
|in the same direction||score 1|
|in opposite directions, but not against each other||score -1|
We can calculate the competition scores between every possible pair of pairs, and for a balanced movement these scores should all be equal or approximately so. To see how this scheme works out, look at the 7-table Mitchell, switched on the last round.
|Round 1||1A 8||2B 9||3C10||4D11||5E12||6F13||7G14|
|Round 2||1B14||2C 8||3D 9||4E10||5F11||6G12||7A13|
|Round 3||1C13||2D14||3E 8||4F 9||5G10||6A11||7B12|
|Round 4||1D12||2E13||3F14||4G 8||5A 9||6B10||7C11|
|Round 5||1E11||2F12||3G13||4A14||5B 8||6C 9||7D10|
|Round 6||1F10||2G11||3A12||4B13||5C14||6D 8||7E 9|
|Round 7||9G 1||10A 2||11B 3||12C 4||13D 5||14E 6||8F 7|
(In this matrix, a group of symbols such as 2D14 under Table 2 in Round 3 means that Pair 2 is NS, Pair 14 is EW, and they play the fourth set of boards (D being the fourth letter of the alphabet)).
|The competition score between Pair 1 & Pair 2 works out thus:|
|On boards B,C,D,E,F||Pair 1 & Pair 2 both play NS:||Score 5|
|On boards A and G||Pair 1 & Pair 2 play in opposite directions:||Score -2|
|Total competition score||= 3|
|A further example: Pair 3 and Pair 8|
|On board E||they encounter one another:||Score 6|
|On boards B and F||they play in the same direction:||Score 2|
|On boards A,C,D,G||Pair 3 is NS and Pair 8 is EW:||Score -4|
|Total competition score||= 4|
Continuing in this way for all the pairs of pairs, we arrive at
The total competition score is equal to 42 x 3 + 42 x 4 = 294, as we know it must be.
A measure of the spread of competition scores (the smaller the better) is given by the standard deviation of these numbers, which is calculated as the square root of (SS - S x S/N)/N
|SS means||the sum of squares of the scores||(in our case 42 x 9 + 42 x 16 = 1050)|
|S means||the total competition||(in our case 294)|
|N means||the number of pairs of pairs||(in our case 91)|
|The standard deviation works out at 1.05 and cannot be further reduced by switching more or fewer boards|
For more complicated movements, a computer program is needed to calculate the standard deviation, and to find the switching regime which minimises it.
In the case of the unswitched Mitchell, the competition score is 7 between any two NS pairs, 7 between any two EW pairs, and 0 between any NS and any EW pair. Hence the two groups of players are uncoupled from each other, but there is more intense competition within the NS field and within the EW field. The total amount of competition (294) is the same whether or not you apply arrow-switching, but arrow-switching spreads the competition more thinly and enables you to issue a single list of results.