An analysis of movements

gordonrainsford

One of our members Richard Hilton, who has spent a number of years analysing pairs movements, has provided us with the attached paper representing his conclusions.

16248

As far as I can understand this, it leaves out almost all of the movements we normally use - Mitchells, skip Mitchells and Web Mitchells.
But it does seem to imply in the introduction that these movements are all good.
Alan

TawVale

Richard should be thanked on behalf of all those who have said "that movement seemed unfair".
It appears that we have a mathematical proof.

gordonrainsford

@16248 said:
As far as I can understand this, it leaves out almost all of the movements we normally use - Mitchells, skip Mitchells and Web Mitchells.
But it does seem to imply in the introduction that these movements are all good.
Alan

I think he will be looking at these as his next project. His main purpose seems to have been to point out that half-tables have their own properties and movements specifically created or chosen for them will improve their "fairness". That's certainly something I haven't seen addressed before.

Paul_Gibbons

Thanks to Richard for this. I haven't had the opportunity to go through the appendix in detail. As I have not been able to get Jeanie to work I have been planning to write some code to establish the fairness of movements. I will consider whether I should use Richard's method or the simpler formulae previously advocated. I am particularly interested in 2-session movements as our county events these days have too many entries for an all-play-all over two sessions and I am aware that some of the movements we currently use are unfair.

To go back to Richard's single-winner 1-session movements it is interesting that the weave comes out better than the share and relay for 8 tables and perhaps we could emphasise this more to club TDs,

However there do seem to be a couple of omissions of very common movements or I may be missing something.

With 7 tables I have always thought the Full Howell (Blue EBU26 F 7, Green EBU24A F 7, Saffron EBU24B T 7) was a perfect movement and this does not seem to be mentioned and the effect of a missing pair analysed.

With 7.5 tables, wanting to play 27 boards (so that sitting-out pairs play 24 boards), I and others frequently use an 8-table hesitation Mitchell with an arrow switch on the last round except at the switching table and with the missing pair being NS at Table 1 to avoid board sharing rather than the pairs suggested. How fair is this and how does this compare with other options for 7.5 tables?

gordonrainsford

Richard has kindly provided me with some further information to help me answer a correspondent about the relative merits of a Hesitation Mitchell or a two-winner 11-table 8-round Mitchell, as follows:

10-table skip Mitchell, playing 8 rounds: 7.504
10-table skip Mitchell, playing 9 rounds: 5.098
11-table Mitchell, playing 8 rounds: 9.459
11-table Mitchell, playing 9 rounds: 7.658

He says: "These imply that 2-winner incomplete Mitchells are significantly less equitable than any reasonable 1-winner movement, due to the effect that a significantly strong or weak pair can have on the players in the opposite direction, depending on which they play against and which they miss. In a 1-winner movement this effect is partially ameliorated by the direction-switches.

"Full Mitchells are, of course, perfectly equitable."

Peter

I'm probably missing something but looking at this Richard says:

"(A) With an even number of pairs: Writing T = the number of tables, so 2T = the number of pairs and R = the number of rounds = the number of boards,
the “top” pair’s total score will be 2xRx(T-1) leaving a total of 2xRx(T-1) squared to be shared between the remaining (2T-1) pairs i.e. and average of 2Rx(T-1) squared / (2T - 1) per pair

I may be wrong but I'd have thought the remaining total to be shared would be:
Rx2xSUM(1 to (T - 2)) = rx2x(T-2)x(T-1)/2 = Rx(T-2)x(T-1) assuming SUM(1:N) = Nx(N+1)/2

giving a slightly different average of Rx(T-2)x(T-1)/(2T-1)

I'm not at all sure I'm right and I have no idea what impact it would have if I was bur can someone point out where
I'm going wrong and put me out of my misery?

Thank you

ManchesterRambler

can someone point out where
I'm going wrong and put me out of my misery?

Total MPs available on each board = 2T(T-1)
Total MPs available on R rounds (boards) = 2RT(T-1)
MPs taken by top pair = 2R(T-1)
Remaining MPs = 2RT(T-1) – 2R(T-1) = 2R(T-1)(T-1)

ManchesterRambler

I think you have forgotton to include the MPs available to the pairs playing in the same direction as the pair who scored Nil against the top pair which is 2RT(T-1).
2R(T-2)(T-1) + 2RT(T-1) = 2R(T-1)(T-1)

Peter

@ManchesterRambler said:
I think you have forgotton to include the MPs available to the pairs playing in the same direction as the pair who scored Nil against the top pair which is 2RT(T-1).
2R(T-2)(T-1) + 2RT(T-1) = 2R(T-1)(T-1)

Yes I have :) Thank You!