Home Scoring and other IT questions

An analysis of movements

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.

Comments

  • 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

    Alan

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

  • @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.

  • 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?

  • 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."

  • edited December 2019

    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

    Peter Bushby Suffolk

  • 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)

  • 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)

  • @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!

    Peter Bushby Suffolk

Sign In or Register to comment.