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Talk:Fairshare between two and more

From Rosetta Code

Perl 6 count of how many turns each person gets[edit]

Whilst important to some degree, the sequence minimises any advantage that going first/going earlier might give. I've blogged twice, here, and here about it and the sequence appears many times in science and maths. (Try this paper (PDF), for example. --Paddy3118 (talk) 23:54, 1 February 2020 (UTC)

I have to say, I kind of missed the point of the task initially so was not really sure what it was demonstrating. The actual algorithm was simple, the reason for it escaped me. After reading your links, the lightbulb lit. I removed the "number of times each person goes" which was kind-of pointless, and added a "fairness correlation" calculation showing the relative fairness to the Perl 6 entry. --Thundergnat (talk) 13:43, 2 February 2020 (UTC)
Great :-)
--Paddy3118 (talk) 18:26, 2 February 2020 (UTC)
I tried to clearify things to me, like Paddy3118 described in his links.Without different values, it makes no sense.
Rest removed, explanation by Paddy3118 (Horst.h) Horst.h (talk) 11:42, 26 June 2020 (UTC)

Fairness example and cycles[edit]

I saw Horsts' Perl program above, and recognized that the idea of fairness is hard to bring across so I thought I might do an example by hand.

The set of numbers in this case are not a linear progression, so we (maybe), see the emergence of Thue-Morse as being the most "fair" calculated as the spread in final amounts per person.

For all cases we will have have three people A B and C to choose the best at their turn, from the same, ever decreasing pots of money.

18 then 27 Fibonacci numbers[edit]

Numbers: [2584, 1597, 987, 610, 377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1]

  Order: ABC_ABC_ABC_ABC_ABC_ABC : Simple Repetition
    A gets: 2584 + 610 + 144 + 34 +  8 +  2 = 3382
    B gets: 1597 + 377 + 89 + 21 +  5 +  1 = 2090
    C gets: 987 + 233 + 55 + 13 +  3 +  1 = 1292
  Maximum difference in amounts = 2090

  Order: ABC-BCA-CAB_ABC-BCA-CAB : Simple Rotation
    A gets: 2584 + 233 + 89 + 34 +  3 +  1 = 2944
    B gets: 1597 + 610 + 55 + 21 +  8 +  1 = 2292
    C gets: 987 + 377 + 144 + 13 +  5 +  2 = 1528
  Maximum difference in amounts = 1416

  Order: ABC-BCA-CAB-BCA-CAB-ABC : Thue-Morse Fairshare
    A gets: 2584 + 233 + 89 + 13 +  5 +  2 = 2926
    B gets: 1597 + 610 + 55 + 34 +  3 +  1 = 2300
    C gets: 987 + 377 + 144 + 21 +  8 +  1 = 1538
  Maximum difference in amounts = 1388

Numbers: [196418, 121393, 75025, 46368, 28657, 17711, 10946, 6765, 4181, 2584, 1597, 987, 610, 377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1]

  Order: ABC_ABC_ABC_ABC_ABC_ABC_ABC_ABC_ABC : Simple Repetition
    A gets: 196418 + 46368 + 10946 + 2584 + 610 + 144 + 34 +  8 +  2 = 257114
    B gets: 121393 + 28657 + 6765 + 1597 + 377 + 89 + 21 +  5 +  1 = 158905
    C gets: 75025 + 17711 + 4181 + 987 + 233 + 55 + 13 +  3 +  1 = 98209
  Maximum difference in amounts = 158905

  Order: ABC-BCA-CAB_ABC-BCA-CAB_ABC-BCA-CAB : Simple Rotation
    A gets: 196418 + 17711 + 6765 + 2584 + 233 + 89 + 34 +  3 +  1 = 223838
    B gets: 121393 + 46368 + 4181 + 1597 + 610 + 55 + 21 +  8 +  1 = 174234
    C gets: 75025 + 28657 + 10946 + 987 + 377 + 144 + 13 +  5 +  2 = 116156
  Maximum difference in amounts = 107682

  Order: ABC-BCA-CAB-BCA-CAB-ABC-CAB-ABC-BCA : Thue-Morse Fairshare
    A gets: 196418 + 17711 + 6765 + 987 + 377 + 144 + 21 +  8 +  1 = 222432
    B gets: 121393 + 46368 + 4181 + 2584 + 233 + 89 + 13 +  5 +  2 = 174868
    C gets: 75025 + 28657 + 10946 + 1597 + 610 + 55 + 34 +  3 +  1 = 116928
  Maximum difference in amounts = 105504

Thue-Morse is best in this case.

Hmmm, I feel a blog coming on...

--Paddy3118 (talk) 03:50, 26 June 2020 (UTC)

Fairness and randomized rewards[edit]

I needed to further investigate Horst's work above finding repeated rotations being best, being best, and me finding that for Fibonacci (also above), Thue-Morse wins over repeated rotations.

I wrote something that for the rewards just took some random integers and ordered them highest to lowest; then used different orderings of three people selecting their best reward on the from that reward order. The winning ordering would have the least difference between peoples total reward, and crucially multiple orderings can have the win if they all have the same minimum spread of peoples winnings.

Results
## ALL SIMULATIONS BEING OF 3 PEOPLE TAKING TURNS TO TAKE EVER DECREASING AMOUNTS OF CASH FROM A REWARD_LIST.

Give summary stats on 250_000 repetitions of:
    New reward_list = an ordered, random, selection of between 3 and 84 ints from the range 999_999 .. 0:
    (Note: reward_list length is always a multiple of 3)
        Each of the following 4 methods of ordering used to order this reward_list [
            "Randomised: Gen from repeat(shuffle('ABC')) eg ABC CAB ABC ABC BCA ...",
            "Simple Repetition: Gen from repeat('ABC') eg ABC ABC ABC ...",
            "Simple Rotation: Gen from repeat(rotate('ABC')) eg ABC BCA CAB, ABC BCA CAB, ... ",
            "Thue-Morse Fairshare: Start _sequence_ not replicated later, but 'balanced' ABC's",
          ]
        The ordering(s) with the LEAST spread of winnings per person wins the round.

SUMMARY
  Wins for All 250_000 repetitions
    order_by_thue_morse : 116_022
    order_by_rotation   : 116_007
    order_by_randomise  : 51_744
    order_by_repetition :  8_905

  Wins for All 88_099 repetitions where: len(reward_list) div 9 == 0
    order_by_thue_morse : 37_591
    order_by_rotation   : 37_471
    order_by_randomise  : 13_037
  Wins for All 115_767 repetitions where: len(reward_list) div 9 == 3
    order_by_rotation   : 42_194
    order_by_thue_morse : 42_080
    order_by_randomise  : 22_588
    order_by_repetition :  8_905
  Wins for All 88_812 repetitions where: len(reward_list) div 9 == 6
    order_by_thue_morse : 36_351
    order_by_rotation   : 36_342
    order_by_randomise  : 16_119

  Wins for All 26_679 repetitions where: len(reward_list) div 27 == 0
    order_by_thue_morse : 11_310
    order_by_rotation   : 11_209
    order_by_randomise  :  4_160
  Wins for All 62_423 repetitions where: len(reward_list) div 27 == 3
    order_by_thue_morse : 20_059
    order_by_rotation   : 19_973
    order_by_randomise  : 13_486
    order_by_repetition :  8_905
  Wins for All 35_168 repetitions where: len(reward_list) div 27 == 6
    order_by_thue_morse : 14_211
    order_by_rotation   : 14_137
    order_by_randomise  :  6_820
  Wins for All 34_589 repetitions where: len(reward_list) div 27 == 9
    order_by_rotation   : 15_083
    order_by_thue_morse : 14_936
    order_by_randomise  :  4_570
  Wins for All 26_598 repetitions where: len(reward_list) div 27 == 12
    order_by_rotation   : 11_068
    order_by_thue_morse : 10_888
    order_by_randomise  :  4_642
  Wins for All 26_654 repetitions where: len(reward_list) div 27 == 15
    order_by_thue_morse : 11_045
    order_by_rotation   : 10_976
    order_by_randomise  :  4_633
  Wins for All 26_831 repetitions where: len(reward_list) div 27 == 18
    order_by_thue_morse : 11_345
    order_by_rotation   : 11_179
    order_by_randomise  :  4_307
  Wins for All 26_746 repetitions where: len(reward_list) div 27 == 21
    order_by_rotation   : 11_153
    order_by_thue_morse : 11_133
    order_by_randomise  :  4_460
  Wins for All 26_990 repetitions where: len(reward_list) div 27 == 24
    order_by_rotation   : 11_229
    order_by_thue_morse : 11_095
    order_by_randomise  :  4_666

Thue-Morse wins for this case of random rewards, but not by much!

--Paddy3118 (talk) 12:04, 26 June 2020 (UTC)

A simple rotation of n slots/people is balanced (= A..Z were all once in every position 1..n ) after n rotations like Thue-Morse, but Thue-Morse changes than the order of the permutations of ABC so A isn't always the first fetcher after n*n.That's more fair.Still more work to calculate and for big n the balance n*n is hard to reach.After that point Thue-Morse makes the difference over simple rotation. --Horst.h Horst.h (talk) 06:39, 27 June 2020 (UTC)