Round-robin: Frequency of Ties

[willemien]

Could not resist it
Here the raw data for p=6 no jigo

Very nice. Is there a clever way to count the number of combinations without exhaustive enumeration?

No sorry on purpose i did not do anything clever, not even one fixed game
I just let my pc ramble trough all 32.768 enumerations.
and then sort enumeration to graph and so.

I think that makes the data more reliable.

Ties breaker:

Two players with same score.

This tie is always broken (there is always an non jigo result between the players)


Three players with the same score

again a brute force approach
I did select the enumerations where

- A B and C have the same (total) number of wins
- A doesn't have the same number of wins as D E or F
- A has more than 3 wins
(so not where 4 or 5 tie nor where the number of wins of A B and C is below 3)

in this overview a tie is broken iff A has a different number of wins as B only counting the results between A B and C

Code: Select all

Wingraph    #Combinations    ABC ties    broken between abc   broken  all
533310           240            12          none                0
444210           240            12          none                0
444111            80             4          none                0
433320          1680            84          36/84             720
433311          1680            84          36/84             720
333222          2640           132          72/132           1440

In three way ties ABC is one of the 20 possibilities (is that correct? 6!/(3!* 3!)= 720/(6*6) = 20)

The broken all is just the broken broken between abc x 20 (so it is compareble with the # combinations)

The results of graph 333222 are complicated.Here only the results between A B and C is checked
IT SAYS NOTHING ABOUT THE OTHER TIE

For 444111 we can say more, because no combination is broken with DC or IDC also the other tie is not broken.
(it looks like you only get the 444111 result if the 3 stronger beat all the weaker and both groups tie under themselves)

[willemien]

willemien, a tie on place 3 can occur only if place 3 is not skipped due to a tie on place 1 or 2. It is common sense to skip place numbers when players share a better place. - The correct frequency is 24/64. So your text corrected:

Code: Select all

p=4

Wins #PlayerCombinations TiesOnPlaces #PlayersInTies
3210         24          -            0
3111          8          2            3
2220          8          1            3
2211         24          1+3          4
----------------------------------------------------
             64

4 different distributions
3 with ties

Place FrequencyOfTies
1     32/64
2     16/64
3     24/64

Place FrequencyOfTopMostTies
1     32/64
2     16/64
----------------------------
      48/64

But the you are wrong with the frequency of ties for the second place
that should then be 8/64 (graph 3111 only)

[RobertJasiek]

Thanks, corrected:

Code: Select all

p=4

Wins #PlayerCombinations TiesOnPlaces #PlayersInTies
3210         24          -            0
3111          8          2            3
2220          8          1            3
2211         24          1+3          4
----------------------------------------------------
             64

4 different distributions
3 with ties

Place FrequencyOfTies
1     32/64
2      8/64
3     24/64

Place FrequencyOfTopMostTies
1     32/64
2      8/64
----------------------------
      40/64