Round-robin: Frequency of Ties

[RobertJasiek]

The following numbers come closer to what I am interested in:
FrequencyOfTies, FrequencyOfTopMostTies, FrequencyForPlayerToBeInTie.

Code: Select all

No jigos

P=2

Wins #PlayerCombinations
10            2

1 different distribution
0 ties

Code: Select all

P=4

Wins #PlayerCombinations TiesOnPlaces #PlayersInTies
3210         24          -            0
3111          4          2            3
2220          4          1            3
2211          6          1+3          4
----------------------------------------------------
             38

4 different distributions
3 with ties

Place FrequencyOfTies
1     10/38
2      4/38
3      6/38

Place FrequencyOfTopMostTies
1     10/38
2      4/38
----------------------------
      14/38

FrequencyForPlayerToBeInTie
(4*3+4*3+6*4)/(4*38) = 48/152

[willemien]

Sorry but there are statistical errors on your overview for 4 players.

I am not at home but provisionally corrected it should be something like:

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 32/64

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

FrequencyForPlayerToBeInTie

THIS NEEDS MORE INVESTIGATION
(8*3+8*3+24*4)/(4*64) = 144/256 ??

I don't know where your factors 4 4 and 6 come from (I guess from combinatorix) but they are not the right weighting factors to use here. You ned to go back to base facts. not just start from halfway calculations.

Always remember

There are Lies, Dammed Lies and Statistics.

[pwaldron]

Let us ignore specific players but consider result tables in general regardless of player names.

...

Depending on P, how many different result distributions (tables) do exist?

Your recent calculations don't treat the tournaments in a general way: 3210 is structurally the same as 0123.

However, if you do distinguish those two cases, then a tournament of P players will have 2^(P*(P-1)/2 different outcomes if there are no jigos, and 3^(P*(P-1)/2) if there are. Willemien has shown the 64 possible outcomes of an event with 4 players and no jigos.

A hint as you look for your error: simple permutations of the result vector aren't enough for enumeration. Players can end up with the same scores by beating different people. A 2211 result can occur with player D beating any of players A, B or C.

[RobertJasiek]

willemien's text retagged for greater reading pleasure:

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     32/64

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

FrequencyForPlayerToBeInTie

THIS NEEDS MORE INVESTIGATION
(8*3+8*3+24*4)/(4*64) = 144/256 ??

[RobertJasiek]

Oops. Thanks to willemien and pwaldron for pointing out my thinko. I thought: "3111. The 3 wins can belong to any of the 4 players. Then there is only way to get the 111 pattern." But I overlooked that actually the pattern itself is insufficient information. If A represents the 3 wins player, then it might mean either B>C>D>B or B>D>C>B, which are 2 possibilities to be combined with the 4. We get 4 * 2 = 8. I see.

How to spell out 2211?

There are 6 possibilities which two players can be those with 2 wins. Let us look at one of the 6 cases in detail by calling these two players A and B:

A>B + A>C + D>A + B>C + B>D + C>D

A>B + C>A + A>D + B>C + B>D + D>C

B>A + A>C + A>D + B>C + D>B + C>D

B>A + A>C + A>D + C>B + B>D + D>C

There are these 4 subcases. So we get the 6 * 4 = 24.




willemien, how do you get 3 | 32/64 as FrequencyOfTies? Must that not be 24/64?

[RobertJasiek]

Your recent calculations don't treat the tournaments in a general way: 3210 is structurally the same as 0123.

Yes, but I am interested in the combinatorics of tie frequencies rather than the score pattern structure.

However, if you do distinguish those two cases, then a tournament of P players will have 2^(P*(P-1)/2 different outcomes if there are no jigos, and 3^(P*(P-1)/2) if there are.

Right. (Calculation: Number of available results to the power number of all games.)

***

How in general do you get the "Number of ties for first place" if you don't distinguish structurally same score patterns?

[pwaldron]

Yes, but I am interested in the combinatorics of tie frequencies rather than the score pattern structure.

In that case the original problem could be stated better. The 'names' (i.e., labels) of the players do matter, since the tournament with result vector 3210 is not equivalent to 0123. No matter, it's cleared up now.

How in general do you get the "Number of ties for first place" if you don't distinguish structurally same score patterns?

It doesn't show up so well for the n=4 case. In the case of n=6, there are a variety of ways to get the tournament winner with four wins:

444210
444111
443310
443220
443211
442221

In n=6 case, there are 22 nonisomorphic tournament graphs. There are six cases where the winner is tied with four wins, and (if memory serves) three cases where the winner has three wins and is tied. The tie frequency is therefore 9/22. Naturally this isn't the same as the case where player names matter.

Regardless, the result set is small enough that all tournaments with n=6 can be directly enumerated. n=8 should be tractable as well (270 million cases). Beyond that you'll probably have to shift to Monte Carlo sampling techniques to get an estimate of the tie frequency rather than a hard number.

I'm curious whether this actually has any meaning beyond an amusing exercise. I assume it is motivated by the upcoming discussions about the EC, but the fact that players are of differing strengths in real life means the results here have little application.

[RobertJasiek]

Round-robins are thrown into EC system proposals as if they were the solution to everything. While they provide the second-best (double round-robin is yet better) possible pairing quality and give a lot of top-top games for those few inside, the pretty high tie frequencies require abundant usage of tiebreakers for seeding or (if desired) splitting result hairs. The requirement for tiebreakers reduces the fake high quality of round-robin. If non-Europeans (immaterial for EC results) shall be in some of the round-robins (as in some of the proposals), then ties become yet more frequent (a second or third place might mean to win the EC or a preliminary stage as a European) and so becomes usage of tiebreakers.

From experience I knew that round-robins with P=6 have a pretty high tie rate. This thread might confirm profoundly the guess that also for slightly greater P the tie frequency is pretty high.

When we will have understood the basic theory, we can then also try to attack varying player strength models...

[willemien]

Oops. Thanks to willemien and pwaldron for pointing out my thinko.


willemien, how do you get 3 | 32/64 as FrequencyOfTies? Must that not be 24/64?

The problem is that the meaning of 3 ties is ambigious.

And depending on the meaning you get 24/64, 32/64 or 48/64.
(Maybe you can even find a reasonable definition that makes it 8/64 or 16/64 )

Code: Select all

3210 24  Clear 3rd place
3111  8  No distinction between 2 3 and 4
2220  8  No distinction between 1 2 and 3
2211 24  No distinction between 3 and 4

I used it as there is no distinction between 3rd and 4th place (3111 and 2211)
While i guess you only counted 2211 (no distinction between 3rd and 4th place but there is a distinction between 2nd and 3rd)

(While if you mean there is no clear 3rd place the answer is 3111 + 2220 + 2211 = 48/64)


Still puzzeling about FrequencyForPlayerToBeInTie

If it means any player ties the chance is (there is a tie in the result)
3111 + 2220 + 2211 = 48/64

if it means a specific player ties it is (player A ties)

3/4 of 3111 + 3/4 of 2220 + 2211 =
6 + 6 + 24 = 36/64

But with another meaning the result would be different.

Was thinking about writing a program that enumerates all options in a 6player round robin but with all these ambiguies it is better to first to define what really is wanted


Statistics is such a fun.

[willemien]

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

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p=6

Wingraph    #PlayerCombinations    TiesOnPlaces 
543210           720               none
543111           240               (456)   
542220           240               (345)
542211           720               (34)    (56)
533310           240               (234)
533220           720               (23)    (45)   
533211          1440               (23)    (56) 
532221          1680               (345)
522222           144               (23456)
444210           240               (123)
444111            80               (123)   (456)
443310           720               (12)    (34)
443220          1440               (12)    (45) 
443211          2880               (12)    (56)
442221          1680               (12)    (345)
433320          1680               (234)
433311          1680               (234)   (56)  
433221          8640               (23)    (45)
432222          2400               (3456)
333330           144               (12345)
333321          2400               (1234)
333222          2640               (123)   (456)
----------------------------------------------------
22 graphs      32768 combinations            

so now which statistics do you want?

[pwaldron]

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?

[RobertJasiek]

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

Still puzzeling about FrequencyForPlayerToBeInTie

Given all player-dependent win distributions and a particular player. With which frequency will this player "draw" a tie in that he is involved? So I guess that this is what I mean:

FrequencyForPlayerToBeInTie

THIS NEEDS MORE INVESTIGATION
(8*3+8*3+24*4)/(4*64) = 144/256 ??

[RobertJasiek]

so now which statistics do you want?

You tie annotation is more detailed. I was looking only for the first number per bracket but, now that I see your detailed brackets, I have got a taste for the greater detail;) Anyway, you should allow the output of frequency of the leading bracket numbers and their combinatorical summing up, which I can't resist to be about doing:)

How have you calculated the #PlayerCombinations?

[RobertJasiek]

Believing willemien's numbers blindly, we would get these sums for No Jigo, P=6:


No ties:

720/32768


Tie on place 1:

(240+80+720+1440+2880+1680+144+2400+2640)/32768 = 12224/32768


Tie on place 2:

(240+720+1440+144+1680+1680+8640)/32768 = 14544/32768


Tie on place 3:

(240+720+1680+720+1680+2400)/32768 = 7440/32768


Tie on place 4:

(240+720+80+1440+8640+2640)/32768 = 13760/32768


Tie on place 5:

(720+1440+2880+1680)/32768 = 6720/32768

[RobertJasiek]

I have become thirsty of numbers:)

There are more interesting questions: What are the tie frequencies (esp. for place 1) after application of either of these tiebreakers?

- MutualGameScore iff 2 players are tied on a place
- Non-iterativeDirectComparison
- IterativeDirectComparison