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



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?

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.



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.

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...

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 )




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.

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Promotor and Librarian of Sensei's Library

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






so now which statistics do you want?

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Promotor and Librarian of Sensei's Library

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

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:





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:

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?

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

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

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



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)

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Promotor and Librarian of Sensei's Library

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

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Promotor and Librarian of Sensei's Library

Thanks, corrected: