I wrote the other day that in mathematically extreme cases, a team could win fourteen games and miss the eight or that a team could win only ten games and still make the eight.

A pesky poster in the House has disputed my argument.

He’s come up with a scenario that has the top nine teams all having a W/L ratio of 15-7 and the bottom nine teams all having a W/L ratio of 7-15.

“This,” he says, ‘has a team with **15 wins **missing out on the final eight. Prove to me that I am wrong.”

My 14-10 range was based on logic, rather than maths, not there’s not that much difference between the two. My argument went like this:

**If Team One wins 22 games, then Team Two can only win 21**. It’s already been beaten by T1.

**If Team Two wins 21 games, then Team Three can only win 20**, because it’s already been beaten by T1 and T2.

And thus, when you extend this logic sequentially to Team Nine, you have:

T9 is beaten by T1-8 and **can win 14 games and still miss out**.

You can use a similar converse argument to establish that the minimum number of wins to make the eight is ten.

PP’s preposterous proposition that a 15-7/7-15 scenario will allow a team to win 15 games and still miss the eight needs a different approach.

*It’s a challenge.* The relevant factor is that each team plays seventeen others once and five others twice.

For PP’s scenario to work, nine teams must beat five other teams twice. That will give nine teams 10 wins. Each of them will now need five more wins, playing each other once.

That’s 72 games to go with only 36 wins available. Collectively, nine teams will need 45 wins. It’s not going to work. **Winning fifteen and missing out will not work.** *Myth busted, you petty bastard!*

There’s only one scenario that I can think of that disputes my 14-10 range of missing or making the top eight and that only comes in if we consider the draw.

Theoretically, the range can be extended to one win to make the eight – if every other game in the season is a draw. But, remember, we said at the start *that we were excluding draws*. Someone else can do the maths, or the logic, to work out how few games a team can win to make the eight if the draw factor is included.

That’s too complex for me, tonight.

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Stop wasting your time JK. There is no logic in the AFL draw. Not even mathematical logic. However the whole vibe of things tells me that 15 wins gets you in the 8 – even after factoring in Collingwood’s perpetually soft draws.

Error! Oh, no.

“That’s 72 games to go with only 36 wins available… ” SHOULD READ:

” That’s 36 games to go with 72 outcomes but there will only be 36 wins available.”

Luckily, the conclusion is unchanged. Just.

John,

The scenario is easily provable

(more details if you wish)

Your logic flaw is here:

The relevant factor is that each team plays seventeen others once and five others twice. For PP’s scenario to work, nine teams must beat five other teams twice. That will give nine teams 10 wins. Each of them will now need five more wins, playing each other once.All teams play five ‘other’ teams twice and 12 teams once

After the nine teams get their 10 wins (against the ‘lower’ sides) – They play each other once BUT they don’t just play each other.

Playing each other accounts for 8 of the 12 remaining games they have left – but they still have four other games against the lower sides they didn’t play twice

That’s where the extra wins come from

Isn’t there a more basic flaw here?

According to my fixture, there are 23 rounds, not 22.

On what grounds – apart from Blundstone Arena and a few others – can you be

right to postulate a 22-game round home-and-away season?

At a meta-logical, or just downhome abstract, level, can any thesis based on a myth be other than mythical itself? Discuss.

Theoretically you could win 18 games and still come ninth. Each of the top nine beat each of the bottom nine. That would leave the top none with 9-0 and the bottom nine with 0-9. The top nine have to play each other at least once and assuming they all go 4-4 this would mean that after 17 rounds the top nine would all be 13-4.

Now assuming that in the last five rounds each of the top nine only plays teams in the bottom nine and win all of them then each of the top nine would be 18-4 with only percentage separating eighth and ninth.

How ever this scenario is extremely unlikely unless of course Richmond win 18 games.

Ken

At least I have the games and rounds right. Each team plays 22 games spread over 24 weeks. One round is spread over two weeks and, during three rounds, each team has one bye. Thus, we have the peculiar situation of 18 teams playing 22 games each over 23 rounds in 24 weeks. Don’t ask why.

GA Thompson

You are spot-on correct… absolutely correct and I was absolutely wrong.

Dan

You are right, too. You have picked up on GAT’s forensic discovery of the flaw in my logic.

Let me rephrase the problem.

Each team plays each other team at least once. That’s 17 games. To make their 22 games, each team must play five other teams twice.

Let’s assume that the top nine teams beat all the bottom nine teams whether they play them once or twice.

That gives the top nine teams 5×2 wins + 1×4 wines = 14 wins.

For the top nine to then reach the magical 18 wins figure, they must win four games playing against each other.

Teams 1-9 playing each other equals 36 contests which will produce 36 wins and 36 losses. 36 divided by nine = four wins and four losses each.

Thus, 18-4 is perfectly possible. And, conversely, so is 4-18 for the bottom nine teams.

But nine teams at 19-3 is not possible and nor is nine teams at 3-19. In that scenario, there are not enough wins available to be distributed amongst the top nine teams, nor are there enough losses to be distributed amongst the bottom nine teams.

Thus, it is possible in a perfectly symmetrical season to win 18 games and still miss the eight.

In my arrogant first post I said that 14 was the maximum to miss out and I admit that I was wrong. Pesky pernicious PP claimed 15. He was wrong, too. It’s 18 and, I believe, that part of the problem can now be put to bed.

I also said that it was possible to get into the eight with only 10 wins. I suspect that I am wrong there, too, because I may have based that number on the same flawed premise that I used in the first part of the problem.

I’m an adaptive and cooperative person. I’m willing to revisit that second part of my proposition and that is, ask whether 9 or 8 or 7 or 6 wins can get you into the eight.

But not tonight. I’m gutted, tonight.

My flipflop team, Adelaide, came storming home in the last quarter against Richmond at Adelaide Oval in front of 50,000 frantic people to hit the front after being behind all night… to once again let the chance to secure a spot in the eight pass them by. Unbelievably, the other contenders for eighth spot have also lost this weekend and thus Adelaide’s infuriating tease continues along its miserable way for another seven days.

I’ll revisit Part Two of the problem in a few days, when I think my nerves have settled down. Maybe GAT or Ken or somebody else might want to have a go at the maths first.

And, by the way, this is not merely a theoretical exercise. Right now, we are closer to the edge of an historical mathematical extreme than I can remember.

In Round 21, after playing 20 games, Adelaide, Richmond, Collingwood, and Gold Coast are all equal eighth on ten wins. With two games to go, you can fiddle with the Ladder Predictor and add West Coast to the mix, to have five teams finishing with ten wins at equal eighth.

How bizarre is that? My mother said there would be days like this.

There is quire a bit of maths involved in the AFL draw. A couple of Computer Scientists from Finland created another AFL draw and will present their methods next week at

http://www.patatconference.org/patat2014/programme.pdf

You can download the paper from the above web site, or i can email you a copy.

George

I checked out the patat 2014 website but I could only find a PDF for the conference programme. I’d love to read the Finland paper (and you seem to be involved with this, too). Can you email me a copy of the paper to [email protected] . Thanks!

While searching for it, I came across:

Edith Cowan University, Research Online : 2006

Fixture-scheduling for the Australian Football League using a Multi-objective Evolutionary Algorithm

by Luigi Barone, University of Western Australia; Lyndon While, University of Western Australia;

Paul Hughes, University of Western Australia and Philip Hingston, Edith Cowan University.

There’s a lot of mathematics in this paper, too. Which is not a bad thing.

Tonight, still reeling from Adelaide’s loss to North this weekend and squandering, once again, its chance to lock in a final eight spot, I desperately need diversion.

And, thus, I’ll attack the problem of the minimum number of wins a team needs to make the eight.

I start with these premises…

18 teams play 22 games for 396 outcomes which means 198 games with 198 wins and 198 losses. And, remember, we assume NO DRAWS.

Minimum wins for Team Eight requires maximum wins for Teams 1-7 and Teams 9-18.

And, now, the argument.

For Team Eight to have the lowest possible win tally, Teams 1-7 need to beat Team Eight ONCE, five teams out of the eight twice and the other five teams out of the eight ONCE.

That gives the top seven teams sixteen wins each.

Collectively, that’s 112 wins.

Now it becomes a simple problem of the distribution of the remaining wins and losses.

The top seven play each other once on 21 occasions (6+5+4+3+2+1=21) which means that 21 wins and 21 losses have to be allocated to these seven teams.

It doesn’t matter, here, how these wins are allocated amongst the top seven. Remember that we are only interested in the minimum number of wins that Team Eight can have, to make the eight. .

Thus, teams 1-7 have 112 wins playing against teams 8-18 plus 21 wins playing against themselves. That equals 133 wins.

198 minus 133 = 65 wins left to be distributed between teams 8-18.

65 wins divided by 11 teams equals five wins each plus ten spare.

Thus, we say, with tentative confidence, that a team can win six games and still make the eight.

PLEASE TELL ME if my logic is flawed.

In a previous post I said that the range to make or miss the eight was 14-10. I was corrected. I think the range, now, is 18-6, without draws.

With draws, the universe becomes absurd. All 198 contests and all 396 outcomes could be a draw. Thus the true range of making or missing the eight is 18 wins and miss out and no wins and still make it.

That should give Richmond fans some comfort for 2015..

And one last thing… given Adelaide’s flipflop 2014 season, I don’t want them to sneak into the eight next week. They’d be there under false pretenses. I want them to field a team against St Kilda in Round 23 that serves the sole purpose of finding out how best to attack season 2015. And that means thinking about playing Matt Crouch and a couple of the others who have been languishing in the reserves for most of the season… and even giving captain Nathan von Berlo his first run for the year.

That means that teams 9-18 still have 50 wins

Thanks John

I emailed the PATAT papers to you. The paper was presented in England last Monday.

Thanks for your other links I will check them out.

At the moment I’m applying the algorithms to high school timetables.