Hi Rowsus, hope all is well on your end. Was hoping you could help me with some numbers, you seemed like the best person to go to!
Not SC related but has intrigued me nonetheless. Each R1 winner is versing a R1 loser in R2, and was wondering what the probability of this happening is, as I was staggered when I noticed it somehow when looking ahead to the weekend's games.
Am I correct in saying "18 * (18 - 1)" fits in the equation(s) somewhere?
Would like to know!
Cheers in advance.
Hi Max, things are good here, I hope they're good for you too.
Not only are the 9 losers playing the 9 winners, but all the losers are home teams this week!
It reminds me of Round 2, 1977! In Round 2 1977 all the teams that won in Round 1, played a team that also won in Round 1, and obviously, all the teams that lost in Round 1, played another team that lost in Round 1. This guaranteed that the ladder after Round 2 would have 3 teams with 2 wins, 6 teams with 1 win, and 3 teams with no wins. A little fact that smart arse 15 year old me latched onto, and won a tidy little sum betting my friends at school, that that was how the ladder would look after Round 2. The only thing that could stop me winning the bets was a draw. I was in a panic at the end of the last game, when Richmond lead Hawthorn by 1 point, but happily, there was no draw, and I won my bets.
But back to your question.
It's not as easy to answer as people might think!
The number of ways you can make a single round draw with 18 teams is:
17 x 15 x 13 x 11 x 9 x 7 x 5 x 3 x 1 = 34,459,425
Why?
Pick a team at random, and there are 17 teams left you can match it up against.
Pick your next team at random, and now there 15 teams left you can match this one up against.
etc etc until all the teams are gone.
Now, I know this will be possibly refuted by some maths people in SCS, but it works, and it's right! They will say I forgot to factor in the "random" team, but in this case, the random team is irrelevant!
It is because you are picking the teams in pairs, and not making a lineal or random order of the 18 teams.
There are 9! = 362,880 ways in which you can make a draw with 9 losers from one week, playing 9 winners from one week. (For those that don't know 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)
Why?
Pick any loser at random, pick at random one of 9 winners to play them.
Pick another loser at random, and then pick at random one of the 8 winners left
etc etc
This means the chances of this anomoly happening is 1 in 34,459,425/362880 = 1 in 94.96 or around 1.05%
Now, imagine someone came to you, and said BEFORE round 1 was even drawn to be played, that in Round 2, all the home teams would be losers from Round 1, and all the away teams would be winners. If we assume that everything is equal, and there is no outside influence on this, then we have to change the odds by a factor of:
2^9 = 512. So the chances of that happening are 1 in 48,620 (exactly!) or 0.002%