What Are the Odds of Top 8 AFL Teams Not Playing Each Other?

[Slats18]

Just something I've been wondering, the probability that the top half of a ladder in a national sport will have matches against the bottom half. I'll relate this specifically to AFL.

In the AFL, 17 teams make up the sport. The top 8 at the end of the season are the ones that will go through to the finals, while the bottom 9 miss out. What are the odds that, at any given time, the top 8 of the ladder will not play against each other, i.e., the whole of the top 8 play against the bottom 9.

I've done this (roughly) just by converting it to balls in a sack. Take the top 8 to be blue, and the bottom 9 to be red. Then, find the probability that you will pick these balls out of the bag alternating in colour, obviously leaving one blue behind (the team that has a bye).

I've got that it happens roughly once every 24,000 rounds, am I anywhere near the right answer?

 

[chiro]

Just something I've been wondering, the probability that the top half of a ladder in a national sport will have matches against the bottom half. I'll relate this specifically to AFL.

In the AFL, 17 teams make up the sport. The top 8 at the end of the season are the ones that will go through to the finals, while the bottom 9 miss out. What are the odds that, at any given time, the top 8 of the ladder will not play against each other, i.e., the whole of the top 8 play against the bottom 9.

I've done this (roughly) just by converting it to balls in a sack. Take the top 8 to be blue, and the bottom 9 to be red. Then, find the probability that you will pick these balls out of the bag alternating in colour, obviously leaving one blue behind (the team that has a bye).

I've got that it happens roughly once every 24,000 rounds, am I anywhere near the right answer?

You can't use a binomial model because every draw you make effects the probability.

Lets assume you can only use alternating example (for example you can't choose four reds in a row and then four blue in a row, they all have to alternate).

So let's start with selecting blue first.

Probability of first picking blue is 8/17. This means 9 red and 7 blue left.
Probability of second picking red is 9/16. This means 8 red and 7 blue left
Probability of third picking blue is 7/15. This means 8 red and 6 blue left
Probability of fourth picking red is 8/14. This means 7 red and 6 blue left
Probability of fifth picking blue is 6/13. This means 7 red and 5 blue left
Probability of sixth picking red is 7/12. This means 6 red and 5 blue left
Probability of seventh picking blue is 5/11. This means 6 red and 4 blue left
Probability of eighth picking red is 6/10. This means 5 red and 4 blue left.

Multiply all of those and you have the probability of first picking a blue team and then alternating between red and blue selections.

You have to now figure out the probability if you start with a red instead of a blue and alternate in same kind of way I have done above, and add those two probabilities together to get the total probability of your alternate selection.

Please note that this only counts selections that alternate in the way you have described.

If you have any questions about doing the alternate probability starting with red feel free to ask if you get stuck.

 

[Slats18]

That's what I did, but only one way, a.k.a only starting with blue, I condensed it down to (9!8!)/(17!). So if I start with red, I would get the same answer basically, so the whole answer is just half of what I said, i.e., roughly 1 in 12000 chance, does that sound more like it?

 

[chiro]

That's what I did, but only one way, a.k.a only starting with blue, I condensed it down to (9!8!)/(17!). So if I start with red, I would get the same answer basically, so the whole answer is just half of what I said, i.e., roughly 1 in 12000 chance, does that sound more like it?

Could you show some calculations of how you arrived at that answer? That way I can tell you whether you are right or wrong. It's hard to know and help you otherwise.

 

[Slats18]

Sure. It's just the same way you went about it.

8/17 * 9/16 * 7/15 * 8/14 * ... * 1/3 * 2/2

= (9*8*7*6*5*4*3*2)(8*7*6*5*4*3*2*1)/(17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2)

= (9!8!)/(17!)

And then, since that was only one way to begin the alternation, beginning at the other way gives the same answer, hence, multiply by 2.

 

[chiro]

Sure. It's just the same way you went about it.

8/17 * 9/16 * 7/15 * 8/14 * ... * 1/3 * 2/2

= (9*8*7*6*5*4*3*2)(8*7*6*5*4*3*2*1)/(17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2)

= (9!8!)/(17!)

And then, since that was only one way to begin the alternation, beginning at the other way gives the same answer, hence, multiply by 2.

I'm not sure why you went to 2/2. We are only dealing with eight trials here for each starting color (one starting with a blue, and one starting with a red). Also we have only way of selecting alternate balls (starting with blue) and we don't have to take into account different combinations because there is only one combination for alternate balls starting with a fixed color (red or blue). So we calculate the probability starting with blue and then probability starting with red and then add those two together.

Using the example I posted earlier we have the probability that we start with a blue selection is:

(8 * 9 * 7 * 8 * 6 * 7 * 5 * 6) / (17 * 16 * 15 * 14 * 12 * 11 * 10) = 5080320 / 980179200 = 63/12155

Now your answer (9! x 8!) / 17! = 1/24310

I'm going on the assumption that after we choose eight balls, we are finished the experiment altogether.

 

[Slats18]

Why are we only dealing with 8 trials? That accounts for 8 of the balls being drawn out of the bag, so there is still 4 of one colour and 5 of another in the bag, which we have to perform trials for as they might not necessarily be drawn alternatively.

 

[chiro]

Why are we only dealing with 8 trials? That accounts for 8 of the balls being drawn out of the bag, so there is still 4 of one colour and 5 of another in the bag, which we have to perform trials for as they might not necessarily be drawn alternatively.

Yeah sorry I didn't read your whole first post, so I misread that you need only 8 trials instead of the rest.