After the quarter-final draw for the FA Cup had been made, I
stumbled across comments on Twitter about the hot ball-cold ball theories of
draw-fixing. The “top four” teams in the
competition all avoided each other. The
best argument against fixing comes from the fact that different combinations of
players, ex-players and celebrities are used by the media in each round. When you consider how many people by now
would have to be willingly complicit in the arrangements, and to have kept
their silence all these years, the idea becomes very, very implausible. Nevertheless there are some people who take a
lot of convincing, so let’s look at the maths.
Credit to my son Chris who took a look at my first attempt, helped me
spot a reasoning error and gave me an idea for a better way of presenting the
arguments visually.
What are the chances
of the top four teams in a QF draw all avoiding each other?
Let’s imagine a completely fair draw in which the best four
teams are given RED balls and the others are BLUE. We don’t care about the home-and-away order,
just the pairings. We start with eight
balls in the bag – and so there are eight possibilities for the first ball
out. Then there are seven different
possibilities for the second ball drawn out, so even at this point there are
8x7=56 different possible outcomes.
There are 6 alternatives for the third ball, and so on, so
the total number of different draw sequences for the eight balls is 8x7x6x5x4x3x2x1,
known as 8! or “factorial-8”, and this has a value of 40,320.
We will need this number later on.
Now, let’s go back to the start of the draw, and the first
two teams drawn. The first could be
either a red (top four) or blue (not top four), and likewise the second. But the relative chances for the second ball will
depend on what happened for the first.
The chances are shown in the following table. The first ball has equal (4 out of 8) chances
for red and blue. Now there are only seven
balls in the bag. If the first one was
red, then the chance of the second one being red is now 3 out of 7. We are interested in the chances of getting
one blue and one red, in either order.
The combined chance is calculated by multiplying together the chances in
the first ball and second ball columns, and we end up with a number out of
56. The first tie would be red v red
(top four v top four) on 12 out of 56 pairings, red v blue on 32, and blue v
blue on 12.
First Ball Out
|
Second Ball Out
|
Outcome
|
First Tie
Combined Chance
|
Red (4/8)
|
Red (3/7)
|
Red v Red
|
12/56
|
Blue (4/7)
|
Red v Blue
|
16/56
|
|
Blue (4/8)
|
Red (4/7)
|
Blue v Red
|
16/56
|
Blue (3/7)
|
Blue v Blue
|
12/56
|
Notice that the sum total of all the combined chances is
56/56 – which must be true because there are no other possibilities. The chance of one top team and one lower team
– in other words one blue and one red in either order – is 16/56 plus 16/56,
which is 32/56 or 4/7. This means that 4
out of 7 of the 40,320 possible draws would have pitched a top team against a
lower team. The number is 23,040. We are no longer interested in the other
17,280 possibilities because the first tie out of the bag would have brought up
either two top teams (red) or two lower teams (blue) to play each other. If the latter, then it would be guaranteed
that at least one of the other three ties would be red v red. So only 23,040 or the 40,320 possibilities
keep the “perfect separation of the top four teams” alive as a possibility.
Let’s assume that first tie was either red v blue or blue v
red, and we go on to draw two more balls for the second tie. Again, we are interested in the case where we
again draw a red and a blue, in either order.
However, this time we start with only six balls left in the bag (three
red, three blue) so the fractions are different, as shown here. Same logic, just different values for the
fractional chances.
Third Ball Out
|
Fourth Ball Out
|
Outcome
|
Second Tie
Combined Chance
|
Red (3/6)
|
Red (2/5)
|
Red v Red
|
6/30
|
Blue (3/5)
|
Red v Blue
|
9/30
|
|
Blue (3/6)
|
Red (3/5)
|
Blue v Red
|
9/30
|
Blue (2/5)
|
Blue v Blue
|
6/30
|
Again, the total of the combined chances adds up to a
certainty, or 30/30. The red-blue or
blue-red combination that interests us occurs in 18/30, or 3/5 of the 23,040
draws we are still considering out of the possible 43,020. We are therefore now down to 13,824.
Let us now assume that BOTH of the first two ties have been
red-blue or blue-red, and go on. Now
there two red and two blue in the bag, and the next table shows the possible outcomes
for the 5th and 6th balls drawn.
Fifth Ball Out
|
Sixth Ball Out
|
Outcome
|
Third Tie
Combined Chance
|
Red (2/4)
|
Red (1/3)
|
Red v Red
|
2/12
|
Blue (2/3)
|
Red v Blue
|
4/12
|
|
Blue (2/4)
|
Red (2/3)
|
Blue v Red
|
4/12
|
Blue (1/3)
|
Blue v Blue
|
2/12
|
The chances of another red-blue combo for the third match is
8/12 or 2/3 of the 13,824 we were still considering of the original 40,320, and
that is 9,216. There is nothing else to consider for the
fourth match – if the first three are red-blue pairings, then the fourth match
is bound to be the same.
So, it can be seen that in a perfectly fair 8-ball quarter
final cup draw, 9,216 of the 40,320 possible draw sequences
would keep the “top 4” teams in the draw away from each other.
That’s 22.9%.
I’d bet that would be in
line with historical analysis if anyone has the time!
The same maths would apply to any arbitrary four : four
split of the eight teams. For example,
if there were four “northern” teams and four “southern” teams, the probability
of all four ties being north v south confrontations would be the same 22.9%. It would mean that the northern teams had
avoided each other, and so had the southern teams.
All of this also explains why competent mathematicians are
deeply interesting people and well worth your time ;)
Extra bit:
Why are we
sometimes multiplying fractions and sometimes adding them?
In simple terms, when
we want to find the probability of two independent events happening,
multiplication of the two chances is needed.
The key word is AND. This applies
when one event AND the second event are to happen independently. So for a red ball to come out first of all
(4/8) AND then followed by another red ball (now 3/7), then the calculation is
4/8 x 3/7 which is 12/56.
However, when we
wanted to find the total chance of a red-blue OR a blue-red combination, the
key word is OR. We need to add together
the two separate chances for the events.
Blue-red is 16/56 and so is red-blue.
16/56 + 16/56 = 32/56 which is 4/7.
Links to other Modus
Hopper Random posts on cup draws:
Scottish Cup
Semi-Finals and Old Firm Conspiracies
Champions League
Quarter Final Draws
Updated Version
Links to Modus Hopper
Random posts on football games involving dice and chance:
Wembley
Soccerama
Soccerboss
