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Shot Stopping and an Elite Defense

While reading a forum discussion on the topic of Manchester United’s next goalkeeper I came across an argument that went something like this (and I’m paraphrasing):

Manchester United have a world class defense that is able to limit the opposition to only a few shots per match and therefore it is not important that United’s goalkeeper is a world class shot stopper, it is far more important that he is good at organising his defense, claiming loose balls, catching and punching crosses and distributing the ball with throws and kicks.

To investigate whether shot stopping is less valuable for a team which concedes fewer shots I did some simple calculations:

x = number of shots on target faced
z = save percentage
y = goals conceded

To determine goals conceded we can use the equation:

y = x – zx

If we increased z by 1%:

y(2) = x – 1.01zx

We can then combine the two and look at the percentage change in goals conceded, %y, and ask what happens to %y when we increase z by 1%?

%y = 100(((x – 1.01zx) – (x – zx))/(x – zx))
%y = 100((x – 1.01zx – x + zx)/(x – zx))
%y = 100(-0.01zx/(x – zx))
%y = -zx/(x – zx)
%y = -zx/x(1 – z)

%y = -z/(1 – z)

So %y depends on z. At z = 0.6 a 1% increase in z leads to a 1.5% decrease in y whereas at z = 0.8 a 1% increase in z leads to a 4% decrease in y. The more relevant point is that %y is independent of x, in other words the value of increasing save percentage by 1% isn’t affected by how many shots on target you face.

The reverse of the original argument, that preventing shots becomes more valuable for a team that concedes fewer shots, isn’t true either. A 1% decrease in shots on target faced always leads to a 1% decrease in goals conceded.

%y = 100(((0.99x – 0.99zx) – (x – zx))/(x – zx))
%y = 100((-0.01x + 0.01zx)/(x – zx))
%y = (-x + zx)/(x – zx)
%y = -(x – zx)/(x – zx)

%y = -1

Having an elite shot stopper is extremely valuable regardless of whether you have a world class defense or not.

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Categories: Theory
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