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How does the Expected Goals method deal with two consecutive high quality shots in quick succession?
THREAD:-
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THREAD:-
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Consider Man City’s penalty against Watford.
Sterling takes and misses the penalty worth 0.78(xG).
Immediately after he taps in the rebound worth 0.90(xG).
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Sterling takes and misses the penalty worth 0.78(xG).
Immediately after he taps in the rebound worth 0.90(xG).
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But it is illogical to award Man City 1.68(xG) for this situation, as it’s impossible for them to score more than one goal from a single attack.
So how does xG deal with this problem?
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So how does xG deal with this problem?
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The key lies in working out the probability that the attack DOESN’T result in a goal.
In other words, how likely is it that Man City don’t score either of these chances?
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In other words, how likely is it that Man City don’t score either of these chances?
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The maths behind this particular situation is as follows.
(1 - 0.78) x (1 - 0.90) = n
0.22 x 0.10 = 0.02
So there’s a 2% probability that Man City don’t score either chance.
Therefore:
1 - 0.02 = 0.98(xG)
So Man City should be awarded 0.98(xG) for this attack.
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(1 - 0.78) x (1 - 0.90) = n
0.22 x 0.10 = 0.02
So there’s a 2% probability that Man City don’t score either chance.
Therefore:
1 - 0.02 = 0.98(xG)
So Man City should be awarded 0.98(xG) for this attack.
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Or take this Aston Villa goal against Chelsea.
Kortney Hause has two shots, one worth 0.51(xG) and one worth 0.62(xG).
Therefore:
(1 - 0.51) x (1 - 0.62) = 0.19
1 - 0.19 = 0.81(xG)
So Aston Villa should be awarded 0.81(xG) for this attack.
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Kortney Hause has two shots, one worth 0.51(xG) and one worth 0.62(xG).
Therefore:
(1 - 0.51) x (1 - 0.62) = 0.19
1 - 0.19 = 0.81(xG)
So Aston Villa should be awarded 0.81(xG) for this attack.
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