Calculate odds from statistics on multi-factor events

Question

Let's take a soccer example:

• Player A scores 60% of the time
• Player B scores 30% of the time
• Player C scores 10% of the time
• ...

this is against an 'average' goalie.

and

• Goalie K catches the ball 80% of the time
• Goalie L catches the ball 30% of the time
• Goalie M catches the ball 10% of the time
• ...

this is against an 'average' player

We know that, on average, 72% of all penalties result in a goal (Average Player vs. Average Goalie).

Given each player's individual stats, what would be a reasonable estimate of the chances player A has to score a penalty vs goalie K, B vs. L, A vs. L, B vs. K, etc?

Answer 1

As the comments say, there is not enough information about the interaction between individuals

Here is a possible approach which gives answers (probably wrong but perhaps good enough for your purposes)

Suppose

• the probability of a goal with an average kicker and average goalie is $a$, so the probability of a save is $1-a$, and the odds of a goal are $\frac{a}{1-a}$

• the probability of a goal with a particular kicker and an average goalie is $k$: then the odds of a goal are $\frac{k}{1-k}$

• the probability of a save (rather than of a goal) with a particular goalie and an average kicker is $g$: then the odds of a goal are $\frac{1-g}{g}$

then you could guess that, with that particular kicker and particular goalie, the odds of a goal might be close to $\dfrac{\frac{k}{1-k}\frac{1-g}{g}}{\frac{a}{1-a}}$ and the corresponding probability of a goal $\dfrac{k(1-g)(1-a)}{(1-k)ga + k(1-g)(1-a)}$

The following table then gives the calculated probabilities of goals associated with your example plus an example average kicker and goalie (numbers with four decimal places are rounded)

                I       J       K       L
    a=0.72      g=0.8   g=0.3   g=0.1   g=0.28  

A   k=0.6       0.1273  0.5765  0.84    0.6
B   k=0.3       0.04    0.28    0.6     0.3
C   k=0.1       0.0107  0.0916  0.28    0.1
D   k=0.72      0.2     0.7     0.9     0.72

Note that both B v. J and C v. K, each with $k=g$, give probabilities using this approach of $1-a$ rather than the naively intuitive $0.5$