Conditional Probability- Correct Score of a soccer match given player scores first

Question

I have two probabilities- the probability that a soccer match finishes 3-1 is 16%, and the probability that player X scores first is 11.8%. I'd like to know what the probability of the player scoring first and the match finishing 3-1 is? I know I can use Bayes' Theorem here, but when I get to a certain point I'm not sure how to obtain the probabilties of P(the player X scores first | game finishes 3-1), P(the game finishes 3-1| the player X scores first) or P(the game finishes 3-1| player X doesn't score first) without some form of past data regression...

This is a question that I've been grappling with for awhile, any help is appreciated!

Answer 1

There's not enough information. Think of these events in terms of Venn diagrams. You're given two shapes. The area of the first is 16 and the second 11.8, what's the area of the intersection? Clearly there isn't enough information. I.e. I can construct different probability spaces in which the intersection has different area.

Specifically, consider the space where at the beginning of the game we flip a weighted coin and with 11.8% chance let X score a free goal, otherwise let Y score a free goal. Afterwards, indepedent of the first goal scored, we flip another weighted coin and with 16% chance we score goals to finish 3-1, otherwise we leave it as 1-0. This satisfies your description, and the probability that both events take place is, due to indepedence, the product of the probabilities.

Consider a second scenario where after the first flip, if X scores we also ensure the game is 3-1. If Y scores we flip a weighted coin with probability $p$ to ensure a 3-1 finish, otherwise leave it as 1-0. Almost the same, but we've added dependence. Now to satisfy the initial requirements, we need the probability of a 3-1 finish to be 16%, so $0.118 + 0.882p=0.16 \implies p=\frac 1{21}$. But in this case the probability that both events occur is 11.8%, clearly larger than in the first scenario.

Answer 2

Technically, you do not need Bayes theorem. Bayes theorem deals with conditional probabilities. You aren't asking about a conditional but an intersection. Let $S$ be the probability that the game ends with a score 3-1. Let $X$ be the probability that player X scores first. You are asking for the "probability of the player scoring first and the match finishing 3-1, or $P(X \cap S)$. It stands to reason that the variables are not independent, and without any other data (like conditional probabilities which we could dissect and use with Bayes theorem) the problem cannot be solved.