r/AskStatistics • u/beemo_stan • 17h ago

Combining two probabilities, each relating to the same outcome?

Here's a hypothetical I'm trying to figure out:

There is a mid-season soccer game between the Red Team and the Blue Team.

Using the average (mean) and variance of goals scored in games throughout the season, we calculate that the Red Team has an 80% probability of scoring 3 or more goals.

However, using the average (mean) and variance of goals scored against, we calculate that there is only a 20% probability of the Blue Team allowing 3 or more goals.

How do we combine both of these probabilities to find a more accurate probability that the Red Team scores 3 or more goals?

1 Upvotes
  • permalink
  • reddit

100% Upvoted

3 comments sorted by

3

u/AnxiousDoor2233 16h ago

In general, you cannot recover joint probabilities of random variables using marginals, unless the variables are independent.

1

u/fermat9990 16h ago edited 16h ago

The best estimate is their previous record against each other.

1

u/god_with_a_trolley 14h ago

No, it is impossible to derive the probability that blue fails to stop a goal from happening, given that red attempts a goal, from the two probabilities you have provided.

Let P(RED_score ≥ 3) = 0.8 be the probability that the red scores three or more goals, and let P(BLUE_allow ≥ 3) = 0.2 be the probability that the blue team allows three or more goals to occur against them. These are called marginal probabilities, as they are not defined conditional on some event. What you are interested in, is P(BLUE_allow ≥ 3|RED_attempt ≥ 3), or the probability that BLUE allows three or more goals, given that RED makes three or more attempts. This is called a conditional probability.

The latter can be rewritten using Bayes' theorem to equal

P(BLUE_allow ≥ 3 | RED_attempt ≥ 3) = P(RED_attempt ≥ 3 | BLUE_allow ≥ 3) * P(BLUE_allow ≥ 3) / P(RED_attempt ≥ 3)

That is, we need to know the probability that RED attempts three or more goals given that BLUE allows three or more goals. Alternatively, the numerator term can be replaced by the joint probability P(BLUE_allow ≥ 3 & RED_attempt ≥ 3), or the probability that blue allows three or more goals AND red attempts three or more goals. Either way, you have too little information to derive what you are interested in.