Calculating probability that soccer player scores in a certain game with binomial distribution

Question

Since Ronaldo just came to play in Italy but didn't manage to score on his first 2 matches, I wanted to give an estimate of the probability that he scores tonight. In order to achieve this (in a quite naive way) I made some assumptions: -The strength of opposing team has no influence on his odds of scoring. -His odds of scoring at least one goal or more in a certain match remain constant in the season and is $p = \frac{20}{38}$ obtained by imagining that he probably will be able to score in 20 different matches (I don't care how many goals he scores, just if he scores or not).

We can now treat this distribution as a binomial where X = "Ronaldo scores on certain match" which is a Bernoulli varibale with $P(X) = \frac{20}{38}$.

My doubt now is: to calculate the odds of him scoring today in his 3rd match , given the fact that he didn't score in the first 2, should I use conditional probability given the fact that I know he didn't score in the first 2 matches or not?

In the beginning I thought of calculating the odds he would score tonight as (having $X_3$ the binomial variable Bin( 3 matches , $p = \frac{20}{38}$) ): Probability($X_3 = 0$ ). But then I thought that I might have to calculate it using conditional prob: $P(X_3 = 0 | X_2 = 0)$ where $X_2$ is the same as $X_3$ but after 2 matches(trials). Thank you very much and sorry for my language but of course I am Italian :)

Answer 1

The probability he scores tonight is just the same as the probability he scores in any game $(\frac{20}{38})$. This is due to the independence of events in the Binomial model.

If you were asking "The probability he scores his FIRST goal in the THIRD game" then you would have to take into account the probability of him not scoring in the first two. But as it is, it doesn't matter.

Just like the probability that your third dice roll is a 6 isn't affected by the previous rolls.

EDIT: After comment confirming that you want "The probability he scores his FIRST goal in the THIRD game"

To calculate this use the independence of events;

$P(X_1 = 0 ) = \frac{18}{38} = P(X_2 = 0)$, this gives

$P(X_1 = 0 \bigcap X_2=0) = \frac{18}{38} \frac{18}{38} = \frac{81}{361},$

$P(X_1 = 0 \bigcap X_2=0 \bigcap X_3 =0) = \frac{18}{38} \times\frac{18}{38} \times \frac{20}{38}= \frac{810}{6859}.$

Answer 2

Let $A_i$ denote the probability that he scores in $i$-the match.

Then:$$P(A_3\mid A_1^{\complement}\cap A_2^{\complement})=P(A_3\cap A_1^{\complement}\cap A_2^{\complement})/P(A_1^{\complement}\cap A_2^{\complement})=P(A_3)P(A_1^{\complement})P(A_2^{\complement})/P(A_1^{\complement})P(A_2^{\complement})=P(A_3)$$

the second equality on base of independence. So whether you use conditional probabilities (LHS) or not (RHS), the outcome is still the same.

Does this answer your question?