How to calculate probabilities of a football team scoring a certain number of goals

Question

If a team has 50-50% chances of scoring over or under 1.5 points, how do you calculate what are their chances of scoring exactly 0,1,2 or 3 points? And how do you calculate the chances of scoring over/under 2.5 points? (The team can only score 0,1,2,3,4 and so on, no decimals)

Later edit: The chances for team to score 1.5 points comes from a match were the expected total number of goals is 2.5 (50%-50% chances of both teams to score over or under 2.5) and team A is better than team B with 0.5 points. If this helps.

Answer 1

If all you know is that the chances of scoring under or over $1.5$ points, then there is no way to know what the chance of scoring exactly $0$ points is.

For example, the two cases:

• Chance of scoring $0$ is $0.5$, chance of scoring $1$ is $0$, chance of scoring $2$ is $0.5$
• Chance of scoring $0$ is $0.25$, chance of scoring $1$ is $0.25$, chance of scoring $2$ is $0$, chance of scoring $10$ is $0.5$

both result in the team having a 50-50 chance of scoring over or under $1.5$ points.

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Basically, you want to calculate $p_0, p_1,p_2,p_3\dots$ while all you know is that

1. $p_0+p_1=\frac12$
2. $p_2+p_3+\cdots = \frac12$.
3. $p_i\geq 0$ for all $i$.

There are infinitelly many solutions to the above set of equations.

Answer 2

Actually, you can't. You need extra information to know that.

If I call A : "the team score > 1.5" and B : "the team score <1.5", you will have this tree of probability :

    /   \  
   B     A    
 /  \   /  \   
0   1   2  3  

So you know that the branches of the first level are 50% each, but as far as you do not know anything on the second level, you can't answer the problem.

With this extra information, you will have :
P(team scores 0) = P(B) * P(0|B)
P(team scores 1) = P(B) * P(1|B)
P(team scores 2) = P(B) * P(2|B)
P(team scores 3) = P(A) * P(3|A)

Answer 3

As others have pointed out, we don't have enough information to solve this problem without additional assumptions.

So let's assume the number of goals scored follows a Poisson distribution with mean $\lambda$. To determine $\lambda$, we are given that $P(X \le 1) = 0.5$, where $X$ is the number of goals. This translates to the equation $$P(X=0)+P(X=1)=e^{-\lambda} + \lambda e^{-\lambda} = 0.5$$ We can find an approximate solution numerically, with result $\lambda = 1.67835$.

Then to find the probability of other numbers of goals, use the Poisson density function, $$P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$$ for $x=0,1,2,3, \dots$.

Here are a few values. $$\begin{matrix} x &P(X=x) \\ 0 &0.19\\ 1 &0.31\\ 2 &0.26\\ 3 &0.15\\ 4 &0.06\\ 5 &0.02\\ \end{matrix}$$