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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.

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.

Basically, you want to calculate $p_0, p_1,p_2,p_3dots$ while all you know is that

1. 
   $p_0+p_1=frac12$


2. 
   $p_2+p_3+cdots = frac12$.


3. 
   $p_igeq 0$ for all $i$.

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

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)

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,
$$P(X=x) = fraclambda^x e^-lambdax!$$
for $x=0,1,2,3, dots$.

Here are a few values.
$$beginmatrix
x &P(X=x) \
0 &0.19\
1 &0.31\
2 &0.26\
3 &0.15\
4 &0.06\
5 &0.02\
endmatrix$$