Probability of more than 1 goal at the end of a match

I will have to make some simplifying assumptions in order to give a reasonable answer. I will have to assume that:

1. the score in each half follows the same distribution, and

2. the score in the second half is independent of the score in the first half.

In real life, I doubt this is entirely true. Maybe if one team scores in the first half, it’s more likely that there’s a goal in the second half because the team who’s behind is playing more aggressively. But I think I need to make this assumption to answer your question based on the information given.

Let’s define two random variables: $$X_{1}$$ = the number of goals in the first half and $$X_{2}$$ = the number of goals in the second half. Again, I’m assuming that the X's are independent and identically distributed.

We can reverse-engineer the distribution of the X’s from the given information. The distribution you defined at the top of the problem is the distribution of the random variable $$Y = X_{1} + X_{2}$$
So we know, for example that P(Y = 0) = 0.4. There’s only one way for Y to be equal to 0, both $$X_{1} = 0$$ and $$X_{2} = 0$$
$$P(X_{1} = 0, X_{2} = 0) = 0.4$$
Because the X’s are independent…
$$P(X_{1} = 0) * P(X_{2} = 0) = 0.4$$
Because the X’s are identically distributed…
$$P(X_{1} = 0)^{2} = P(X_{2} = 0)^{2} = 0.4$$
Therefore…
$$P(X_{1} = 0) = P(X_{2} = 0) = \sqrt{0.4} \approx 0.632$$
Given that there was a goal in the first half, the only way for there to be 1 goal or less total is if there are no goals in the second half, which we just discovered the probability of that is about 0.632.

Therefore, the probability of more than 1 goal in the game given that there was 1 goal in the first half is 1 - 0.632 = 0.368.