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

Probability of a match ending with
- 0 goals is 40%,
- 1 goal is 45% and
- more than 1 goal is 15%.

Now at half-time, the score is 1-0.
What is the probability of more than 1 goal at the end of the match?

A. Is it still 15%, or
B. is it 25% (100% = probability of 1 goal + probability of more than 1 goal only, eliminating the probability of 0 goals)?

Or is there anything I missed out?

Thanks for the help.

  • Mathe Mathe

    There's something troubling with this question, and that is your implicit assumption that being 1-0 at half time maintains the original probabilities in some way. It's possible that being 1-0 at half time makes the teams behave differently. The leading team being more defensive now, making the probability of new goals very unlikely. Or maybe it drives the losing team into more aggressive tactics, increasing the probability that one of the teams will make a new goal.


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…
$$P(X_{1} = 0)^{2} = P(X_{2} = 0)^{2} = 0.4$$
$$P(X_{1} = 0) = P(X_{2} = 0) = \sqrt{0.4} \approx 0.632$$
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.
  • Thank you for this very detailed answer. I really appreciate it.

The answer is accepted.
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