# Probability of Team A defeating Team B in the first half, given probability for full game

This is probably an easy question, but it's really confusing me.

Suppose that, given that the result is NOT a draw, Team A has a probability of beating Team B of 60%.

I want to ignore every external factor (players tired, different strategies, etc) and focus only on the math. If the first half was also not a draw, is the probability of Team A defeating Team B in the first half also 60%, or less than 60%?

My intuition would, at first, say 60%, because if they play that first half an infinite number of times, then I guess A would win 60%.

However, I also think that it would make sense to be less than 60%, because, since it's a smaller time frame, Team B would have a better chance of getting 'lucky', and therefore, it's probability of winning would increase.

Which of my reasonings is correct, if any?

• Even upon adopting gross simplifying assumptions the underlying question appears unanswerable, because the answer depends on the probability distribution of the point differential in any given half. That suggests the answers to the questions you did ask are "neither reason can be correct." That uncertainty, together with the need to invent strong assumptions, make this question a poor fit to the format of this site. Perhaps you could offer more specific circumstances for consideration? – whuber Aug 23 '16 at 21:28
• Check out Gerd Gigerenzer's work on fast and frugal heuristics. Specifically, he develops a simple model predicting NBA team wins based on metrics like yours and compares it to a full information Bayesian model. His FFH model outperformed. Here's one summary of that work...library.mpib-berlin.mpg.de/ft/wb/WB_Fast_2006.pdf – Mike Hunter Aug 23 '16 at 22:16

The answer clearly requires a lot of assumptions, but here is a try. Let $Y_1$ and $Y_2$ be the score differential achieved in the first and second halves (A minus B), respectively. Then the overall score differential is $Y=Y_1+Y_2$, and team A wins if $Y > 0$. You told us that $P(Y>0)=0.6$.
Your assumptions of "ignoring every external factor" could be translated as having two independent and identically distributed halves, so let $Y_1$ and $Y_2$ follow the same distribution with mean $\mu>0$ and standard deviation $\sigma$. Then $Y$ has mean $2\mu$ and standard deviation $\sqrt{2}\sigma$.
This means that after the first half, the value of 0 is $z_1=\frac{\mu}{\sigma}$ standard deviations from the mean, while after the entire game it is $z=\sqrt{2}\frac{\mu}{\sigma}$ standard deviations from the mean. At this point it is still difficult to compare $P(Y_1>0)$ to $P(Y>0)$, because the distribution of $Y_1$ and $Y$ could have different shapes (and the value of $z$ does not drive the tail probabilities of every distribution), but in general we would expect that the larger the mean/SD value, the higher the probability of being above 0. So we would expect lower probability of winning in the first half.
We can work out the probabilities explicitly if we assume that the score difference is approximately normally distributed. Since $0.6 = P(Y>0) = P(\frac{Y-2\mu}{\sqrt{2}\sigma}>\frac{-2\mu}{\sqrt{2}\sigma})=\Phi(\sqrt{2}\frac{\mu}{\sigma})$, where $\Phi$ is the normal cdf, this implies $\sqrt{2}\frac{\mu}{\sigma}=0.253$, and $\frac{\mu}{\sigma}=0.179$. So $P(Y_1>0) = P(\frac{Y_1-\mu}{\sigma} > -\frac{\mu}{\sigma})=\Phi(0.179)= 0.571$, that is the team wins the first half 57.1% of the time.
If the two halves are identically distributed, but positively correlated, the effect gets weaker, because the standard deviation of $Y$ becomes $\sqrt{2(1+\rho)}\sigma$, where $\rho$ is the correlation coefficient. As $\rho \rightarrow 1$, we have $z = \frac{\sqrt{2}}{\sqrt{1+\rho}} \rightarrow z_1$.