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We know that in a football game, Team A wins against Team B with probability $= 0.8$

We also know in a game, at the end of half-time, they tied.

Then what's the probability A wins against B? Is it still $0.8$ or we should consider conditional probability?

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We need to produce a not too unreasonable probabilistic model. Here is one possibility. Let $X$ (respectively, $Y$) be the numbers of goals scored by Team A (respectively, Team B) in a full game. Let us assume, unreasonably, that $X$ and $Y$ are independent, and have Poisson distribution. For definiteness let the parameters be $4$ and $2$ respectively.

Then the second-half results $U$ and $V$ have Poisson distributions with parameters $2$ and $1$ respectively. It is not hard by a computation to show that $\Pr(X\gt Y)$ is substantially bigger than $\Pr(U\gt V)$.

Thus under this kind of model, given that they are tied at the half, the probability that Team B manages to win or tie is substantially greater than the unconditional probability that B wins or ties.

Remark: A similar phenomenon happens in playoffs. If you have a strong team A and a weak team B, then the probability that A wins a "best of three" playoff is quite a bit bigger than the probability A wins a one-game "sudden death" playoff.

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What's the $$Pr(S>T)$$? – i3wangyi Jan 25 '14 at 19:11
Sorry, typo, it was supposed to be $\Pr(U\gt V)$, where $U$ and $V$ are as defined above (second half goals). – André Nicolas Jan 25 '14 at 22:56

The probability remains the same no matter how many whole games or fractional games they play. So given that they are tied at the half, means that the probability is 0.8 that Team A will win the 1/2 game that is the second half.


Think about it this way... if Team A wins 100 games in a row, what is the probability that they will win the 101st? It is 0.8. Since there is no declared dependence between games, each game, or portion of a game is an independent statistical event. The fact that they are tied at the half makes the halves independent. The same applies to the overtime situation. The tie makes the overtime period an independent statistical event.

Now if Team A is losing by some amount at the half, then that makes them dependent with respect to the game score. The probability is still 0.8 that Team A will win the second half. But is is lower that they will win the game, because they have to win the second half by enough margin to overcome their deficit and win.

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At first time, I think so. But is it saying, even they got tied for the 90 min, they will be given a overtime. A still wins B with Prob = 0.8 – i3wangyi Jan 25 '14 at 5:33
Yes, with the given information, that is all we can conclude. See my edit. – Tpofofn Jan 25 '14 at 13:51

I would say that the probability remains the same. For one, we don't know the probability of winning one half of the game and it shouldn't make a difference anyways.

Thinking about it in a real life scenario, switching sides at half would change the probability of winning due to sun, wind, etc. Therfore you couldn't say infer the probability of winning the first half of the game.

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I'm not certain but I would say that the probability of A wins is 0.8. Let's break the problem down.

\begin{align} \begin{aligned} p(A = win) &= 0.8 \\ p(A = lose) &= 1 - ( p(A = win) + p(A = tied)) \\ p(A = tied) &= 1 - ( p(A = win) + p(A = lose)) \end{aligned} \end{align}

So, the probability of getting tied is considered from the beginning. So, there is no need for a conditional probability. Also, what do you mean by saying "We also know in a game, at the end of half-time, they tied."? if you know for sure this is the case then there is no need to do the math. I found what you are saying rather ambiguous unless you mean by winning as a general case because the A team is stronger. If this is the case, then I would say exactly what I said at the beginning of my post.

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