Half Time / Full Time Probability

[scotchraptor]

If the probability of a team winning at Half Time is 40% and of them winning the match is 60% what is the probability of them _both_ winning at Half Time and winning the match ?

Multiplying the probabilities doesn't work, and I'm a bit lost.

[tachyon1]

If the probability of a team winning at Half Time is 40% and of them winning the match is 60% what is the probability of them _both_ winning at Half Time and winning the match ?

Multiplying the probabilities doesn't work, and I'm a bit lost.

Here's a basic method.

Take the goal expectancies of each side.

Work out the poisson probabilities of each team scoring 0,1,2,3 etc goals in the first half and therefore correct score probabilites for the first half.(Teams will 'score' 44% of their goal expectation in the first half and 56% in the second).

Do the same for the second half.

Multiply together all the first half/second half correct score combination probabilites that lead to your desired double result.

Add them together.

Example for Draw/Draw double result.

Goal expectancy team A=1.8,team B=0.8.

1)Probability of first half ending 0-0=0.3185
2)Probability of first half ending 1-1=0.0888
3)Probability of first half ending 2-2=0.006

4)Probability of second half alone ending 0-0=0.233
5)Probability of second half ending 1-1=0.105
6)Probability of second half ending 2-2=0.012

Combos that lead to a draw/draw are 1&4,1&5,1&6,2&4,2&5,2&6,3&4,3&5 and 3&6.

Multiply and add them and you get the probability of this match up going draw/draw as about 15%.

For winning at HT/winning at FT you'll need the probs of a team leading 1-0 at HT and then drawing 0-0 in the second,leading 3-0 at HT and losing 0-1 in the second etc,etc...It's about 38% for a team having a 60% of winning the game.

T

[scotchraptor]

Interesting, but I don't have the goal expectancies - just the HT and FT probabilities.

I'm puzzled that I can't derive the HTFT from the HT and FT probabilities.

What's missing from the HT and FT probabilities that HTFT needs ?

[voros]

The reason why multiplying them won't work, is that you don't necessarily have independence between the two probabilities. I'll explain:

Let's assume:

P(H) = Probability of winning at halftime = 0.4
P(F) = Probability of winning at fulltime = 0.6

If P(H) and P(F) are independent of one another, then:

The probability of winning at halftime and fulltime P(H&F) = P(H)*P(F) = 0.24.

To think of it in another way, if we assume the chances of getting a yellow card in the first half and the chances of winning the game are independent (IE, one has no effect on the other), the chances would clearly be 0.24, it would happen 40% of the time you win the match.

However in the case you describe:

P(F|H) = The probability of winning at fulltime given you were winning at halftime.

We can't say for sure that P(F|H) = P(F), indeed from what we know of the sport that seems unlikely.

The real formula is P(H&F) = P(F|H) * P(H)

You need the number for either:

P(F|H) = Probability of winning at fulltime given you're winning at halftime

or

P(H|F) = Probability of winning at halftime given you're winning at fulltime.

And you're formulas would be:

P(H&F) = P(F|H) * P(H) = P(H|F) * P(F)

And obviously if you know one of P(F|H) or P(H|F) you can then calculate the other easily.

I hope you can follow that. In short, unless you assume the two are independent, you need more information.

[voros]

I'll give a quick example using real numbers.

From 2003-2006 in international soccer (a few games are missing due to lack of halftime scores), for a randomly selected team:

P(H) = 0.296
P(F) = 0.382

If those are independent then:

P(H&F) = Probability of winning at halftime and fulltime = 0.296 * 0.382 = 0.113

The actual numbers however are:

P(H&F) = 0.242

Why? Because they are not independent. From the numbers:

P(F|H) = Probability of winning at fulltime given you're winning at halftime = 0.818
P(F) = 0.382

Since those are not even close, clearly P(F) and P(H) are not independent.

P(H&F) = P(F|H) * P(H) = 0.818 * 0.296 = 0.242

Which is the same as we have above.

[EvanJ]

If I didn't make any mistakes, here are some facts about the English Premier League so far this season:

239 games played
130 games played with one team leading at halftime
98 times a team led at halftime and won (75.38 percent of games not tied at halftime)
32 times a team led at halftime and did not win (24.62 percent of games not tied at halftime)

Voros, you worked with a randomly selected team and I think that some teams may be better at protecting halftime leads than others, so whether a sample is done on the national team or club level I think the sample should include many teams. If I was doing this at a national team level I might do something like take 100 World Cup Qualifiers from various confederations.

[woodlands]

98 times a team led at halftime and won (75.38 percent of games not tied at halftime)
32 times a team led at halftime and did not win (24.62 percent of games not tied at halftime)


Let me see if I follow. If correct, you're saying that 3 out of 4 times the team leading at the half will go on to be victorious.

What are the probabilities of winning, losing, and drawing the game...for a team that is losing at halftime by 1 goal...when the total goals scored by the opposition (in the first half) is 3 goals or fewer?

[clavell]

Interesting thread. "In general" we can say about football matches that 50% are home wins, 25% draws and 25% away wins. Of course are leagues when draws are more frequent like France, Italy etc.
When the full time result is home win what are the probabilities for half time results that leaded to home win.
Example for home wins taked like a whole, the half time percenteges for home to be a% for draw b% and for away team leading at half time c%. Sure a+b+c=100
Whar are in general a, b ,c ?
Same question for full time draw and for full time away win ?
Sorry if i was a little coplicated. Thanks.