Ok so I need help working out a probability relating to some soccer stats (could apply to any sport).

Team A (home team) is playing Team B (away team).

Team A's home games have had over 2.5 goals 89% of the time.
Team B's away games have had over 2.5 goals 65% of the time.

How then can we work out the probability, as a percentage, of this game having over 2.5 goals?

Sorry if it's obvious and I'm missing something!

Just average the two

.78

It would be .77, but that's beside the point. An simple average is rarely the right answer here. There is no possible answer given the data you have provided. You need much more info to start tackling this, and I'd probably start with some league-wide data.

1-(1-0.89)*(1-0.65) = 0.9615 = ~96%

Juk

That's essentially using an "OR" argument that either team can make it happen independently. Not sure if that applies in a game between two teams.

For example, if in 50% of both the home team's home games and the away team's away games 2.5+ goals are scored, this formula would say that the prob is 75% that 2.5+ goals would be scored if they play each other.

I'm not sure either, but with all the OP gave to go on it doesn't seem too bad (eg: models that assume independent Poisson distributions for each team, etc).

Juk

Many of you are probably familiar with the "Log5" method of estimating the probability of an occurrence in such a situation.

In a baseball application, you would ask what would a .325 batter be expected to hit off of a pitcher who batters hit .290 off of, in a league that hits .250 on average? That is, you have a better-than-average batter facing a worse-than-average pitcher. So you expect the batter's average facing this pitcher to be quite high (higher than .325 for sure).

The so-called log5 formula tells you what to expect.

E = (BatAvg)(PitAvg)/(LgAvg) / [(BatAvg)(PitAvg)/(LgAvg) + (1-BatAvg)(1-PitAvg)/(1-LgAvg)]

or in the numerical case described above:

E = (.325)(.290)/(.250) / [(.325)(.290)/(.250) + (1-.325)(1-.290)/(1-.250)]

= .371

In OP's soccer example, to use the Log5 method, you'd also need to know how often teams in the league scored 2.5+ goals in a game. Let's say it is 55% just so we can work through the case numerically. (If this is a real-world question, OP should obtain the data for the league-wide pct and plug that into the formula below.)

E = (.89)(.65)/(LgPct) / [(.89)(.65)/(LgPct) + (1-.89)(1-.65)/(1-LgPct)]

and plugging in LgPct=55%, we get:

E = (.89)(.65)/(.55) / [(.89)(.65)/(.55) + (1-.89)(1-.65)/(1-.55)]

= .92478

Hope this helps.

I thought I was pretty good with Excel formulas until I tried working this one out.

Can you offer any suggestions?

Thank you.

I am not sure what you are asking.

If you are looking for explanations of where the formula comes from, you can do an internet search for "Log5 method". Several articles have been written over the years investigating the underpinnings of the formula as well as the accuracy of its predictions.

Thanks for the rapid response.

Before posting, I did a google search and even tried some online Log5 calculators but they didn't help in answering my question.

Which is:

I used Excel to input the the numbers (.89, .69, .55) into cells and then tried creating a formula to automatically compute the answer but the figure that you came up with (.924) is proving elusive.

Make sense?

Oh, I see now.

The answer I posted above was WRONG. I must have had a typo somewhere in the formula I entered.

E = (.89)(.65)/(.55) / [(.89)(.65)/(.55) + (1-.89)(1-.65)/(1-.55)]

= .94836

Thanks for pointing out the error.

I had no clue there even was an error

I am getting answers of like -7.09 so I must be entering the formula incorrectly into excel. Going to keep pounding away at it to see if I can get to your # (.948).

If you have any idea what I may be doing wrong, please let me know.

Thanks.

Great post btw.

Changed to .65.