A simple math question

Could it be (0,82 x 0,71) / 100 = 58 %

82+71 divided by 2 shirley?

If you completely ignore the fact that one team scoring could influence the other team scoring (I'm assuming they're playing each other), then it's (0.71*0.82)*100= 58% like you say.

Well the probability's 0.5822, which is 58.2%. [/pedant]

tamboy is an idiot, 4th poster is right

You're the 4th poster Carsten

LOL Antoine.

5th poster is right

If you completely ignore the fact that one team scoring could influence the other team scoring (I'm assuming they're playing each other), then it's (0.71*0.82)*100= 58% like you say.
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But is this something you actually has to take in consideration, one temam scoring influencing the possibility of the other team scoring?

Of course this is important in in-play betting. But if you only bet before kick-off I can't see how it would matter? You're betting on the chances of each team scoring one goal or more, based how many goals you anticipate each team to score, which again is based on a funcktion of how many goals you anticipate the game to produce and the relative strength relationship between the two teams.

If you can't do this accurately because one goal changes the premises, you can't bet on over/unders either. Because a goal can influence the chance of the other team scoring.

Well, the thing is, by multiplying the odds together, you're treating the events as being completely independent of one another, but I would say treating them as independent is a reasonably fair assumption.

If you could come up with probabilities of either side scoring given that the opposing side has/hasn't scored, I could do the maths for you, but does it make a lot of difference to the probability on average?

Well, the thing is, by multiplying the odds together, you're treating the events as being completely independent of one another, but I would say treating them as independent is a reasonably fair assumption.

If you could come up with probabilities of either side scoring given that the opposing side has/hasn't scored, I could do the maths for you, but does it make a lot of difference to the probability on average?

The events are far from independent, and the fact that they are not makes a massive difference to the odds.

Teams have a tendency to raise their scoring rates if they fall behind, meaning that the probability of a draw is much higher than it would be if the two teams' scoring rates were independent.

Yeah, fair point Contrarian. I suppose it has a bigger impact than I thought.

When I asked 'would it make a lot of difference to the probability on average', I meant, 'how much do you think the probability of a team scoring chances if the opposition has scored?'. I was thinking it was minimal, but I guess you're right.

changes*

Well, the thing is, by multiplying the odds together, you're treating the events as being completely independent of one another, but I would say treating them as independent is a reasonably fair assumption.

If you could come up with probabilities of either side scoring given that the opposing side has/hasn't scored, I could do the maths for you, but does it make a lot of difference to the probability on average?
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Ok. I assume you and Contrarian are correct on this. The odds are not independent from each other. According to book I have with Kevin Pullein, the prospect of a team scoring the next goal is higher when it i losing than when it's drawing and higher when it is drawing than when it's losing. The chance of a team scoring the second goal in a match is raised by 6 % if the team is under with one goal, compared to leading with one goal.

From that one may conclude that the chance of the losing team (being under 0-1) scoring a goal is raised by 3 % compared to when the result is 0-0.

So if I conclude that the chance of team A scoring is 84 % and the chance of team B scoring is 68 %, then the chance of both teams scoring is 57 %, ignoring that the events are dependent on each other.

How do I do the math? How do I take in consideration the 3 %?

I would to it this way: As the odds on correct score being 0-0 is 19,00, I conclude that there is a 95 % chance that the game will produce one goal or more. So there is a 95 % chance that I will have to take the 3 % in consideration. 95 * 0,03 = 2,85.

But I also have to take in consideration that the events only get dependent on each other after a goal is scored, and that may take some time. Maybe 30 minutes in average? Correcting for that I get around 2 % ((2,85/3)*2).

If I the take the originally 57 % and correct it to 55 %. Am I then doing something correct?

If I the take the originally 57 % and correct it to 55 %. Am I then doing something correct?
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I mean correcting from 57 % to 59 % of course

According to book I have with Kevin Pullein, the prospect of a team scoring the next goal is higher when it i losing than when it's drawing and higher when it is drawing than when it's losing
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A little correction again: According to book I have with Kevin Pullein, the prospect of a team scoring the next goal is higher when it i losing than when it's drawing and higher when it is drawing than when it's WINNING.

p.s. *maths

Bayes Theorem is required to do this accurately.

P(A/B)*P(B scoring 1st) + P(B/A)*P(A scoring 1st)

[b]Wedged Joined: 19 Nov 05
Replies: 3536 04 Sep 10 01:49   


p.s. *maths

Seconded. Damn the American saturation.

Losers is right about Bayes theorem, but having tried very hard, I'm still struggling to calculate this properly. I'll let you know when I've worked it out though.

There is a shortcut that makes this a doddle to calculate.


Happy hunting