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Corners formula

I worked out a formula that predicts the number of corners in a footbal match.
The formula depends on the odds for 1-2-X and on the national / divisional averages.

First I make the odds into probabilities.
If the (decimal) prices for home win-draw-away win are a,b,c respectively then:

prob. home win = H = k / a
prob. draw = D = k / b
prob. away win = A = k / c

where k is a constant = 1 / ( 1/a + 1/b + 1/c )

The formula for the mean number of corners expected for the case of the premier is:

M = 8.19 + (3.55 * H) ^ 0.76 + (1.58 * A) ^ 6.98 - 0.03 ^ 1.28 * (1 - D ^ 1.28 )

For other divisions I multiply M by an adjustment factor, using the data for the national averages from windrawwin.com.

The probability for any number of corners is then computed using the Poisson distribution (Poisson distribution - Wikipedia).
The probability spectrum is not however exactly Poisson, it is somewhat flattened and the Poisson values have to be raised to the exponent 0.88
(that is make p = p(Poisson) ^ 0.88 then normalize to 1).

I know that the number of corners also depends on the systems of the teams.
With high defenses (crowded midfield) there are usually few corners, with low defenses more corners.
Also the fluctuation of the scoreline. With an early 3-0 lead, everyone goes to bed and there won't be many corners.
But it's a starting point the way I did it.

Also a friend tells me this:
In the live betting look for cases when the home team concedes an early goal to an inferior or somewhat inferior team - then there will be an avalanche of
corners. Intuitively that sounds true.

Do you know anything more that might help ?
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