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Author Topic: Sky ratings as per JFC  (Read 1919 times)

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Offline carey

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« 2021-Jun-12, 11:50 AM Reply #25 »


Fours

a brilliant contribution by mr figjam.

Online fours

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« 2021-Jun-12, 12:03 PM Reply #26 »
carey,

Not much competition from you so relatively speaking I guess it is.

Fours

Offline carey

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« 2021-Jun-13, 09:16 AM Reply #27 »
Four lots of 100,000 iterations.
about 1 second for the lot.
think i have it correct now.
find it an interesting exercise, but not sure as to its value for my stuff(will do some tests though)



whoops just realised it was set for 10,000, but process the same.

Online fours

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« 2021-Jun-13, 09:19 AM Reply #28 »
  :lol:   :lol:   :lol:

The cartoon was appropriate.

Fours

Offline carey

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« 2021-Jun-13, 09:44 AM Reply #29 »
   :lol:     :lol:     :lol:  

The cartoon was appropriate.

Fours

there you go mr figjam, another brilliant posting.
is your best friend your mirror?

Online jfc

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« 2021-Jun-13, 12:19 PM Reply #30 »
Four lots of 100,000 iterations.
about 1 second for the lot.
think i have it correct now.
find it an interesting exercise, but not sure as to its value for my stuff(will do some tests though)

whoops just realised it was set for 10,000, but process the same.

That output looks like you're doing the correct thing.

And it helps to answer your and Bascoe's question about how many iterations are needed.

Look at the 97 Rater in the 1st lot.

It's 5th Ranked, so it's reassuring that it's shown most likely to finish 5th.

But you would then expect the frequencies to drop monotonically.

And your data shows drop, rise, drop for 6,7,8.

That rise is an anomaly.

Such anomalies are easy for a program to spot and tally.

Your program could count the anomalies, and if it reported "too many " you could increase the iterations until they come down below "too many".

Offline carey

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« 2021-Jun-13, 01:49 PM Reply #31 »
That output looks like you're doing the correct thing.

And it helps to answer your and Bascoe's question about how many iterations are needed.

Look at the 97 Rater in the 1st lot.

It's 5th Ranked, so it's reassuring that it's shown most likely to finish 5th.

But you would then expect the frequencies to drop monotonically.

And your data shows drop, rise, drop for 6,7,8.

That rise is an anomaly.

Such anomalies are easy for a program to spot and tally.

Your program could count the anomalies, and if it reported "too many " you could increase the iterations until they come down below "too many".

thanks jfc, i never noticed that.
initially i was interested in what i would get if the ratings were the same.
now that basic code written, there is lots of ways to make it better i think.

Online jfc

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« 2021-Jun-14, 07:12 AM Reply #32 »
For Normal Distributions, for a pair with STD of 7, their Difference Distribution will have:

Variance = 7*7 +7*7 = 98
STD  = 98^0.5 ~= 9.9
Mean = Difference of Means

When Mean = 3 the H2H probability for the lower Rater = NORMDIST(0,3,TRUE) = 0.3809
So Higher Rater probability = 0.6191

Compare this with Carey's Results for the 100 versus 97 Rater:

.2047/(.2047 + .1072) = 0.6563

Why the difference! A fluke?

Look at the Ratio of Win/H2H for all 8 runners

1
0.993404983
0.958059344
0.948141545
0.943267599
0.926475246
0.90978778
0.908626482

Clearly not a fluke.

Carey's independent results confirm what I've believed for a long time.

H2H probabilities are different from Win Ratios.

But the Don Scott table assumes they are the same.

And that's one reason why he's very wrong.

Online jfc

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« 2021-Jul-07, 02:16 PM Reply #33 »
In some sports other distributions than Normal are used.

The Poisson Distribution is often used for Soccer - usually mistakenly.

But sometimes it is very helpful.

And there are some sexy tricks for exploiting Markets.

If you study the actual formula you should find that you can calculate Lambda from To Score Odds.

Imagine that $2.97 is the True Odds To Score.

Then:

Lambda = LN(1/(1-1/OddsToScore))

~= .410528

And once you have Lambda you can work out the Odds for any Odds Range.

As I write on Betfair the presumed True Odds for Harry Kane to score against Denmark happens to be ~=$2.97.

His Lambda implies To Score 2+ should be ~$15.529.

But in fact Betfair is showing vastly different Odds!

Seems there's something rotten in the Rate of Denmark!

Online jfc

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« 2021-Jul-09, 09:49 AM Reply #34 »
Currently the Correct Score Market for the Euro Final shows:

$7.4 0-0
$8.0 0-1 England
$8.6 1-0 Italy

These figures can magically produce the Lambdas (aka Expected Goals):

England =7.4/8 = .925
Italy = 7.4/8.6 = .860

Armed with these numbers any number of goodies can be produced.

A key one is the Extra Time Distribution, which is crucial to attacking many Markets.

Now, if you go to the To Score Market you should find that the Lambdas for English Players total significantly more than 0.925.

Figure out that anomaly and you might find that quite rewarding.



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