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11-08-2001, 12:49 AM
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#1
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Registered User
Join Date: Oct 2001
Posts: 85
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Although I'm not trying to get so deeply into a mathematical discussion as to confuse anyone, I thought that I would also offer up a useful formula for all of those who enjoy modeling and are eternally unsure if their race sample size is large enough to base future expectation with confidence...namely, to wager, or not to wager.
The following formula will approximate the minimum number of races necessary to accomplish the above stated goal with a 90% confidence level:
WIN % x LOSS % x 1.645 x 1.645 / E / E
where E = WIN % - (2 x ROI) / AVERAGE MUTUEL
If you require a more stringent confidence level, then replace 1.645 with 1.96 (for 95% confidence), or with 2.575 (for 99% confidence).
As an example, a 20% win rate, coupled with a 1.20 ROI and an average mutuel of $15 would result in the following:
E = .2 - (2 x 1.2) / 15 = .04
.2 x .8 x 1.645 x 1.645 / .04 / .04 = 270.6
Therefore, after modeling, this can be used as a tool to determine which models can be expected to produce long term results, and which models need further testing before such assumptions can be made.
Have fun!
THE GENERAL
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11-08-2001, 07:17 AM
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#2
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Registered User
Join Date: Oct 2001
Location: near Lone Star Park
Posts: 5,147
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General,
That should be VERY useful. Thanks, you must be a five star general.
Ranch West
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11-08-2001, 10:03 AM
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#3
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Back from the Abyss
Join Date: Mar 2001
Posts: 254
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G.S.,
Thank you for the formula !!
However I would like to point out a problem with your example. You wrote
" As an example, a 20% win rate, coupled with a 1.20 ROI and an average mutuel of $15 would result in the following:"
There is a relationship between win rate, ROI and average mutuel, which this example violates.
Here are some definitions and the simple formula:
HIT Frequency (H) is how often the particular wager in questions pays off in the form of a percentage (%). This could be a Win wager, a WPS wager, Triple Box Wager, on and on. We originally used the term WIN Frequency, but people confused that with a Win Wager Frequency. Using HIT Frequency avoids that confusion.
Avg Odds (O) is the Odds to 1 (expressed in decimal form) for the respective Wager in question.
ROI (R) is Return On Investment, in other words profit or loss on the respective wager. We shall normalize this value such that an ROI of $1.00 means you break even. We settled on this to keep away from negative numbers for loss situations, and to make the math simpler. For example, an ROI of $1.38 means 38cents profit on a dollar. An ROI of $0.78 means 22cents loss on a dollar.
With the definitions established, here is the simple but very useful formula presented in 3 formats:
H = HIT Frequency
O = Avg Odds
R = ROI
R=H(O+1) When you want to find ROI as a function of HIT Frequency and Avg Odds
H=R/(O+1) When you want to find HIT Frequency as a function of Avg Odds and ROI
O=(R-H)/H When you want to find Avg Odds as a function of HIT Frequency and ROI
Therefore, for a Hit Frequency of 20% and an ROI of $1.20, the average odds must be 5.00 which is an average mutuel payoff of $12.00 assuming the mutuel payoff is for $2.00.
So to run this example through your formula again with the correction:
E = H - [(2 x ROI) / AvgMutuel]
E = .20 - [(2 x 1.20) / 12] = 0
If my relationship for H, R, and O are correct, I believe E comes up zero for all cases.
Is there a flaw in the E formula?
Do I have a misunderstanding?
Can you help me out on this?
Thanks, FH
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11-08-2001, 10:07 AM
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#4
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Registered User
Join Date: Oct 2001
Location: near Lone Star Park
Posts: 5,147
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FH,
Thanks, I found that out, too, but I don't think I could have expressed it as well as you did.
Thanks,
Ranch West
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11-08-2001, 10:49 AM
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#5
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Registered User
Join Date: Oct 2001
Posts: 85
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Sorry for the confusion guys; the confusion resulted from a matter of semantics, rather than a flaw in the formula; I should have been clearer on the meaning of the value for E.
What I should have stated was that you must first set a minimum acceptable profit margin, which must be less than the actual profit margin which has been produced by the modeling.
Therefore, I have edited the original example to eliminate the confusion and provide a clearer explanation of the use of the formula.
Take care.
THE GENERAL
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11-09-2001, 01:36 PM
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#6
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Senior Member
Join Date: May 2001
Location: Phoenix, AZ
Posts: 428
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General,
This type of analysis would be accurate if the only variance was in win percentage and the payoffs were constant. That is clearly not the case in the real world. Payoffs for any selection method have a very large variance which makes for a much larger required sample size. It's not clear to me how to combine both types of variance into a useful formula, but maybe someone here with a better knowledge of statistics than myself could comment on this.
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