What is the strike rate?

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Atho55
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Since April this year I have been paper testing a system that looks to allocate a series of Back or Lay instructions per course by BSP Rank on the premise that for every race at that course the instruction is followed and the results aggregated to give a course total then a daily total. There are 9 instructions per course per day.

It looks like this....

Prediction Win%.jpg
Count is the No of instructions placed, Countif are the ones that have a positive (>£0.00) outcome. Rating is just where the course stands in a £ league table. No is just record No. So Cranbourne, 9 instructions, 4 had returns >£0 meaning 5 lost.

Count now has a total of 9959. I know what Countif is and can work out the strike win % of the Counts but am not sure if using Wins > £0.00 is the correct way to do it.

Basically, I know how it`s doing but looking to compare it to the % of how it ought to be doing.
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alexmr2
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I have no idea what you are doing but I would assume any result which is above £0.00 is positive? So the strike rate as a % is

number of positive results/(number of postive results + negative results)
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ShaunWhite
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Atho55 wrote:
Sun Aug 02, 2020 9:25 pm
Basically, I know how it`s doing but looking to compare it to the % of how it ought to be doing.
Ideally you'd use some sort of standard deviation from a normalised distribution.

Comparing results with expectation isn't a simple comparison and imo it's the reason some people give up on winning strategies too soon or persist with bad ones too long. Let's say you have a system which you believe should have an EV of +2% and a strike rate of 30%.....after 250 events how far from the theoretical result does the actual result have to be to give you confirmation it's working? The idea of running a strategy for a few weeks and then judging it by the PL alone is naive, it's the deviation from expectation that's the correct measure, and to do that you need compare your actual results with a normalised distribution of your expected results.

If you're still not sure about that then simplify everything to a coin flip, 3 wrong calls in a row seems quite normal, 20 would be freakishly bad, so somewhere on the scale of 3 to 20 is the point when you decide you're being duped. So how bad does your +2%EV 30%SR strategy have to be before you pull the plug?

As rule of thumb if you're within +/-1 SD then press on, once you get to the 2nd SD and especially the 3rd, you have to start wondering if outcomes with such a small chance of occuring means that your initial expectations must have been wrong.

Your approach seems to be based on looking for deviations from expected results (eg a BSP rank winning or losing more often than expected) so you're probably already up to speed with seeing if these are statistically significant or within a normally expected deviation. Afterall almost nothing occurs exactly as often as it 'should' so following just any old anomaly will see you go broke really quicky, so just use a similar method for looking at your results.
Atho55
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Joined: Tue Oct 06, 2015 1:37 pm
Location: Home of Triumph Motorcycles

Thanks for the responses but had to move one myself with no replies in first few days.

This is a classic case of when skiving off school leads to a poor understanding of Maths so to keep it simple for myself I have used 50% as being a benchmark. There are 9 selections per meeting. I could get 9 winners or 9 losers as they are exclusive events. Equal No of chances to win or lose hence 50%. Just going for did it win or did it lose.

On this set which looks at 9 ranks the strike rate is this
Strike Rate.jpg
The methodology to solve the underlying problem of where it works best and potentially a strike rate value to weed out the poor performers is being tested.

Course Win %.jpg

The biggest drawback for the Aus races is that they are all jumbled together as in Trots/Flat and NSW/Non NSW tracks so that mess to sort out if it looks to have legs
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sionascaig
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If you are treating the races as independent events & measuring win/loss rate with a probability of p (50% in this case), that is the binomial distribution you want...

https://en.wikipedia.org/wiki/Binomial_ ... ility%20p)

Expected value = n*p, where n is the number of observations
Var = np * (1-p)

The article also tells you how to calculate confidence intervals for your results (straightforward example for 95% in link)

The bit about estimation of parameters (p in this case) is also worth a read, i.e. is it really 50%?...
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ShaunWhite
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Atho55 wrote:
Sun Aug 09, 2020 10:28 am
This is a classic case of when skiving off school leads to a poor understanding of Maths so to keep it simple for myself
Same, B grade O-Level, fortunately they let you leave at 16 and that gives you 60yrs to catch up on the stuff you want to learn. I'm no spring chicken either Athos and learning gets harder but when the state stops educating you that's when the schoolwork starts.
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