Should have said, half the winnings went to the betting bank, and other half to the next three races.

Sounds like a sharp punter. Free betting money for himself !

G'day Young Buck,

Your Chi test result should have read 1.84812E-06. The E-06 means you need to slide the full stop to the left 6 times, so it should really end up reading 0.00000184812, which according to AngryPixie's interpretation means that there's only about 1/5,000th of 1% chance that your results were thanks to luck, and therefor you should feel confident in your system future success.

Personally, I find this Punters Chi-square test fascinating stuff...I have absolutely no idea it it's predictions will ultimately reign true or not...but it's fascinating nonetheless...

__________________
...time held me green and dying, though I sang in my chains like the sea. - Dylan Thomas

Thanks for that S&S!

I'd heard of the Chi test but put it in the 'too hard' basket. It's encouraging to hear those figures in your favour...

__________________
All generalizations are dangerous.

You do need quite a few samples as the score will bounce around a bit till you do.

Here's something else I've posted recently. Bit more complicated but a worthwhile exercise all the same.

When is enough, enough??
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How long should you test that fantastic new selection method?
How many results should a commercial tipster provide before you consider parting with your hard earned cash?

With some easily obtainable data, and a bit of simple maths, it’s not that hard to find out. All you need to get started is the average SP of all selections, and the percentage of winners. Easy stuff.

Suppose a tipster has provided the following data about his service:

Total Bets: 300
Winners: 60
Average decimal SP all selections: $5.50

We can easily work out the percentage of winners (W%) by dividing winners by total bets:

60/300 = 0.20 (20% winners)

From here it’s simple to work out the percentage of losers (L%) by subtracting W% from 100% like this:

1.00 – 0.20 = 0.80 (80% losers)

Using the W% and L% figures we can work out whether the selections provide us with a positive expectation. That’s the profit we can expect to make for each $1.00 bet. Here’s the equation:

Exp = ((DECSP - 1) * W%) - L%

Let’s put our figures in:

Exp = ((5.5 – 1) * 0.20) – 0.80
Exp = (4.5 * 0.20) – 0.80
Exp = 0.90 – 0.80
Exp = 0.10 (10% profit for each dollar bet)

Now we know our expectation but the bad news is that for mathamatical reasons we can't use that figure in our calculations. We calculated it because we need to choose a figure between that and 0% (breakeven) to use as a minimum acceptable profit MINAP. For this example I'll be happy with a profit of 0.05 (5%). We're going to use MINAP in our next calculation. A word of caution here though. The closer the minimum is to the maximum, the more test bets you'll need!!

Simple so far right. The next bit is a little harder.

As part of our final calculation we need to work out one last figure we'll call ERR. This is the figure that represents the difference between the W% and the win percentage that gives a 5% profit. Here's the equation:

ERR = W% - ((1 + MINAP) / DECSP)

With our numbers it looks like this:

ERR = 0.20 - ((1 + 0.05) / 5.5)
ERR = 0.20 - (1.05 / 5.5)
ERR = 0.20 - 0.1909
ERR = 0.0091

Almost there but one more thing. The final calculation allows us to assign a confidence level (Z) to the final result. You don't need to work this out. For this example I'll use a 90% confidence level. Using the list I've provide below we see that this confidence level is assigned the number 1.65.

Here we go with the final calculation. To work out the minimum number of test that should be conducted we use the following equation:

TESTS = W% * L% * (Z / ERR) * (Z / ERR)

Now with our numbers:

TESTS = 0.20 * 0.80 * (1.65 / 0.0091) * (1.65 / 0.0091)
TESTS = 0.20 * 0.80 * 181.3186 * 181.3186
TESTS = 5260.22

There you have it. The minimum number of tests we should conduct to have 90% confidence of a 5% profit is 5261!! Sobering isn't it

APPENDIX
===
1) Other confidence levels (Z):

60% = 0.84
70% = 1.04
80% = 1.28
90% = 1.65
95% = 1.96
98% = 2.33
99% = 2.58
99.9% = 3.29

2) These calculations assume that the average SP is an accurate representation of a selections chance of winning.
3) I was originally introduced to this technique by American writer/punter Dick Mitchell in one of his books. I've localised the calculations for you.

What if we just want to breakeven?
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Using the same basic figures above, how many tests if we just want to breakeven? You'd still need 1316 tests to have 90% confidence of not losing any money.

ERR = 0.20 - ((1 + 0) / 5.5)
ERR = 0.20 - (1 / 5.5)
ERR = 0.20 - 0.1818
ERR = 0.0182

TESTS = 0.20 * 0.80 * (1.65 / 0.0182) * (1.65 / 0.0182)
TESTS = 0.20 * 0.80 * 90.6593 * 90.6593
TESTS = 1315.05

If you only wanted to be 60% confident of not losing any money you've almost done enough tests.

TESTS = 0.20 * 0.80 * (0.84 / 0.0182) * (0.84 / 0.0182)
TESTS = 0.20 * 0.80 * 46.1538 * 46.1538
TESTS = 318.67

For 99.9% confidence of breaking even, a minimum of 5229 tests should be conducted.

TESTS = 0.20 * 0.80 * (3.29 / 0.0182) * (3.29 / 0.0182)
TESTS = 0.20 * 0.80 * 180.7692 * 180.7692
TESTS = 5228.4

__________________
Pixie

"It's worth remembering that profit isn't profit until it's spent off the racecourse." -- Crash

A blast from the past!

does anyone else use the Archie Score method?

I using test it on profitable systems but the other day come across a system that had 7000 bets and a loss of $600.

When I tested the Archie Score it was 56.278

What does this mean for a unprofitable system?

I guess I'm flying solo here.

Is woof43 still around?