mathimatical varience of a system

[tilson]

I believe the correct formula to determine the varience of a system would be as follows.
the square root of wins time the square root of loses time the square root of total events = varience.
In other word if you did 200 races and had 50 wins it would be square root of 25% times square root of 75%times square root of 200
so in this example the resultent number is 9....this number 9 should reflect that if you used the same methodology of selection and did a set of 200 10 more time your number of winners should reliably be predictable as in the 50 range plus or minus 9 winners.

[Rick Ransom]

The variance of a proportion is n * p * (1-p), where n is the number of observations and p is the probability of the desired result (winning in this case). In your example, n = 200 and p = 0.2, so the variance is:

200 * 0.25 * 0.75 = 37.5

Now, what you really want to calculate in order to determine how close you should come in another sample is the standard deviation, which is the square root of the variance. That would be about 6.12. According to the theory you should be within plus or minus one standard deviation about 68% of the time and within two standard deviations about 95% of the time. So it's more like approximately 44-56 68% of the time and 38-62 95% of the time.

OK, now all of that assumes that p is constant. That would be true if you were rolling fair dice or drawing cards from a well shuffled deck, but probably not exactly true in horse racing for reasons I've mentioned in other threads. So, while it's a good guidline, you should generally be a little more conservative because if the true proportion of wins is variable there will be more extreme values than this.

I still think the real problem is than nobody ever considers the variance of win payoffs when they do these kinds of calculations, and that would make the above calculation pretty meaningless in order to try to predict future ROI.