Often I see people ask how many losers can be expected before a win (a worst case scenario). The maths I have discussed above provides the answer. I will just give mainly the final result although it is not that hard and is taught in high school.

Let C=the prob of at least 1 win from n races, called the certainty, expressed as a decimal.
Let q=the prob of a LOSS in a race, expressed as a decimal.

Now to be 100% certain you need an infinite number of races so this theory expects you to decide on an acceptable certainty usually 90% or 95%.

Thus C=1-prob of all losing
that is C=1-q^n

If you solve this the number of expected outs is

n=log(1-C)/log q

You can use log or ln on your calculator but it has to be the same throughout the calculation.

For example, with the previous place percent example and chosing 95%,

n=log(1-0.95)/log0.78
n=log0.05/log0.78
n=12.06 so you can be 95% sure of at least one winner in about 12 races.

To be 99% sure it takes a run of about 18 or 19 races.
gnny