You just had to ask, didn't you Barny

Now my head hurts.

__________________
Cheers.

bhagwan, I haven't as yet read the technical posts from gunny72 (sighted them, yes!) ..... what can I tell you bhagwan ????

Sorry just won't do it I'm guessing.

I'll grab a red, try to free my mind up and I'll slip in a post to explain it all to you. The meaning of life is 47, so at least I have a reference point!

cheers
Barny

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

Hello, I'm new to the forum. I was interested in gunny72's win % and place % system, where there is one rating after the 2 strike rates are taken into account. It looked very interesting, have there been ongoing results kept since June 2011?
I'm probably posting in the wrong forum sorry. Just finding my way here.