18.11% is the probability of getting 3 successes from 5 trials where the probability of success on one trial is 35%.

Look up the binomial theory of probability. This theory assumes that the probability is the same in each trial and is not affected by any previous trial.

gunny

Nice work Gunny , But Angry Pixie sorted it for everone quite a few posts back.

mooee, you got 1.81% originally are you were disappointed I didn't send you a thanks immediately ..... lol

18% or roughly 1 in 5 is pretty good odds for what I'm looking at.

I'ld rather not talk about it.
But I knew sonething wasn't right.
50 to 1 sounded way to low.

Anyway , now that you have the correct answer , have you found a use for it?

mooee, I take it that you know whay your on about here .....

The odds of three winners in a row with each winner being assessed at 35% would be ????? 35% x 35% x 35% = 4.29% So the odds of three winners in 5 selectionas has to be greater. Now, you got 1.81% ..... we accept at the moment 18.1% which is ten times your result. Is it possible that your result was correct, except for the decimal place ???? ..... 18.1% seem too liberal for 3 / 5 @ 35% ????

I'm looking into this further mooee, maybe you weren't too far off !!!!!!!

My answer of 1.81% was the Calculation of having 3 winners and 2 Losers

It is the Odds of EXACTLY 3 Winners out of EXACTLY 5 Races.

There is nothing further to investigate.

I can explain why moeee gets 1.8% instead of 18.1%.

He has correctly calculated the value for 3 particular wins out of 5 (eg WWWLL) only but this can occur in 10 possible ways WWWLL or LWWWL or LLWWW or WLWLW or LWWLW etc so his result must be multiplied by 10 to get the correct 18.1% for any three wins from 5 trials.

All system enthusiasts should come to grips with the binomial theorem of probability as it is a invaluable tool.

gunny

Hi gunny, how is it possible that 3 out of 5 = 1.81% yet 3 out of 3 is 4.29%. 3 from 3 is harder yet ???

cheers
Barny

1.8% is the probability of 3 wins from 5 races with the wins occurring in a specific order, ie, one of the 10 possibilities I pointed out previously, not just any 3 wins from 5 races for which the prob is 10*0.35*0.35*0.35*0.65*0.65=18.1%.

There is only one way to get 3 wins from 3 races and that is WWW and this has a probability of 1*0.35*0.35*0.35=4.9%. Hence it is much more difficult to get 3 wins from 3 races than 3 wins from 5 races.

For 5 races where W represents the probability of a win in one particular race and L represents the probability of a loss in that same race the following terms give the probabilities of the number of wins shown below

0 wins 1*L^5
1 win 5*w^1*L^4
2 wins 10*W^2*L^3
3 wins 10*W^3*L^2 (this gives 0.181 or 18.1%)
4 wins 5*W^2*L^3
5 wins 1*W^5

You should be able to see the pattern for W and L. The numbers 1 5 10 10 5 1 come from Pascal's Triangle, which is too difficult to format here but you can look it up. The sum of all the individual probabilities is always 1 for a particular number of races.

For 3 trials the numbers are 1 3 3 1 and the calculations are

0 wins 1*L^3
1 win 3*W^1*L^2
2 win 3*W^2*L^1
3 win 1*W^3 (this gives 3 wins from 3 races)

As an illustration take the top ranked place percent horse which Knowles claims has a 21.7% (say 22%) win rate (though I dispute this figure for Saturday racing.) Suppose you want the probability of at least 1 win from 5 races using this factor. You could add the probs for 1, 2, 3, 4, 5 wins or simply take the probability of all losses away from 1.

Therefore W=0.22 and L=0.78 and Prob of at least one win from 5 races = 1- prob of all losing = 1-0.78^5 = 0.71=71%

Similarly, the probability of at least 1 win from 3 races = 1-0.78^3=0.52=52%.

This is very important maths for punting so I hope this helps even if you use a spreadsheet. (Not that I have the holy grail)

gunny

Sorry, error in 4 wins from 5 starts example table. it should be 4 wins 5*W^4*L^1
gunny