Question for the mathematically minded
 

At a Win% S/R of 35% what are the odds of this system returning 3 wins in 5 selections ??

17.5%

Thanks you so much bhagwan for the prompt and unequivical response to what I had thought a fairly complex probem regarding probability that I had put forward for discussion. I've no doubt given your high standing here that you did your DD before answering so as to maintain your hard earned integrity.

I'm humbled to have been given a response even though it totalled only 4 characters. I'll cherish this tenuous link for many years to come.

No need to read any further bhagwan .....

Anyone who has any idea of probability, I would love to hear from regarding the probability of an event happening where there is a single 35% success rate. The event is what are the odds of 3 "WINS" in 5 attempts where each attempt shows 35% success.

1.81%

A "Thank you Moee" would have been nice.

You Either ignore help or bag the helper Barny.
No wonder there has been no other responses after 100 views!

Have a nice day :)

this may help in future

It says in that mathematical calculation that if one had 10 tries of 5 bet lots, then the chance of hitting 3 from 5 is 18.11%

So we had a difference .61%

If you just have one go of 5 bets, then your chance of hitting 3 wins is just 1.8%

We were assuming that you were going to have more than just one go at 3 wins from 5 bets.

You wished for a prompt reply to the question.
You received a succinct response.
Its the answer you don't like, even though its correct in essence.

I'm sorry you feel offended by that answer.
We were trying to help you with a prompt answer without the complicated waffle.

Both answers are correct.
Why not just take the answer that upsets you the least.

Now that you have the percentages , all you have to do now is work out how to make a quid out of it .

There is an answer to that also.

bhagwan & mooee,

Don't get your knickers in a twist eh? I've got some info on probabilities and will have a butchers at that. bhagwan, my response was tongue in cheek, humourous, s'pssoed to be funny.

BTW, you both missed the 35% S/R of each.

mooee, you were the one who wanted facts and figures to support a theory of mine not so long ago, BUT now it's OK for you to pluck a number out from wherever whithout any back up? A touch hypocritical don't you think?

Hi woof43, just opened up your file ..... thanks heaps.

I don't pluck numbers buddy.
You didn't ask for backup , you asked for the answer.

So you post and let everyone know what you think is the answer and the backup.
Surely you didn't come up with 35% as the answer - without any backup?

mooee, my 35% has NOTHING ro do with the answer ..... it's a part of the QUESTION. Your answer appears to be correct. Thank you for your effortd. I'll give you a winning system for nothing - 4 sts this time in - PP fav - last two sts 21 or 11 - 250+ or - dist change - 21 days or less - c or t - last st at metro track

BTW there is only one "s" in disappointed ...........

you're welcome

And there only one "s" in efforts and zero "d"

Give the System to someone else.
I don't need help to lose, I can do that just fine without your help.

Barny

Drop this into Excel to give you the probability of getting exactly 3 successes from 5 trials with a 35% chance of success on each trial.

=BINOM.DIST(3,5,0.35,FALSE)

You'll get 18.11%

If you wanted to know the chance of getting up to 3 successes (not exactly 3 successes) replace the FALSE with TRUE.

Thanks AngryPixie, will do.

If it's of interest to you ..... You might like to run that little system through your DB. I don't need your findings.

Gees.This reminds me of Mistermac on another site.

Thanks Barny. Don't keep a database. Always seemed a bit to much effort. ;)

Excellent work Angry Pixie.
Are you a Winning Punter? - You should be.

Thanks. No worries.

Sorry Barny, I took it the other way.

bhagwan, why would I have a crack at you. You've posted more stuff on this site than I've had hot dinners. The 35% S/R is to do with my selection of mules based on a fixed Win% of higher than 35% with the expectation the win S/R will, over time drop to 35%. I'm looking at ways, other than chase losses, to improve the returns. One good thing that I just didn't factor in, are the multiple selections in some races (somteimes 4+) which need to be pruned to one selection, or depending on the day and the race, may be left out altogether. At the weekend there was one race where there were 4 selections, so the best result would have been a 25% S/R which is much lower than the lowest 35% forecast. Multiple selections, even a couple in a race, make the "Initial Win S/R%" less likely to drop to 35%.

cheers bhagwan
Barny

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

You just had to ask, didn't you Barny

Now my head hurts.

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

Number of run of outs
 

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.