In Victoria there is a bet where you need to find the jockey at each meeting to score the highest points on his efforts,with a 3,2,1 reward for finishing in the placings for the whole card.
Anyone know how to come up with a market in this form of punting?

Don't try that one with a calculator or a pencil and paper - you need to program a formula into a computer and give it a price/ rating for every runner in every race. You'd need to be a decent mathemetician to work out the formula and a decent price assessor to work out the prices.

Yeah, I don't know how you would come up with a market Mo.

It's pretty tricky. If I think one of the top jockeys has a great book of rides, I'll sometimes have a go, but so many things can go wrong. I backed a double of Beadman and Oliver a few weeks ago and Oliver was never in doubt but Beadman had to win the last race on an $8 shot to get up. Nerve racking!!!

I remember one day backing a jockey who I thought was a certainty. He started with a couple of wins and I thought he was home, but then he injured his foot before the 3rd race, was stood down for the rest of the day and got beaten.

With the speed of computers,Filante,I don't know if the permutations are a huge obstacle.
The formula would also apply to the Brownlow Medal.
I will work on it.
Trouble with the Brownlow is injuries and suspensions.
Jockeys shouldn't be too tricky.

Trying to come up with a formula regarding the Jockey Challenge I narrowed down the possibilities by coming up with a basic set of hypothetical,but unreal,set of races.

Let there be 3 races,each of which have 4 starters,ridden by only 4 available jockeys.
Say,Beadman(B),Cassidy(C),Nikolic(N) and Oliver(O).
And just for the sake of experimentation,let Beadman always get the $2 favourite,Cassidy gets the $4 chance,Nikolic gets the $5 chance and Oliver has the $20 chance.
So we know the chances of our jockeys are winning the race,but what are their chances of running 2nd and 3rd and unplaced.
Well in my more patient years,I came up with a formula which would calculate the probabilities of every possible trifecta combination,after inputting all the starters prices.
From that I could then work out the place chances and even the price of finishing in a particular position.
Now of course I don't recall how I done this,but the program is still there and I have used it to come up with the appropriate prices for the hypothetical races today.

The percentage possibilities for race 1(which will also apply to all our hyperthetical races.) as regards finishing position are:

----1st2nd3rd4th
B - 50 32 15 03
C - 25 33 32 10
N - 20 28 38 15
O - 05 08 15 72

Just side tracked for a moment,notice how the 20-1 shot has the same chance of running 3rd (15%) as has the Even money favourite.Yet we can understand 20-1 shots running third but jump up and down when an evens favourite can only sneak 3rd.

GEE!.I was doing allright for a while but now it looks as though I've come to a huge abyss.And to cross the abyss it seems to be you need to know every possible permution for the placings in said hypothetical race!

Goodbye cruel woorrrrrlllllldddddddd!

I'll be back!

Roger Dedman gives a trifecta formula in his book...I think it's called "Practical Punting - A Mathematical Approach" (...or something like that).

The difficulty in assesing this form of bet could turn out to be a good thing. The people working at the tab are probably struggling just as much to line up the form. Hence there should be opportunities for false odds, this as you all know is the fundamental way of profiting off the punt.

Just saw this thread for the first time

OK I am going to think out loud on this one, ie on this page.

To use Mooee's last example, jockeys are called A,B,C, they ride in the two races on the $2.0, $2.50 and $10 chances each, in each race, adding to 100% in each race. (We have a friendly bookie)

OK.

There is 36 possible finishes.... I think, ie 6 possible ways in the first leg, six in the second. That's easy enough. So next we calculate the probability of each one coming up. Hang on, I'll go write an excel thing for that.

OK so what I did was write an excel thing working out the likelihood of each of the 36 combinations, and - most importantly - WHO WOULD WIN THE JOCKEY CHALLENGE given that that was the combination that came in.

I'll list here the 36 combinations, followed by their decimal chance of occurring, and who would win if that occurred.

a b c a b c 0.16 A
a b c a b c 0.16 A
a b c a b c 0.16 A
a b c a b c 0.16 A
a b c a b c 0.16 A
a b c a b c 0.16 A
a c b a c b 0.01 A
a c b a c b 0.01 A
a c b a c b 0.01 A
a c b a c b 0.01 A
a c b a c b 0.01 A
a c b a c b 0.01 A
b a c b a c 0.111111111 B
b a c b a c 0.111111111 B
b a c b a c 0.111111111 B
b a c b a c 0.111111111 B
b a c b a c 0.111111111 B
b a c b a c 0.111111111 B
b c a b c a 0.004444444 B
b c a b c a 0.004444444 B
b c a b c a 0.004444444 B
b c a b c a 0.004444444 B
b c a b c a 0.004444444 B
b c a b c a 0.004444444 B
c a b c a b 0.00308642 C
c a b c a b 0.00308642 C
c a b c a b 0.00308642 C
c a b c a b 0.00308642 C
c a b c a b 0.00308642 C
c a b c a b 0.00308642 C
c b a c b a 0.001975309 C
c b a c b a 0.001975309 C
c b a c b a 0.001975309 C
c b a c b a 0.001975309 C
c b a c b a 0.001975309 C
c b a c b a 0.001975309 C


total decimal chance of each jockey winning the challenge is:

A - 1.02
B - 0.69
C - 0.03
Total - 1.72

And, I suppose, therefore the chance of each winning the jockey challenge is:

A - 1.02/1.72 = 0.59 (59%) odds $1.68
B - same maths - $2.48
C - same maths - $56.63

Does this seem right? As I said, this is just 40 mins of me thinking as I went.

Duritz

aah ******** just realised some cmobinations have doubled up

heaps, in fact. ignore answers. post again shortly.