Jockey Challenge

[moeee]

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?

[Filante]

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.

[Sportz]

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.

[moeee]

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.

[moeee]

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!

[Filante]

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

[moeee]

He was in fact my mathematics teacher back in my RMIT days and I always thought it was him I'd often see on the racecourse later on.
Yeah I got the book and still working on the Challenge challenge.

[moeee]

I said I'd be back

Look,this is ridiculous.
We have our percentages and we know our points.
Let's just simplify it like this to see if it leads us anywhere.

B - 50% of 3 + 32% of 2 and 15% of 1 = 2.29
C - .25 x 3 + .33 x 2 + .32 x 1 = 1.73
D - 0.6 + .56 + .38 = 1.54
O - .15 + .16 + .15 = 0.46

Now I don't know what we got but who cares.
Common table salt is made up of 2 very deadly elements,yet sure makes my fish'n'chips taste good!

So Beadman has 2.29
and Cassidy has 1.73
and Nikolic has 1.54
and Oliver has 0.46

I often use a formula like X=Y^Z
where y is a special number and z is our rating.
Well Y=0.884 and again I feel like jumping over the cliff.BByyyyeeeeee!

[mugwump]

I've tried this form of betting a few times. Successfully backed Gauci on one occasion. Trouble is, the TAB don't post the results anywhere, and when I asked them about this they said 'you have to work it out for yourself'. Also, they don't pay until the next day.

[moeee]

So lets make 2 hypothetical races with Beadman and Cassidy and Evans the only riders.
Beadman rides the Evens favourite in both races with Cassidy riding the 6/4 shots and Evans has the 9-1 chances.

So this is what it can all boil down to.

BCE 40
BEC 10
CBE 33.3
CEB 6.7
EBC 5.6
ECB 4.4
for each of the 2 races giving us 36 permutations.

After going through the 36 combinations,in this case the Jockey Challenge price is the same as the price of the starters.

Does that seem right?

Tell you what I did learn but!
If you can't work out whether you will be getting value on your bet,DON'T BET!

Can't wait till footy starts!.

[Sportz]

Trouble is, the TAB don't post the results anywhere, and when I asked them about this they said 'you have to work it out for yourself'. Also, they don't pay until the next day.

Well, I listen to Radiotab and they always announce the winner after the last race. And I'm sure you should be able to get your winnings then too. I usually bet on the computer and when I've backed the winner, the money goes into my account not long after the last race.

[Sportz]

Think you're getting yourself confused Mo. If not, you're certainly doing a good job on the rest of us!

[Mr ed]

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.

[Duritz]

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

[Duritz]

aah shit just realised some cmobinations have doubled up

[Duritz]

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

[Duritz]

No! Actually they were OK the first time!!!

Those odds SHOULD be the true odds.

Any thoughts?

[Duritz]

actually no they ARE wrong. let me try again.

BRB

[Duritz]

OK here's the changes

a b c a b c 0.16 A
a b c a b c 0.04 A
a b c a b c 0.133333333 A
a b c a b c 0.026666667 B
a b c a b c 0.017777778 A
a b c a b c 0.022222222 A
a c b a c b 0.04 A
a c b a c b 0.01 A
a c b a c b 0.033333333 A
a c b a c b 0.006666667 A
a c b a c b 0.004444444 C
a c b a c b 0.005555556 A
b a c b a c 0.133333333 A
b a c b a c 0.033333333 A
b a c b a c 0.111111111 B
b a c b a c 0.022222222 B
b a c b a c 0.014814815 B
b a c b a c 0.018518519 A
b c a b c a 0.026666667 B
b c a b c a 0.006666667 A
b c a b c a 0.022222222 B
b c a b c a 0.004444444 B
b c a b c a 0.002962963 B
b c a b c a 0.003703704 C
c a b c a b 0.022222222 A
c a b c a b 0.005555556 A
c a b c a b 0.018518519 A
c a b c a b 0.003703704 C
c a b c a b 0.002469136 C
c a b c a b 0.00308642 C
c b a c b a 0.017777778 A
c b a c b a 0.004444444 C
c b a c b a 0.014814815 B
c b a c b a 0.002962963 B
c b a c b a 0.001975309 C
c b a c b a 0.002469136 C


and therefore on the same maths as before

decimal chances

a - 0.72
b - 0.60
c - .13

and odds should be

A - $2.37
B - $2.85
C - $13.30

Any thoughts on the NEW answers???

(I must say my confidence is now shot)

[Duritz]

LOL sorry posted the OLD combinations. Here's the new ones:

a b c a b c 0.16 A
a b c a c b 0.04 A
a b c b a c 0.133333333 A
a b c b c a 0.026666667 B
a b c c b a 0.017777778 A
a b c c a b 0.022222222 A
a c b a b c 0.04 A
a c b a c b 0.01 A
a c b b a c 0.033333333 A
a c b b c a 0.006666667 A
a c b c b a 0.004444444 C
a c b c a b 0.005555556 A
b a c a b c 0.133333333 A
b a c a c b 0.033333333 A
b a c b a c 0.111111111 B
b a c b c a 0.022222222 B
b a c c b a 0.014814815 B
b a c c a b 0.018518519 A
b c a a b c 0.026666667 B
b c a a c b 0.006666667 A
b c a b a c 0.022222222 B
b c a b c a 0.004444444 B
b c a c b a 0.002962963 B
b c a c a b 0.003703704 C
c a b a b c 0.022222222 A
c a b a c b 0.005555556 A
c a b b a c 0.018518519 A
c a b b c a 0.003703704 C
c a b c b a 0.002469136 C
c a b c a b 0.00308642 C
c b a a b c 0.017777778 A
c b a a c b 0.004444444 C
c b a b a c 0.014814815 B
c b a b c a 0.002962963 B
c b a c b a 0.001975309 C
c b a c a b 0.002469136 C

The odds end up the same, I used the new maths for those, so the prices are still what I said, ie

A - $2.37
B - $2.85
C - $13.30

[Duritz]

one thought - why doesn't the decimal chance total add up to 1.00?

[Duritz]

I am talking to myself here, and worse still, I don't know the answers. I do love maths sometimes though. Great challenge for the mind. Really makes it stretch.

[Duritz]

Actually, they do add up to one. The chances of each combination do. Problem is there were some dead heats, which I didn't account for.

BRB

[Duritz]

OK with dead heats accounted for, these are the decimals:

A - 0.53
B - 0.41
C - 0.06

and odds therefore are a matter of dividing those into 1, because those add up to 1.

For those who couldn't be bothered, however, here is one I prepared earlier:

ODDS

A - $1.89
B - $2.45
C - $15.75

OK - I reckon that is RIGHT. I reckon, given those outcomes, those are the TRUE ODDS of each of those jockeys WINNING the jockey challenge.

SIGH!

Thoughts, anyone?

I am going to get a glass of red.

[Sportz]

Having a good conversation with yourself???

[mugwump]

Now that you've had your Cab Sav, can you put it all in a nutshell so I can win some money?

[Duritz]

Mugwump yeah not really. Would need to write the program bigger. Wrote it merely for the possibilities only of this example.

Oh yeah, and it's shiraz.

Duritz

[moeee]

Think you're getting yourself confused Mo. If not, you're certainly doing a good job on the rest of us!

Don't you worry Buddy.
I'll make up for the mess during the Footy season.

[moeee]

one thought - why doesn't the decimal chance total add up to 1.00?

Most likely because of a slight hiccup!
Or the vine expelling some of those fermented gasses!

[moeee]

ODDS

A - $1.89
B - $2.45
C - $15.75

OK - I reckon that is RIGHT. I reckon, given those outcomes, those are the TRUE ODDS of each of those jockeys WINNING the jockey challenge.

SIGH!

Thoughts, anyone?



Yes my good man.You are spot on.
Prices being quite close to the original starting prices of the mounts.
And that,boys and girls is how it is done.
So with 8 races of 12 starters each,the permutations are somewhat daunting.
Can I borrow your calculator Filante?

[Duritz]

Yeah they are. I have a pretty good and simple working model for how to price them though, I won't detail it right now, will tomorrow if I get time.

Duritz.

[moeee]

Please do!
Make great usefulness perhaps in predicting the Brownlow winner.

[Duritz]

What I do is this, when working out a jockey challenge market:

In each race, award points from 6 to 1, in order of favouritism, so 6 for the fave, 5 second fave etc, AND give each rider a point just for having a ride. There is room for leeway, like if one of the faves will be Might and Power in the QE Stakes, ( ie 1/10 odds) you can give him more than six. The six through one is a guide to the "normal" race, adjust if it's an abnormal one.

Add em up.

Divide any riders points by the total, multiply by 100, that's their percentage chance.

It's crude, but it's pretty accurate.

Duritz.