Place markets math

Is it not 1.42 * 1.66 = 2.36 or 42.4%.

If you only lay for the win , you only have one div to pay.
If you lay for the place you have 3 divs to pay.
I myself like to just lay one for the place and it has to be a real shortie.
e.g a even money shot to win lay the place @ $1.20 etc.
In most cases where you lay 2 to 3 for the place you are going to finish about square.
in some cases they will run 1 2 or 2 3 . and if you try the the rest
they are well over for you to make a profit.
So just lay the Shortie one out for the place.....

But they add up to 2 because the book for the place market for 2 places is 200% (?)

Two events Andriy,yes I think you are right.

Traditionally you draw a probability tree with the outcomes all worked out.

In this case you can test your maths by doing all
the possible combinations.

AB or BA	0.7 X 0.6	0.42
AC or CA	0.7 X 0.4	0.28
AD or DA	0.7 X 0.3	0.21
BC or CB	0.6 X 0.4	0.24
BD or DB	0.6 X 0.3	0.18
CD or DC	0.4 X 0.3	0.12
TOTAL		1.45

But this is clearly not right because we end up with a total of 1.45.

Why?

Because the probability of AD is not the same as DA, so it is not correct to multiply 0.7 by 0.3
and say 0.21 is the chance of A and D being in the first two.

Look at all the possible outcomes and see the ones where A is placed. It is these that
come to 70%

AB

AC

AD

BA

BC

BD

CA

CB

CD

DA

DB

DC

Or for all the combinations:

Result	A placed	B placed	C placed	D placed
AB	Y	Y
AC	Y		Y
AD	Y			Y
BA	Y	Y
BC		Y	Y
BD		Y		Y
CA	Y		Y
CB		Y	Y
CD			Y	Y
DA	Y			Y
DB		Y		Y
DC			Y	Y

So how about this approach:

Given A is placed the chances of B, C or D being placed must add up to 1.
B - 0.6
C - 0.4
D - 0.3
Adds up to 1.3, so their relative chances are...
B 6/13
C 4/13
D 3/13

so....
Given A is placed, prob B placed = 0.7 x 6 / 13

etc...

A B = 0.323
A C = 0.215
A D = 0.162

Given B is placed we can then expand to get

B A = 0.3
B C = 0.171
B D = 0.129

and so on...

C A = 0.175
C B = 0.15
C D = 0.075
D A = 0.124
D B = 0.106
D C = 0.071

All of which add up to 1.
So we add them up to get the following:

AB = 0.623
AC = 0.39
AD = 0.286
BC = 0.321
BD = 0.235
CD = 0.146

That is: the chances of A and B coming 1st and 2nd in or 2nd and 1st are 62.3%

However, you should be able to go backwards from these figures, but if you do, you get

A 0.65
B 0.58
C 0.42
D 0.33

So we probably need a real mathematician to show us where I've gone wrong.

YOU WILL GET DIZZY TRYING TO DO THIS,AND THEN YOU WILL STILL LOSE

The answer is that you don't have enough information in the place probabilities alone to determine the joint probabilities of each pair.

You could add an assumption in order to get some numbers. For example, you could say that the probability of A placing doesn't depend on the other horse which places. If we write P[A] for the probability of A placing then the assumption can be written mathematically as

P[A given B] = P[A given C] = P[A given D]

Bayes theorem tells us

P[A and B] = P[A given B] x P[b]

...and after a bit of algebra we have

P[A and B] = P[A] x P[b] / (2 - P[A])

This is numerically the same as Eddie the Eagle's approach. There are two problems with the results of this assumption though: firstly if you add up all of the joint probabilities of each pair of horses the sum comes to 1.075 in the example case. So either the assumption is wrong, or the example doesn't match the probability sets you'd see in real life. I haven't tested this against market prices but I suspect that the back/lay spread would give you a range of solutions.

The second problem is that the assumption is not symmetrical. That is, if you choose to rebalance using horse B instead of A, i.e. your assumption is

P[B given A] = P[B given C] = P[B given D]

then you get a different result.

DBZE, you are correct.   If my assumption was correct you would get the same result starting with assuming B had place and then work out the probabilty of A placing too.
  Doing it this way you get a 0.3 ( 30 % ) chance or the decimal odds of 3.33.
  It wasn't that easy then..

Thanks Eddie - glad I managed to explain myself. I agree, it does look like to approach gives useful bounds on the probabilities.

Dunlaying - I'm not sure I follow your figures. If the total probabilities add up to 1 then are they the probabilities of winning instead of placing? Also, what do we know about the conditional probabilities?

To get the 0.311 i quoted above i used the following method, considering that the relative probabilities would stay the same between 2nd/3rd/4th once winner was selected.

Prob = P(A 1st, B 2nd)+ P(B 1st, A 2nd)

= [0.35*(0.3/(0.2+0.3+0.15)] + [0.3*(0.35/(0.35+0.2+0.15)]
= 0.161 + 0.15
= 0.311

Agree Andriy - I wonder if one of the posters is fine-tuning their bot after trying to back VDV at 29.

I think the assumption that win probability "exactly correlated" with place probability is too strong. For example, if there is a strong odds-on favourite you could imagine that the other win odds for the other runners would look very different to the place odds.

The win odds alone are not sufficient to determine the place odds, however there must be lots of people out there with statistical models to bridge the gap.

YOU CAN WRITE AS MANY NUMBERS AS YOU WANT ,BUT YOU STILL WONT WIN

My gut feeling is that the chance of any runner finishing in the forecast is variable (not fixed)and changes according to whichever runner joins it in the forecast but then again what do I know!!!

WHEN YOU HAVE FINISHED YOUR PROBLEM AND YOUVE GOT YOUR ANSWER ,YOU WILL HAVE ANOTHER PROBLEM WHEN YOU FIND OUT YOU STILL CANT WIN