Odds trading question - Help needed please

Not including commission id say around 1.82

what it doesnot contract

if

then theres no opportunity to trade out

Totally off topic, but I was at Coalville yesterday.

there is software out there that u can programme in percentage win.
ie say 10%
as soon as you bet a selection say at evens,
the software put in a bet at 1..80 to lay

Why not use .....


http://www.oddschecker.com/betting-tools/hedging-calculator

Betfair lay does this for you:
add the % profit you want to your back stake in the stake lay box
put up the odds you got to your back bet
press the black down arrow till it matches the back bet profit

The maths behind greening up is relatively simple, just a balanced equation. Things might start to look complicated when you start taking into account commission rates etc but as long as you do it step by step it's not too hard.

THe basic maths behind it is

New_Odds x New_Stakes = Original_Odds x Original_Stakes

to get the New_Odds we simply divide both sides by New_Stakes

New_Odds  = Original_Odds x Original_Stakes/New_Stakes


Because your New_Stakes will simply be a Percentage of the Original_Stakes we can say

New_Stakes =  Original_Stakes x (100 + Percentage)/100

If we want to take Commission into account we adjust that to

New_Stakes =  Original_Stakes x (100 + Percentage)/100 x -(100/(Commission-100))


So replacing New_Stakes in our equation we have

New_Odds  = Original_Odds x Original_Stakes/(Original_Stakes x (100 + Percentage)/100 x -(100/(Commission-100)))

To simplify that equation fully we end up with a simple equation of


New_Odds = − Original_Odds(Commission−100)/(Percentage + 100)

If you don't want to take into account your commission rate you can simply use

New_Odds = (100 x Original_Odds)/(Percentage + 100)

Help me with this Ghetto

Three horse race A 50% B 40% C 10% winning chances

What are the odds on each finishing 1st or 2nd

Thanks

You need to look up the  Harville method which basically treats it as a set of races, so If horse A wins you now simply treat it as a separate race between B and C to come first (i.e. second to A). Then recalculate assuming C has won the race and therefore no longer in the separate race to be second to C. The problem is it overestimates odds on chances and also fancied runners probably because they may not be trying as hard to place as they were to win.

Thanks Ghetto

Got a fix
A = 88.88 chance of being in the first 2
B = 84.45
C = 26.67
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