Illogical inverse relationship between letting positions ride and greening to SP

We've gone to the dogs.
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jamesedwards
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Joined: Wed Nov 21, 2018 6:16 pm

I'm working on a new Greyhound lay automation. I've got one version letting the lay bet ride, and another that greens up to SP. Both are placing the original lay at the same stake and price.

1000 races in and I'm seeing an inverse relationship between letting the lay bet ride, and greening up to SP. I can't get my head around why that might be.

Any ideas?

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eightbo
Posts: 2343
Joined: Sun May 17, 2015 8:19 pm

nothing obvious i can determine, i'll go with spurious correlation for now as you still have periods where it's not correlated.
if there was some sort of fast player spiking price at off then you would imagine BSP would be significantly lower on those dogs which you layed that actually won, however as you're taking a profit on lay bet to back bet BSP it suggests drifters are often winning

ive got a lot of 1,000 mkt live samples of gh mkts and it's not enough to draw much of a concrete conclusion about anything.
on another note takesp is just gonna make P&L worse as anywhere you force a bet into the mkt just for the sake of reducing liabilities and the more size you're just making your price worse. of course when you're playing for tiny amounts hedging is possible without giving too much profit back using BSP but you'd likely be better off to take the variance on the chin and have higher underlying +EV on every market, scale up bet size over time. there's only so many pounds you can put through those markets anyway so the max drawdown should be absorbable for you (if not just limit stakes as needed until you can afford to up it)
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matekus
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Joined: Thu Apr 30, 2009 2:46 pm

At a very granular level, the following scenario (or equivalent) is unfolding.

- Lay one unit of stake at decimal odds β (liability β − 1).
- Let α be SP
- Let Y = 1 if the selection wins, 0 otherwise.

The back that greens the book is β/α. Both outcomes then pay the same amount:

{g: 'Green-At-SP'}
g = 1 − β/α

Let-It-Ride, same opening lay, held to settlement:

{r: 'Let-It-Ride'}
r = 1 − βY

which is +1 if the selection loses and −(β − 1) if it wins. Subtracting gives the residual:

r − g = β(1/α − Y)

The right-hand side is exactly the settlement of a fresh lay of stake β/α struck at SP. So, on every race,

Let-It-Ride = Green-At-SP + Lay-At-SP

All divergence between the bots lives in that residual (lay at SP, held to the result).
Last edited by matekus on Fri Oct 02, 2026 5:19 pm, edited 1 time in total.
elofan0
Posts: 466
Joined: Fri Jan 13, 2017 4:44 pm

Chat Gpts view , the two lines measure different things, even though both start with the same lay.
- Let it ride (blue): profit depends on whether the dog wins or loses.
- Green at SP (orange): profit depends on whether its odds drift or shorten before the start. A lay followed by a back at higher odds produces a trading profit. support.betfair.com
For example, laying £10 at 3.00, then fully hedging at SP:
SP Back stake to hedge Green result*
4.00 — drifted £7.50 +£2.50
2.00 — shortened £15.00 −£5.00


*Before commission. Calculation: back stake = lay stake × lay odds ÷ SP.
That explains how they can move in opposite directions: a dog can shorten, giving the orange line a loss, but then lose its race, giving the blue line a profit. A drifting dog can give a green profit and still win, causing a loss on the unhedged lay.
The screenshot alone cannot establish why this happened across those 1,000 races. The blue line also has much larger swings, which makes the apparent inverse relationship more noticeable.
To investigate it properly, split the results into three groups: drifted, shortened and unchanged. For each group, compare lay odds, actual SP, whether the dog won, and both profits. Also check that every SP hedge fully matched.
That will show whether the difference comes from price movement, race outcomes, or the automation’s hedge execution. An opposite-looking graph by itself does not prove either strategy has a reliable edge.
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