Arbing between 2-way and 3-way markets ?

Snake46 · Nov 8, 2007

Hi,

This is a question for anyone who knows a little bit more about arbitrage (scalping) than most...

I'm wondering if there's a way to get an arb out of matching a 2-way and a 3-way market.

For example, Pinnacle has 2-way markets for Boxing, whereas Betfair has 3-way markets (Opponent1, Opponent2, Draw).

If it's possible, could someone give an example (with numbers please), and even better if you have the formula that will determine the stakes to put on each outcome in order to have the risk-free profit.

Thanks.

Kornholio · Nov 8, 2007

scalpulator.com

Kornholio · Nov 8, 2007

Pretty simple, a 3 way arb obviously has to add up to more than +600 for all bets, to be profitable. Type in the odds, it tells you what to bet on each.

Woody0 · Nov 8, 2007

Kornholio said:
scalpulator.com

I don't think you can use this because some 2 ways return stake on a draw, if I understand the question correctly.

I don't know boxing, but the 2 way and 3 way markets also exist for hockey with 2 way -0.5 and +0.5 bets. The -0.5 the same as a simple win bet while the +0.5 bet is a combination win and draw bet. These two bets can be simply entered into the scalpulator, I guess.

More difficult is the two way ML bet which has no action on a draw. To compare with a 3 way you need to compare with both the win and a simultaneous bet on the draw to return stake.

I guess there's a formula but I always work these out by hand. Generally I find the bookie gets a higher overall vig on 3 ways and scalps are rare to non-existent.

Snake46 · Nov 8, 2007

Woody0, you understand exactly what I mean...

I still don't see how to work out the math though ...

Woody0 · Nov 8, 2007

Snake46 said:
If it's possible, could someone give an example (with numbers please), and even better if you have the formula that will determine the stakes to put on each outcome in order to have the risk-free profit.

Well to take an example Pinnacle has the following for Saturday’s match using their decimal odds:<o

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Cotto 1.649<o

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Mosley 2.440<o

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while Betfair has the following bet offers:<o

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Cotto 1.71<o

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Mosley 2.64<o

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Draw 23<o

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To get the amount needed for a maximum arb divide 100 by the decimal odds. For Pinnacle this gives 100/1.649 [$60.64] + 100/2.44 [$40.98] = $101.62. Thus betting $101.62 gives a guaranteed return of $100.<o

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For Betfair the same calculation gives 100/1.71 [$58.48] + 100/2.64 [$37.88] + 100/23 [$4.35] = $100.71 for a guaranteed return of $100, ignoring commission.<o

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Now to calculate the arb we bet one contestant at Pinny and the other at Betfair and also bet the draw at Betfair to return the stake on the Betfair contestant.<o

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1. Betting Cotto at Pinnacle:<o

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100/1.649 [$60.64] + 100/2.64 [$37.88] + $37.88/23-1 [$1.72] = $100.24<o

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2. Betting Mosley at Pinnacle:<o

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100/2.44 [$40.98] + 100/1.71 [$58.48] + $58.48/23-1 [$2.66] = $102.12<o

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Now if PA is the Pinnacle odds of contestant A and BB is the Betfair odds of contestant B and D is the Betfair odds for a draw then the bet amount formula is as follows:<o

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100/PA + 100/BB + (100/BB)/D-1<o

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which if less than 100 yields a profit.<o

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Adding in the effect of Betfair commission percentage C gives the following modified formula:<o

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100/PA + [100/BB]X100/[100-C] +[(100/BB)/D-1]X100/[100-C]<o

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Hope this helps.<o

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Snake46 · Nov 9, 2007

Woody0 said:
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Now if PA is the Pinnacle odds of contestant A and BB is the Betfair odds of contestant B and D is the Betfair odds for a draw then the bet amount formula is as follows:<o></o>
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100/PA + 100/BB + (100/BB)/D-1<o></o>
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which if less than 100 yields a profit.<o></o>
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Adding in the effect of Betfair commission percentage C gives the following modified formula:<o></o>
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100/PA + [100/BB]X100/[100-C] +[(100/BB)/D-1]X100/[100-C]<o></o>
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Hope this helps.<o></o>

And if we generalize even more, let's say that:

O2 = the odds for a contestant (A) at the 2-sided book
C2 = the commission for the 2-sided book
O3 = the odds for the other contestant (B) at the 3-sided book
D = the odds for the draw
C3 = the commission for the 3-sided book

We have:

SA = Stake for contestant A at 2-sided book = [100/O2]X100/[100-C2]
SB = Stake for contestant B at 3-sided book = [100/O3]X100/[100-C3]
SD = Stake for Draw at 3-sided book = [(100/O3)/D-1]X100/(100-C3)

Thus, if (SA + SB + SD) < 100, we have an arb.

Thanks Woody0, I'm sure a lot of arbers will find this very interesting.