Sports Arbitrage Formulas

[qcan]

Hi,

I am looking for a couple of formulas that can be used to calculate
an
arbitrage in sports wagering.


Example 1.


An even split based on a 1000 dollar total wager divded between both
sides.


- Side 1 odds = +110
- Side 2 odds = +150


I know that one would have to wager $ 543.48 on side 1, in order to
get a return $ 1141.31
I also know that a wager of $ 456.52 on side two in order to get a
retrun of $ 1141.30


In other words a 14.13 % profit would be generated evenly no matter
which side won.


Example 2.


The same as example 1 except I want to maximize side 1 while breaking
even on side 2. To clarify - this would mean that all the profit
would
go to side 1 should side 1 win. If side 2 wins - I would break even.


Using 1000 dollars again (actually 1140 dollars so that it's even),
with the same odds as the first exmaple - I know that


- $ 684.00 would have to be wagered on side 1 and $ 456.00 would
have
to be wagered on side 2.


If side 1 wins - I would get back a total of $1436.40 (meaning a $
296.40 profit)
If side 2 wins - I would get back exactly the total money wagered on
both sides (1140 dollars) and break even.


I have no idea how to express these two examples into a formula.
Further, the formula(s) might be different when the favorite is
actually below -100.


I hope I made some sense. I will try and clarify should there be any
questions.


Anyone ?


Thanks !!

 

[dCromley]

- Side 1 odds = +110
- Side 2 odds = +150
I know that one would have to wager $ 543.48 on side 1,
in order to get a return $ 1141.31

[I doubt this is the group for this]
I see noone answered, so I'll bite --
how do you arrive at 1141.31?

 

[qcan]

qcan  wrote:

- Side 1 odds = +110
- Side 2 odds = +150
I know that one would have to wager $ 543.48 on side 1,
in order to get a return $ 1141.31

[I doubt this is the group for this]
I see noone answered, so I'll bite --
how do you arrive at 1141.31?

Hi,

(thanks for biting). Yes, you are probably right that this may not be
the proper group to post this in. I thought it would be a good start
as I am looking for an Excel formula that can hopefully do this
(actually I am sure it can - I just don’t know how to do it).

I arrived at $ 1141.31 by using another stand-alone program called
“Arbscalc”. A trial version is available @ http://www.downloadatoz.com/developer-cheong-koo.html

The program seems to precisely calculates how much must be optimally
wagered in order for the return on both sides to come out even (for
example 1).

In other words.

- If you wagered $ 543.48 and won based on odds of +110 that means for
every dollar you wager, you get back $ 1.10.
- $ 543.48 multiplied x 1.10 = $ 597.83
- You get back $ 597.83 + your original wager of $ 543.48 for a total
of $ 1141.31

The first example was easy to explain. I can’t explain example two,
although by looking at the results and working backwards, I know it’s
correct.