What are the real odds?

If Team A is paying 1.8 and team B is paying 2.1 is there a way of roughly determining what odds the bookmaker has placed on Team A winning? (estimate within a few % points and without knowing the house edge).

Is there a formula/method that can be used for any set of odds?

Thanks
Paul

For my American readers, let's explain that the wins are expressed the European way. In this example, a bet of $1 on team A would pay $1.80 if it won, which includes the original wager.

Let's assume that the European bookmaker decides what are the fair odds on both teams, and then subtracts a certain amount from both, to allow for profit. That is how lines are set in the U.S., more or less. For example, suppose the bookmaker thinks team A has a 60% chance of winning, and B is 40%. Then fair odds on team A would be 1/.6=1.67, and on team B 1/.4=2.50. Then he subtracts 0.05 from both for a profit margin. So his posted odds would be:

A 1.62
B 2.45

The question at hand is if you know the offered odds only, then what are the fair odds, probability of winning, and house edge?

In this case a=1.8 and b=2.1. Let's call c the commission deducted from each win. We know one team must win so...

(1/(a+c)) + (1/(b+c)) = 1
Find a common denominator...
a+b+2c/((a+c)(b+c)) = 1
Next, solve for c...
c^2 + c*(a+b-2) + ab-a-b=0
Using the quadratic formula...
c=(-(a+b+c) +/- ((a+b-2)^2+4*(a+b))^0.5)/2
Plugging in the values for a and b we get c= 0.061187.
Here are the fair figures for each team:
Team A
Fair odds = 1.861187
Prob. win = .537291
House edge = 3.2875%
Team B
Fair odds = 2.161187
Prob. win = .462709
House edge = 2.8312%

If you can assume the house edge is equal on all bets, and you want to cut right to it, then you can use my futures bet technique, explained in my sports betting appendix 5. In this case the sum of the inverses of the offered odds is 1/1.8 + 1/2.1 = 1.031746. The house edge is 1-(1/1.031746) = 0.030769.

Ugh I really don't like math! I gave up after Calc One (I did try Calc2 it was a disaster)

Nice one - that explains it. I still remember quadratic formula from school but haven't had much need for it in the last 20 years...

Thanks Wizard