Betting lines and implied percentages

[Romanowski]

Key to analyzing a set of betting lines is determining the win probability implied by those lines. This is simple mathematically but understanding the underlying concepts behind it can be quite valuable.

If a money line is offered at -200, then this of course means that for every $200 risked $300 is returned to the bettor if the wager is successful. In other words the amount risked is 2/3 of the total amount returned. Not at all coincidentally, 2/3 is also the implied probability of the -200 line. (By “implied probability” we’re referring to the frequency with which the wager would need to win for the profit/loss expectation to be zero.)

Using the same logic, were a line offered at -110 then for every $110 risked, $210 would be returned. $110 is 52.38% of $210, meaning that the implied probability of a -110 line is 52.38%. As such, if your bets at -110 win with frequency greater than 52.38% you’ll be making money, and conversely if your bets at -110 win less 52.38% of the time you’ll be losing money.

So in general, for a favorite offered at a line of F (F < -100), the implied probability of a win is given by -F/(100-F). (This is just the ratio of dollar value risked to dollar value returned. Remember that F is a negative number.)

The thought process is exactly the same for dogs. A dog offered at +200 corresponds to $300 being returned for every $100 risked, meaning that the dog’s implied probability is just $100/$300 = 1/3.

So in general, for a dog offered at line of D (D > +100), the implied probability of a win is given by 100/(100+D). (Once again -- the ratio of dollar value risked to dollar value returned.)

But up until now we’ve strictly been dealing with the zero-vig case (no juice), where (in the two-outcome scenario) a dog is offered at the negative of where the corresponding favorite is offered. Unfortunately, unless you’re either betting with friends or a “crossed-market” situation exists across multiple sportsbooks this is an uncommon occurrence.

Actually, given a line set there isn’t really a way to determine a single implied probability on the outcome of the event. This is because there’s no way that both sides of a vigged bet can simultaneously be zero expectation (the vig has to come from somewhere).

Therefore, to come up with meaningful results one typically makes the assumption that a player’s expected losses from betting on either side of the event are equivalent. One might argue that in most sports one could comfortably expect the loss on favorites to be greater than the loss on dogs. Of course this is an issue of preferences rather than of probability theory.

So let’s look at a line set of -120/+100. From the equations above we see that the zero-vig implied probability of the -120 line is just 120/(100+120) ≈ 54.55% and that the implied probability of the zero-vig +100 line is 100/(100+100) = 50%. Each of these individually represent what would be the implied probability were the line offered without vig.

Because we’re assuming an equal expectation on both sides, we know that the relative probability levels are going to remain constant. The total of the zero vig lines, 54.55%+50% equals 104.55% and so we divide each zero-vig line by 104.55% to give us the appropriate implied probability of either side of the composite line set:

Favorite probability = 54.55%/104.55% ≈ 52.17%
Dog probability = 50%/104.55% ≈ 47.83%

And of course, almost as if by magic, 52.17% + 47.83% = 100% total probability.

So in general, to determine the implied probability of a two-outcome line set, one first determines the zero-vig implied probability of each line:

[Romanowski]

Favorite zero-vig implied probability
= P(f) = -F/(100-F)

Dog zero-vig implied probability
= P(d) = 100/(100+D)

And then divides each probability by the total of the probabilities to determine the proper line set probability:

Line set favorite probability
P(f)
= -----------
P(f)+P(d)

Line set dog probability
P(d)
= -----------
P(f)+P(d)