(I realize that this is going to be very basic for many readers, but the idea is to start off slowly.)

Key to quantitatively analyzing a set of betting lines is determining the win probability implied by those lines. This is a simple task to accomplish 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.)

So up until now we’ve strictly been dealing with the zero-vig case, 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 (or, I suppose, you’re still insane enough to play at No-Juice Sportsbook), 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 fromsomewhere). 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. (To be sure this can at times be a rather specious assumption. One might well argue that in most sports one could comfortably expect the loss on favorites to besignificantlygreater than the loss on dogs. Of course this is an issue of preferences rather than of probability theory, and were bettors’ preferences ever to shift towards underdogs, one would have to draw different conclusions.)

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 therelativeprobability 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:

And then divides each probability by the total of the probabilities to determine the proper line set probability:Code:Favorite zero-vig implied probability = P(f) = -F/(100-F) Dog zero-vig implied probability = P(d) = 100/(100+D)

It's probably a good idea to note that this is NOT the same as just taking the implied probability of the average of the two absolute values of the constituent lines. In other words, using the -120/+100 example, it's NOT correct to say that the implied line set probability is just the implied probability of -110/+110. Determining implied line set probabilities using this method will overestimate the win probability of favorites. This becomes particularly apparent in the case of large favorites.Code:Line set favorite probability P(f) = ----------- P(f)+P(d) Line set dog probability P(d) = ----------- P(f)+P(d)

The same methodology can also be used when looking at multi-way propositions: Namely, determine each individual zero-vig probability, and then divide each by the total to get the implied probability for each outcome in the line set.