analyzing a set of lines & determining the win probability implied by those lines

[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-300, then this of course means that for every $300 risked $400 is returned to the bettor if the wager is successful. In other words the amount risked is 3/4 of the total amount returned & 3/4 is also the implied probability of the -300 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.) So, by your own analysis, this bet must win > 3/4 to to be +EV.

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.

This however is a zero vig scenario...

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.)

However above 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. In other words its not likely to find fav -150 and dog +150

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.

To come up with real result you makes the assumption that a player’s expected losses from betting on either side of the event are equivalent.

So let’s look at a line set of -110/-110. From the equations above we see that the zero-vig implied probability of the -110 line is just 110/(100+110) ≈ 52.4% Each of these individually represent what would be the implied probability if the line were 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, 52.4%+52.4% equals 104.8% 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 = 52.4%/104.8% = 50%
Dog probability = 52.4%/104.8% = 50%

And of course, almost as if by magic, 50% + 50% = 100% total probability for -110/-110 line set

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:

This can be used for what some call an efficient market in Pinnacle

lets look at this saturday's final four matchup between Duke and WVa

current odds at Pinnacle Duke -2.5 -105 and Wva +2.5 +103

the zero-vig implied probability is:

Duke covers -2.5 ... 105/(100+105) =51.2/(51.2+50.7) ≈50.2%

WVa covers +2.5 ....103/(100+103) =50.7/(51.2+50.7) ≈49.8%

[Romanowski]

and of course Pinnacle just moved to -2.5 -101/+2.5 -105 lol

but anyways, by your own analysis, your edge must exceed what the books implied probability is

the variance of course plays a bigger part with MLs and sports such as baseball and hockey

[uva3021]

Very interesting to see the vig on a line laid out in percentage form in order to see the advantage. Moneylines are a different animal in some cases though. By degrees as moneylines become progressively larger, so does the vig yes? So the actual implied probability appropriated by the line would be shaded considerably toward the team favored to win as the moneyline rises.

[Romanowski]

Originally Posted by uva3021

So the actual implied probability appropriated by the line would be shaded considerably toward the team favored to win as the moneyline rises.

true, but the same theory is applied

take for example

-1100/+800 (a live line I saw for the Duke/Butler game just as Duke appeared to be pulling away in the 2nd half up 7 or so)

what are the implied probabilities there?

Butlers is 1/8 or 12.5 % in a zero vig vacuum or to break even

Dukes 11/(11+1) or 91.7 % to break even

since 12.5% > (100-91.7), Butler had the value

but as we know no value in losing :)

[IrishTim]

If the line is -1100/+800 is the no-vig line not -825? The implied win probabilities are as follows:

Duke: 1100/(1100+1000)= 91.66%
Butler: 100/(100+800) = 11.11%

91.66 + 11.11 = 102.77%

Obviously it is impossible to be greater than 100%, so the "overround" or the book's theoretical hold (theoretical meaning perfectly balanced action) is 2.77%. To adjust for the vig and get the true win probability you divide each implied win percentage by the sum of the implied win percentages including overround:

91.66/102.77 = 89.19%
11.11/102.77 = 10.81%

Then when you add these back up you get exactly 100%.

[Romanowski]

thats right, wasnt taking into account the hold %

[IrishTim]

All good. Everyone makes careless errors. I mistakenly added an extra period into one of the cells on my excel spreadsheet last week = 20k samples with incorrect data.

[Seanie Mac]

Irish Tim coming over to cappersmall

[IrishTim]

I was cordially invited, Seanie. Didn't even know this forum existed. Do you post under a different name at one of the other forums?

[Seanie Mac]

I do

[Seanie Mac]

rome soliciting sharp posters I like it.

will get my dictionary back out if a discussion gets going

[Romanowski]

Originally Posted by IrishTim

If the line is -1100/+800 is the no-vig line not -825? The implied win probabilities are as follows:

Duke: 1100/(1100+1000)= 91.66%
Butler: 100/(100+800) = 11.11%

91.66 + 11.11 = 102.77%

Obviously it is impossible to be greater than 100%, so the "overround" or the book's theoretical hold (theoretical meaning perfectly balanced action) is 2.77%. To adjust for the vig and get the true win probability you divide each implied win percentage by the sum of the implied win percentages including overround:

91.66/102.77 = 89.19%
11.11/102.77 = 10.81%

Then when you add these back up you get exactly 100%.

Originally Posted by Romanowski

true, but the same theory is applied

take for example

-1100/+800 (a live line I saw for the Duke/Butler game just as Duke appeared to be pulling away in the 2nd half up 7 or so)

what are the implied probabilities there?

Butlers is 1/8 or 12.5 % in a zero vig vacuum or to break even

Dukes 11/(11+1) or 91.7 % to break even

since 12.5% > (100-91.7), Butler had the value

but as we know no value in losing :)

just so things dont get confusing here

Tim corrected my example for the Duke/Bulter live line, not accounting for the overround

however the example in the original post is correct, and that is how you can find the zero vig line

from that point you can look further, to get the theoretical hold

I went into more detail on the hold % in this thread Expectations and Theoretical hold

[Ganchrow]

Originally Posted by 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-300, then this of course means that for every $300 risked $400 is returned to the bettor if the wager is successful. In other words the amount risked is 3/4 of the total amount returned & 3/4 is also the implied probability of the -300 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.) So, by your own analysis, this bet must win > 3/4 to to be +EV.

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.

This however is a zero vig scenario...

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.)

However above 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. In other words its not likely to find fav -150 and dog +150

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.

To come up with real result you makes the assumption that a player’s expected losses from betting on either side of the event are equivalent.

So let’s look at a line set of -110/-110. From the equations above we see that the zero-vig implied probability of the -110 line is just 110/(100+110) ≈ 52.4% Each of these individually represent what would be the implied probability if the line were 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, 52.4%+52.4% equals 104.8% 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 = 52.4%/104.8% = 50%
Dog probability = 52.4%/104.8% = 50%

And of course, almost as if by magic, 50% + 50% = 100% total probability for -110/-110 line set

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:

-snip-

I suppose I should be flattered that the bulk of this was lifted virtually word-for-word from my SBR post of June 2006 entitled "An introduction to betting lines and percentages".

See http://forum.sbrforum.com/handicappe...rcentages.html.

No worries, but I would appreciate an attribution next time. :)

[Romanowski]

same deal ganch