1. #1
    matthewmsturgeon
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    Question about EV probability

    Hi guys,


    I am trying to get less causal about my bets. I need some help with calculating EV.


    I realize that I need to shop for the best lines, but for this example I am going to use the numbers from my local book. For comparison, I will also use the lines from Pinnacle as I understand they have some of the most solid lines.


    So if I look at this matchup:

    Pinnacle Break Even % No Vig %
    Diamondbacks -140 58.3% 57.1%
    Mets +129 43.6% 42.8%


    Local book Break Even % No Vig %
    Diamondbacks -143 58.8% 56.7%
    Mets +123 44.8% 43.2%


    Letís say that I want to research the Diamondbacks and that I personally feel like they have a 60% chance of winning (though I donít know how to figure that out in real life yet, feel free to give me some guidance there).


    EV Calculations
    So in calculating EV I need to multiply the probability of winning times the amount won - the probability of losing time the amount lost. The amounts come from the -143 from my local bookmaker, since that is who I am placing the bet with.


    So I have (P X 100) - ((1- P) X 143) I think.


    My question is whose probability should I use? Should I use the No Juice value from my local guy 56.7%, the No Juice value from Pinnacle since it should be more accurate 57.1%, or should I use my personal feeling of 60%?


    Thanks so much.

  2. #2
    Waterstpub87
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    If you want the most accurate price, use the pinnacle line. Tony Soprano probably has a less efficient line.

    If you wanted to measure how much of an edge you would have, you would take your probability and compare against the probability of where you are betting.

    Alternatively, you can look for place where your local is substantially different from pinnacle and bet those games.
    Last edited by Waterstpub87; 08-24-17 at 04:56 PM.
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  3. #3
    Bsims
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    I compute the expected return. First you need to convert the American line to the European version. In this case your local book's lines would be Arizona 1.70 NYM 2.23. Then multiply your estimate times these to get the expected return on each side.

    Arizona .6*1.70=$1.02 NY Mets=.4*2.23=$0.89

    The obvious wager would be Arizona with an expected return on $1.02. Personally, I don't wager on anything with expected return less than $1.05 or more comfortably $1.10. If the expected return goes above say $1.25, I'm leary of the wager. Obviously my calculations are for a game with something askew (maybe a star player out).
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  4. #4
    matthewmsturgeon
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    Thanks a lot guys. This helps. So to recap:

    Getting my probabilities from Pinnacle is probably a good idea since it is likely closest to reality.

    Multiplying the decimal odds * the probability can give me the expected return. I would assume that as described I could use my 60% if I felt good about it. Likewise I could use Pinnacle's 57.1% if I just wanted to stay closer to the experts.

    Question: I have seen this same formula basically used to describe Edge. (decimal odds * probability) - 1 expressed in %. Is that correct?

    Thanks again.

  5. #5
    thom321
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    In short, yes that is correct. Another way to think of it is to convert the decimal odds to an implied probability and compare to the actual probability. The actual probability in this case is whatever your winning model has predicted.

    So decimal odds of 1.70 equals an implied prob of 1/1.7 = 0.588235
    Actual probability 0.6
    Edge 0.6/0.588235 - 1 = 0.02 or a 2% edge.


    Quote Originally Posted by matthewmsturgeon View Post
    Thanks a lot guys. This helps. So to recap:

    Getting my probabilities from Pinnacle is probably a good idea since it is likely closest to reality.

    Multiplying the decimal odds * the probability can give me the expected return. I would assume that as described I could use my 60% if I felt good about it. Likewise I could use Pinnacle's 57.1% if I just wanted to stay closer to the experts.

    Question: I have seen this same formula basically used to describe Edge. (decimal odds * probability) - 1 expressed in %. Is that correct?

    Thanks again.

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