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Implied Probability - How does the bookmaker distributes his margin?

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  • Implied Probability - How does the bookmaker distributes his margin?

    You often see the odds proportional calculation: odds = payout rate / (implied) probability --> (implied) probability = payout rate / odds.
    This leads to equal bookmaker margins (= (1 / payout rate) - 1) and different risk buffers (= 1 / odds - (implied) probability) for all outcomes.

    An example: odds 1.2 / 5.0 --> payout rate = 1.2 * 5.0 / (1.2 + 5.0) = 6 / 6.2 = 0.96774 --> (implied) probabilities 0.80645 / 0.19355 --> expected values (= probability * odds) 0.96774 / 0.96774 --> margins 0.03333 / 0.03333 & risk buffers 0.02688 / 0.00645.

    Impact of an 2% probability error:
    A) +2% / -2% --> expected values: 0.99174 / 0.86774
    B) -2% / +2% --> expected values: 0.94374 / 1.06774
    --> Tipsters who constantly finds this kind of error would make a nice profit 6.774%.


    I never liked this because i don't understand why the bookmaker would use that much higher / lower risk buffers for smaller / bigger odds.
    Of course he'll get much more wagers on lower odds, but since odds are being calculated with the inverse of probability, the potential damage of a faulty probability assumption corresponds with the odds.


    Alternatively the equal risk buffer method: odds = 1 / (probability + (whole market margin / number of outcomes)) --> probability = (1 / odds) - (whole market margin / number of outcomes).
    This leads to different bookmaker margins and equal risk buffers for every outcome.

    In the example: odds 1.2 / 5.0 --> whole market margin = (1 / 0.96774) - 1 = 0.03333 --> (implied) probabilities 0.81667 / 0.18333 --> expected values (= probability * odds) 0.98 / 0.91667 --> margins 0.02 / 0.08333 --> risk buffers 0.01667 / 0.01667.

    Impact of an 2% probability error:
    A) +2% / -2% --> expected values: 1.00400 / 0.81667
    B) -2% / +2% --> expected values: 0.95600 / 1.01667
    --> Even a tipster who finds such error repeatedly won't get rich.


    So, if i would be a bookmaker, i wouldn't use the odds proportional calculation but the equal risk buffer method.

    Statistics suggests that bookmakers use rather the latter. Betting on all pinnacle sports closing odds leads to smaller losses with lower odds and far greater losses with bigger odds.
    Even laws of market economy suggests that the larger the turnover (-> lower odds vs higher odds) the smaller the margin.


    Practical use: Over the years i saw a few betting colleagues mourning that although they beat the pinnacle sports closing line by 5% or more they still were in the reds and how unlucky they felt.
    For me the reason wasn't a lack of luck but the explanation above. Beating a @6+ pinnacle sports closing line even by 5% just equals a probability difference < 0.8% and that's just not enough even for low margins / risk buffers pinnacle sports.

  • #2
    Hi! Interesting article. Are you using this method?

    Comment


    • #3
      Not directly, since for me the ratio best odds / pinnacle odds isn't the initial criteria for a bet, but i know some people who use this strategy.

      In slightly modified form i use this method to calculate 'expected best (playable) odds' from average odds in my database.

      This has some obvious disadvantages, since the actual best odds is almost never exactly the expected one, but there are some major advantages:
      - actual best odds << expected best odds -> value probably gone
      - actual best odds >> expected best odds -> probably missed some crucial (new) information
      - you can test strategies with old data & presume equal odds conditions even when the average margins has changed or the available past odds informations are incomplete / not that good

      Of course you don't have to use / start with the equal risk method to formulate expected best odds, but it's the best / most accurate one i know of.
      Last edited by hapax; 18-04-18, 13:39.

      Comment


      • #4
        Thanks for a reply bro. In your post you give example for a 2 outcomes. What about ?standart 1x2 outcome? Can you give some example and advice. Thanks.

        Comment


        • #5
          Liverpool - AS Roma pinnacle 1.549 / 4.54 / 6.24

          -> whole market margin = 1/1.549 + 1/4.54 + 1/6.24 - 1 = 0.0261 & number of outcomes = 3

          (fair) probability = (1 / odds) - (whole market margin / number of outcomes) & (fair) odds = 1 / (fair) probability

          -> pinnacle (fair) 63.688% / 21.156% / 15.156% (vs with equal margins 62.916% / 21.466% / 15.618%)
          -> pinnacle (fair) 1.570 / 4.727 / 6.598 (vs with equal margins 1.589 / 4.658 / 6.403)

          So in the case you think 'pinnacle knows best' would the expected value (odds * probability) for a Roma win @6.50 (bet365) be above or under 1 right now? The answer differs with the method you use for deducing fair probabilities from pinnacle odds.

          Comment


          • #6
            Originally posted by hapax View Post
            Not directly, since for me the ratio best odds / pinnacle odds isn't the initial criteria for a bet, but i know some people who use this strategy.

            In slightly modified form i use this method to calculate 'expected best (playable) odds' from average odds in my database.

            This has some obvious disadvantages, since the actual best odds is almost never exactly the expected one, but there are some major advantages:
            - actual best odds << expected best odds -> value probably gone
            - actual best odds >> expected best odds -> probably missed some crucial (new) information
            - you can test strategies with old data & presume equal odds conditions even when the average margins has changed or the available past odds informations are incomplete / not that good

            Of course you don't have to use / start with the equal risk method to formulate expected best odds, but it's the best / most accurate one i know of.
            Can you share with spreadhseet?

            Comment

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