Question on push probability from half point calculator

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  • xyz
    SBR Wise Guy
    • 02-14-08
    • 521

    #1
    Question on push probability from half point calculator
    I have a question about how the push probabilities are computed in the half point calculator. Are those based on historical data or some other formula? Thanks.
  • Ganchrow
    SBR Hall of Famer
    • 08-28-05
    • 5011

    #2
    Originally posted by xyz
    I have a question about how the push probabilities are computed in the half point calculator. Are those based on historical data or some other formula? Thanks.
    The figures are based on smoothed historical data.
    Comment
    • xyz
      SBR Wise Guy
      • 02-14-08
      • 521

      #3
      Originally posted by Ganchrow
      The figures are based on smoothed historical data.
      Thanks for the quick response. I plan to keep track of the following way in using the push probabilities from the half point calculator:

      Leaving the middle open. For example, today's NBA game between Detroit and Milwaukee. I would bet on +112 Milwaukee +5.5 and +108 Detroit -6.5. This would leave the middle open at Detroit winning by 6.

      Using the push probability from the half point calculator, the probability of Detroit winning by 6 is 4.16%. Using those odds, the expectation of leaving the middle open would be
      +0.15%. Of course we would like a number bigger than this. But does the idea sound logical to you? Thanks.
      Comment
      • Ganchrow
        SBR Hall of Famer
        • 08-28-05
        • 5011

        #4
        Originally posted by xyz
        Thanks for the quick response. I plan to keep track of the following way in using the push probabilities from the half point calculator:

        Leaving the middle open. For example, today's NBA game between Detroit and Milwaukee. I would bet on +112 Milwaukee +5.5 and +108 Detroit -6.5. This would leave the middle open at Detroit winning by 6.

        Using the push probability from the half point calculator, the probability of Detroit winning by 6 is 4.16%. Using those odds, the expectation of leaving the middle open would be
        +0.15%. Of course we would like a number bigger than this. But does the idea sound logical to you? Thanks.
        I'm getting a different result than you.

        Risk 1 unit at +108.
        Risk 2.08/2.12 ≈ 0.9811 units at +112.

        - Win 1.08 - 0.9811 = 0.09887 units with probability 1-4.16% ≈ 95.84%.
        - Lose 1.9811 units with probability 4.16%.

        This yields an expectation of 95.84%*0.09887-4.16%*1.9811 ≈ 0.01234 units or 0.01234/1.9811 ≈ 0.6229%.

        That said, this is a pretty thin expectation and probably only a shade better than the margin of error. As such I'd be sure to tread cautiously.
        Comment
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