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Thread: Mike H: hypothetical count question

  1. #1
    Mike H
    Guest

    Mike H: hypothetical count question

    Let's say I wanted to count the ten as -2 and the jack, queen, and king as -4. If I ran a simulation with them all counted as -3.5 how much penalty should I subtract to approximate the way I'm actually counting the tens? What if I ran a sim with the red tens as -3 and the black tens as -4? Is there a way to figure this out or could someone give me a ballpark guess as to how much percent to reduce the WinRate and Score?

    Thanks for any help.

  2. #2
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: hypothetical count question

    Probably not a great deal. But, what is the point of using a difficult, high-level count and then compromising it? The point of a high-level count is increased accuracy. But counting tens differently reduces the accuracy.

    > Let's say I wanted to count the ten as -2 and the
    > jack, queen, and king as -4. If I ran a simulation
    > with them all counted as -3.5 how much penalty should
    > I subtract to approximate the way I'm actually
    > counting the tens? What if I ran a sim with the red
    > tens as -3 and the black tens as -4? Is there a way to
    > figure this out or could someone give me a ballpark
    > guess as to how much percent to reduce the WinRate and
    > Score?

    > Thanks for any help.

  3. #3
    Mike H
    Guest

    Mike H: Re: hypothetical count question

    Well actually I was planning on counting the Ten as -1 and J, Q, K as -2. This averages out to -1.75 so I multiplied by 2 for CVData to handle the fraction (-3.5). I generated the indices, divided by 2, and made a new count that counts red tens as -1.5 and black tens as -2. The performance so far looks pretty good, just slightly less than the -3.5 count. I was just wondering how much more accuracy would be lost from counting half the tens as -1.5 versus counting only the Ten as -1? Do you think it would be less than 1%?

    > Probably not a great deal. But, what is the point of
    > using a difficult, high-level count and then
    > compromising it? The point of a high-level count is
    > increased accuracy. But counting tens differently
    > reduces the accuracy.

  4. #4
    Mike H
    Guest

    Mike H: the hypothetical count

    Ace through T,J,Q,K...
    -1,1,1,2,2,1,1,0,0,-1,-2,-2,-2
    On paper this count works pretty darn good. Probably not very practical and not sure how much counting Tens this way would cost. An unbalanced version would count the six as +2.

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