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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.
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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.
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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.
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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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