
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.

Norm Wattenberger: Re: hypothetical count question
Probably not a great deal. But, what is the point of using a difficult, highlevel count and then compromising it? The point of a highlevel 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.

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, highlevel count and then
> compromising it? The point of a highlevel count is
> increased accuracy. But counting tens differently
> reduces the accuracy.

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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