Simulate this count please!

positiveEV

Well-Known Member
2 = +4
3 = +5
4 = +6
5 = +8
6 = +5
7 = +3

Ignore 8

9 = -2
10,J,Q,K = -5
A = -6

Can anyone simulate it on 6-8 decks to see if it's good? I just made that by looking at the advantage and disadvantage of removing a card and then I rounded it at +/- 0.1%, so for each True Count of 10, this count should give you an advantage of 1%.

Thanks in advance :D
 

Canceler

Well-Known Member
Does not compute

You know this count is unbalanced, right? I'm not sure how True Count would work with an unbalanced count. Or would you like to provide the initial running count for 6D?
 

QFIT

Well-Known Member
This is very close to the Griffin 7 Count: 4, 4, 5, 7, 5, 3, 0, -2, -5. BC is a tiny bit higher, PE and IC are a tiny bit lower. There is only a tiny difference between these and UstonSS, a three-level count.



asiafever said:
2 = +4
3 = +5
4 = +6
5 = +8
6 = +5
7 = +3

Ignore 8

9 = -2
10,J,Q,K = -5
A = -6

Can anyone simulate it on 6-8 decks to see if it's good? I just made that by looking at the advantage and disadvantage of removing a card and then I rounded it at +/- 0.1%, so for each True Count of 10, this count should give you an advantage of 1%.

Thanks in advance :D
 
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