| О доказательствах. (Пример одного.) ID:49971 ответ на 49970 |
Чт, 27 июля 2000 15:23 [#] |
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| Garry Baldy |
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Форумы ABC-casino
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Копирую статью, написанную автором нескольких несбалансированных систем. Возможно, это не совсем то, что хотел Пан Вотруба, но дает прозрачное представление о том, как работают несбалансированные системы и как их можно превратить в сбалансированные.
Переводить, гадом быть, ломает.
Удачи.
Garry Baldy.
Unbalanced True Count Proof
by Brett Harris.
Here is how unbalanced systems work : I will use K-O as an example.
Call the 13 card values C(i), then the unbalance per deck is equal to
U = 4* Sum(i=1,13) C(i) = +4 for K-O
Now define the Equivalent Balanced Count B(i) by
B(i) = C(i) — U/52 for each card i=1,13.
For K-O,
B(i) = (-1.077,+0.923,+0.923,+0.923,+0.923,+0.923,+0.923, -0.077,-0.077,-1.077)
for Ace thru Ten
Notice the B(i) is a balanced count, with Sum(i=1,13) B(i) = 0.
Since B(i) is a balanced count, we define the running count as
Rb(n) = Sum{n cards dealt} B(i),
and the true count
Tb(n) = Rb(n) / (N — d), where N is the total number of decks and d is the number which have been dealt (d=n/52)
But by the definition of B(i), we have
Rb(n) = Sum{n cards dealt) [ C(i) — U/52 ]
= [ Sum{n cards dealt} C(i) ] — (U/52)*n
= Rc(n) — U * d, where Rc(n) is the unbalanced running count
So therefore Tb(n) = [ Rc(n) — U*d ] / (N — d)
= [ Rc(n) — N*U + (N — d)*U ] / (N — d)
= [ Rc(n) — N*U ] / (N — d) + U (*)
This equation explains how unbalanced counts work, notice that if the unbalanced running count Rc(n) is equal to N*U then the true count Tb(n) is always equal to U, no matter how many decks (d) have been dealt. The pivot is defined as
pivot = N*U [ = Number of decks * 4 for K-O ],
The equivalent balanced count B(i) for K-O ( or Red7 for that matter ) is very close to Hi-Lo ( actually a little better ) and so when the K-O running count equals the pivot, one is close to a Hi-Lo TC=+4.
Actually equation (*) shows a little known property of ubalanced counting systems. Define an unbalanced true count
Tc(n) = [ Rc(n) — pivot ] / (N — d), then
Tc(n) = Tb(n) — U.
Or to put it another way, if you start your running count at -pivot instead of zero, then it is possible to compute a true count for the unbalanced system, and this true count is always different from the equivalent balanced count by U, and so this may be used in conjuction with published indices for Hi-Lo to use the unbalanced system in the same way as for a balanced system. Note though that the neutral deck no longer has a true count of zero, it has a true count of -U.
Cheers,
Brett.
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