CSMs

RJT

Well-Known Member
#1
Is there currently any information out there on beating CSMs? Even just on the basics? I know that some of the earlier models could be beaten, how about the new ones?
Thanks in advance for any replys.

RJT.
 

Sonny

Well-Known Member
#2
There isn't m uch information on the new ones, but here is a good place to start:

http://www.blackjackinfo.com/bb/showthread.php?t=2312
http://www.blackjackinfo.com/bb/showthread.php?t=2629

The systems that I am aware of either rely on clocking the latency of redistribution or finding a physical anomaly with the machine’s internal mechanics. I personally have not had any success with CSMs and I haven’t pursued it much farther since there are plenty of other games to play in my area. In any case, I believe the edge would be small but the camo would be great.

-Sonny-
 
#3
Here's another link to a classic must read for all AP's- Clarke Cant's "Blackjack Therapy"!

http://www.bjrnet.com/archive/BlackjackTherapy.htm

I wouldn't take his health advice (that's Zen Zone material!) but his BJ theory and inspirational words seem very worthwhile. Chapter 9 is about beating CSM's. His theory applies more or less to all CSM's. How much of an edge you can get with each machine depends on it's specific workings and how it is used.
 

dacium

Well-Known Member
#4
I read that blackjack therapy thing and he seems a little crazy o_O

He talks about CSM with no know latency (direct shuffle after each hand of all cards before the CSM spits any out).

He says that the normal distribution of cards into a random shuffle will provide an average latency before a card is re-dealt. He says that if you count out 3 hands then this count can be divided by so much and that is the true count for next so many hands.

He says that if 52 cards come out of the CSM and you count +10. Then these cards a distributed evenly over 5 decks in the CSM resulting in +2 to each deck so that an advantage exists. But this is totally wrong if the 1 deck is +10 then the average 5 left are at -2 so the shuffle brings everything back to zero.
 

RJT

Well-Known Member
#5
Just to say, thanks for the replys folks. That's exactly what i was looking for - in the process of sifting through it now.

RJT.
 
#6
Hope someone could confirm about these:

For a “one2six” machine, using 4 decks of cards, there are about 40 seats in the wheel. Each seat, it is empty, or some cards inside, usually less than 20 cards.

When a card (card#1) is inserted, the wheel turns randomly, one seat(seat#1) is ready for card#1, then card#1 will go to the bottom of that pack of cards in seat#1.
When card#2 is inserted, same thing happens. If the cards in seat#1 did not go to the exit, there is 1/40 of chance that card#2 will be placed at the bottom of card#1.

Here are my questions:

1. How many seats are there in the wheel?

2. There must be a (programmed) maximum number of cards that one seat can take, how many?

3. When a card is inserted into a seat, will it insert in the middle (or at the top) of that pack of cards? Or it will only be put at the bottom?

Thanks
 
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