In Ken Uston's classic work Million Dollar Blackjack, the author proposes some revolutionary methods of getting the advantage over the house. One of these methods is playing multiple hands in negative counts(aka card eating effect).
On page 153, Uston states as follows:
"You've just bet 4 greens and 2 reds and won the hand. The count has dropped. Spread to 2 hands of 2 red each-or even 3 hands. You've cut your bet way down and your eating up cards in this negative situation. Again, the dealer is pushed more rapidly toward the shuffle and you will tend to be dealt fewer negative hands."
I can see how playing multiple hands per round will reach the shuffle card faster. What I do NOT understand is how
the player will be dealt fewer negative hands for the shoe.
In anything, wouldnt he be dealt MORE negative hands for the shoe by playing multiple spots in negative counts?
Here is my reasoning:
Suppose a counter is playing heads up in a 6D shoe game. There are 72 cards left until the shuffle card comes out. Assuming roughly 3 cards are used per hand, playing one spot would require 12 rounds to reach the shuffle card(3 cards/player + 3 cards/dealer = 6 cards/round). So, the player would be dealt 36 cards playing one spot/round or 12 hands.
Now, what if the player decided to play 3 spots in this negative count? Then it would require 6 rounds to reach the shuffle card(1 round = 3 cards/dealer + 3 cards per player hand x 3 hands = 12 cards/round). In the end, the player would be dealt 54 cards playing 3 spots/round or 18 hands.
So, how can Uston conclude playing multiple spots in a negative count will lead you to be dealt fewer negative hands? 18 hands is more then 12 hands!!! Is there really any advantage to this approach? Thanks for any answers.
-MJ
On page 153, Uston states as follows:
"You've just bet 4 greens and 2 reds and won the hand. The count has dropped. Spread to 2 hands of 2 red each-or even 3 hands. You've cut your bet way down and your eating up cards in this negative situation. Again, the dealer is pushed more rapidly toward the shuffle and you will tend to be dealt fewer negative hands."
I can see how playing multiple hands per round will reach the shuffle card faster. What I do NOT understand is how
the player will be dealt fewer negative hands for the shoe.
In anything, wouldnt he be dealt MORE negative hands for the shoe by playing multiple spots in negative counts?
Here is my reasoning:
Suppose a counter is playing heads up in a 6D shoe game. There are 72 cards left until the shuffle card comes out. Assuming roughly 3 cards are used per hand, playing one spot would require 12 rounds to reach the shuffle card(3 cards/player + 3 cards/dealer = 6 cards/round). So, the player would be dealt 36 cards playing one spot/round or 12 hands.
Now, what if the player decided to play 3 spots in this negative count? Then it would require 6 rounds to reach the shuffle card(1 round = 3 cards/dealer + 3 cards per player hand x 3 hands = 12 cards/round). In the end, the player would be dealt 54 cards playing 3 spots/round or 18 hands.
So, how can Uston conclude playing multiple spots in a negative count will lead you to be dealt fewer negative hands? 18 hands is more then 12 hands!!! Is there really any advantage to this approach? Thanks for any answers.
-MJ
