'Ace Only' Count System

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matt21

Well-Known Member
#1
#1
Hi everyone,
I was considering whether a side-count of aces alone is sufficient to beat a six deck shoe game with 75% penetration & 1-8 bet ramp.

In other words, the only cards I am tracking is the aces that have come into play so I am looking at either an ace-rich or an ace-poor deck. Assuming that all non-ace cards are random, would the ace-side count on its own be good enough to provide a +EV?

Possibly somebody has looked into this before?

Would be great to hear your thoughts- thanks in advance!

Matt21
 
M

MangoJ

Well-Known Member
#2
#2
An Ace as first card gives you ~50% EV. Hence a non-Ace gives you -4.7% EV on average (for an overall -0.5% EV of the game). You want to play an Ace with probability of p >= 8.6% to break even.
If you count Aces with -12, and all other cards with +1, you look for TC >~ +6 to play. (a TC=6 gives you 4 ace in 46 cards)
 
FLASH1296

FLASH1296

Well-Known Member
#3
#3
The answer is NO. Fuggedaboudit. Not even remotely possible.

It is not possible, not even with a Single Deck.

A Single Deck game — with good rules and pen' —

can be beaten with the Ace-Five Count; the oldest count extant.
 
M

matt21

Well-Known Member
#4
#4
MangoJ said:
An Ace as first card gives you ~50% EV. Hence a non-Ace gives you -4.7% EV on average (for an overall -0.5% EV of the game). You want to play an Ace with probability of p >= 8.6% to break even.
If you count Aces with -12, and all other cards with +1, you look for TC >~ +6 to play. (a TC=6 gives you 4 ace in 46 cards)
Click to expand...
right. I guess i could input this counting system into Casino Verite, assign a bet ramp and see what return it comes up with.

How did you arrive at the 8.6% figure?

Practically, rather than add -1 to the running count for each non-ace card, could I just count the aces remaining, so start with 24 at the beginning of the shoe and then divide by the number of decks/half-decks remaining?
 
M

MangoJ

Well-Known Member
#5
#5
matt21 said:
How did you arrive at the 8.6% figure?
Click to expand...
Just a rough estimate on the numbers derived from above. If you fetch 8.6% times an Ace (giving you 50% EV), and 1-8.6% times a Non-Ace (costing you 4.7% EV), you just break even.

Although this counting system is very very basic (and misses the 5), could you please post the numbers from CV ?
 
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