# Ace Tracking Methodology

Discussion in 'Skilled Play - Card Counting, Advanced Strategies' started by stophon, Mar 22, 2009.

1. ### stophonWell-Known Member

Alright so I'm one of those guys who likes to figure stuff out for myself, so here was my attempt at solving Ace Tracking. Please check my methodology.

An ace yields a 51% advantage if it is your first card.
Stophon shuffling data:
Clumps of one: 77.6%
Clumps of two: 20.0%
Clumps of three: 2.4%
Clumps of four or more: 0%

(This data matches up pretty well with some of the ranges Snyder gives on his site for shuffles. I ran 5 "trials" to get those numbers, I'm sure that in a casino you will get 4+ more often, however since the numbers will be very small it shouldn't affect betting that much I don't think.)

Possible ways a key card can occur
K-Key card, O-Other Card, A-Ace
Original Setup
O--------- O
K---- +---- O
A--------- O
O--------- O

Arrangement 1
K--------
----- O-- = P(1 card)*P(A&K not grouped)
A-------- = (.776)(.776) = 60.2%

Arrangement2
K----------
----- O---- = (P 2 cards)*(P K&A not grouped)
----- O---- =(.20)(.776) = 15.5%
A----------

Arrangement3
K --------- = P(grouped) = P(2 cards) + P(3 cards)
A----------= (.20)+(.024) = 22.4%

Arrangment4
K-----------
----- O---- =P(3 cards)*P(not grouped)
----- O---- = (.024)(.776) = 1.9%
----- O----
A----------

(It added up to 100%! But, that doesn't by any means prove that my math was correct. Please point out anything if it is wrong.)

So all of this means that the probability of:
k , A------------22.4%
k , O, A---------60.2%
k, O, O, A-------15.5%
k, O,O,O,A-------1.9%

So our betting plan to get a fully Kelly bet:
f* = Advantage * P(Ace in player's hand)
so if you see the key card come out as the last card then
f* = Advantage * (.602+.019)
f* = 0.51 * 0.621
f* = .317
So to place a full Kelly bet, you would play 31.7% of your bankroll. However this assumes that you will get the same distribution throughout each shuffle. Now how the hell do I acquire the standard deviation charts for a shuffle? (Edit: No it does not assume the same distribution through each shuffle. It just assumes the shuffler on average will drop the same percentage of 1, 2, and 3 card clumps. That good because it means figuring out my E.V. will be much easier.)

P.S. I haven't read tracking books yet, though I searched the forum for all it had and read a bunch of online articles.

2. ### Brock WindsorWell-Known Member

Data

Where does the 'Stophon shuffling data' come from? I'm not following where you are getting the 77% number from.
-BW

3. ### stophonWell-Known Member

My own shuffle trials. I know that is unreliable, but Snyder cites similar data:

"Anthony Curtis also obtained empirical data from four professional Las Vegas dealers to determine how they broke a deck for a riffle and how well their riffling mixed the cards. His data showed that a professional dealer interleaves a single card 66% of the time, 2 cards 26% of the time, 3 cards 5% of the time, 4 cards 2% of the time, and 5 or more cards less than 1% of the time. A previous empirical study of professional dealers published by Richard Epstein in his Theory of Gambling and Statistical Logic (Academic Press, 1977) showed a more thorough mixture, with single cards interleaving 80% of the time, 2 cards 18% of the time, and 3 or more cards only 2% of the time."

4. ### Brock WindsorWell-Known Member

Ok. I follow the theory, but don't plunk your bankroll behind your expected EV. In the real world you will get false keys (except in single deck), the ace will be stripped (especially in single deck), and you will see multiple riffles. All you're showing is the expectation a dealer will interweve a single card in one riffle. Im not sure why you square that number (77%). It means 77% of the time a key card will have a single card following it, and then the target card. (other factors being ignored of course).
BW

5. ### stophonWell-Known Member

I square .776 becuase it is both the probability that the A and the K will be in different clumps and the probability that the shuffler will only release one other card from the other grab.

And what does stripped mean? I also have to factor in the probability that the key card and the ace will get separated when the deck is cut to get my numbers as accurate as possible. And account for whatever this stripping is.

I should be able to use this data to make it into a formula for any number of rifles so that I can get more realistic data for a casino. That might take me a bit though.

Edit: Your right. .776 is not the probability that A&K will not be in the same clump. It's actually (1-0.224).
Edit2: I should be checked for retardation. 1-.224 is .776

6. ### stophonWell-Known Member

Alright I looked up the definition of strip on bj21.

On average, how many times is a SD riffled? A double (for each deck)?
And how many pieces is it divided into as it is stripped?

I realize there is great variance, but what would you estimate as a fair average.

7. ### taipafanMember

Do you know why there is no 0 card interleaved?

8. ### NazgulWell-Known Member

You didn't include the probability that the card underneath the ace has a 3 card clump.

10. ### stophonWell-Known Member

I think I included the probabilities that it will appear in either a 2 or 3 card clump.
K
A
-------O (20%)

or

K
A
O-------(2.4%)

I did assume only one rifle, however this data can continue to be replicated for as many rifles as you so choose (just requires more work, I'm working on a formula. From what I can tell it's a pretty intense quadratic). The formula for the chance your ace does not get lost due to stripping is:
(1-s/52)
Where s is the number of times the deck is cut + the number of chunks that the deck is divided into as it is stripped. In a real casino you can improve on this number if you know approximately where your key card is. However if you just want objective math probability, that's the formula.

Suppose the deck is riffled once, stripped down once into five pieces and cut at the end. You are playing heads up against the dealer and you do not spread to multiple hands.

5 strips + 1 cut = 6
1-6/52 = (46/52) chance the key card and the ace are not seperated

Mathematical EV
----------------
Event Player Adv Probability Expected Value
Player Ace +.51------ .549-------- .2799
Dealer Ace -.34------ .335-------- -.1139
Ace lost___-.01----- .115-------- -.00115
----------
Total .16485 = 16.5% Advantage

I assumed that if the ace was lost due to stripping the chances of it being in that clump of cards was now 0%. I know this isn't true but the odds of it showing back up shouldn't be mathematically significant. I use the hi-lo count, and at a count of -1 (ace lost) I have a 1% disadvantage. Again, not a perfect number (though hi-lo has a good betting correlation) but its fine for what we're doing.

You could also calculate how your EV changes for spreading to two hands. I'll work on getting the data for a more realistic number of riffles.