I've played a bit of blackjack at casinos now and then, just using basic strategy, and I've always won a good bit of money. That's just pure chance, though, because I wasn't using any advantage play. I've always been a little intimidated by the considerable focus required in Hi-Lo card counting, but I wanted to increase my chances a bit, so I decided to play around with some numbers and see if I could come up with a simpler counting system that wouldn't be hard to start trying, even if it didn't offer as consistent an advantage.

Adjusting the count for every single card that comes out of the shoe seems daunting. Instead, I want to try a basic ace count and use that as the basis for my odds. Aces are dealt in about 19% of hands, but they're the single-largest contributor to the chances of being dealt a blackjack (which is in turn the single-largest advantage I have playing against the casino).

In an 8-deck shoe, the casino has an advantage of about 0.65% over a player using basic strategy. The chances of any given hand being a blackjack are 4.7451%, meaning that the 3:2 payout gives the player a 2.37255% advantage. So without the 3:2 payout of a blackjack, the casino would have a 3.02% advantage over the player.

On average, each hand is 2.7 cards, so the chances of an ace being dealt are about 19% per hand. However, each hand that's dealt without an ace in it increases the chances of a blackjack being dealt, because the proportion of aces in the deck has increased. For example, after one hand is dealt without an ace, the shoe still has 32 aces in it but only 413.3 cards, meaning that the chances of a blackjack (2*aces*faces/(cards*cards-1)) go up to 2*32*127.17/(413.3*412.3) = 4.7762%. This is an increase to the player odds of 0.0156%. The odds go up exponentially, but it takes 33 aceless hands before the casino advantage is eliminated:

2*32*100.58/(326.9*325.9) = 6.0424%

With the 3:2 payout, that gives the player a 3.02% advantage, balancing out the casino's advantage.

On the flip side, each dealt ace does the opposite thing: decreasing the population of aces in the shoe and thus decreasing the chances of a blackjack. The odds of a dealt blackjack with one ace missing from the deck after the first hand drop to 2*31*127.48/(413.3*412.3) = 4.638%, decreasing the player blackjack advantage by 0.0535% and thus increasing the casino's net advantage to 0.7035%.

The decrease-in-odds from a dealt ace is 3.44x the increase-in-odds from an aceless hand.

Taken together, this suggests that I should start the shoe with a count of 33 and decrement by 1 for each hand dealt. I should add 3.44 for each ace. If I do this, then I'll have an net advantage whenever the count goes negative.

Of course, adding 3.44 would be very painful, so instead I'll just add 3. But if a blackjack is dealt, I'll add 6. A blackjack decreases the odds of a subsequent blackjack to 2*31*127/(414*413) = 4.6052, decreasing player advantage by 0.07%, 4.49x the increase in odds from an aceless hand. By rounding a blackjack up to +6 and an ace down to +3, it should come out nearly even (aces occur about 4x as often as blackjacks, so those four extra 0.44s in each ace will be roughly balanced out by the extra 1.51 I add to each blackjack).

So, all that math aside, the system itself is pretty simple.

Play basic strategy. Start the shoe with a count of 33 and count down for each hand dealt. Add 3 for each ace you see and 6 for each blackjack dealt. Increase your bet whenever the count is negative. If the count ever exceeds 80, there is very little chance you will recover an advantage before the end of the shoe and you should wong out.

Here's a graph showing the calculated odds with respect to the running count (if I've figured it all correctly):

The system seems easy to remember and easy to implement. It should be fairly simple to count down in my head as each new hand is dealt; I can do that as soon as bets are placed. I don't have to watch for face cards or for low cards, and I don't even have to keep any particular eye out for blackjacks because they will always be announced. I only need to scan the table once after hands are dealt to check for aces and then watch for any additional aces dealt during that play. There's never any significant mental math; just counting backward and adding by 3s and occasionally 6s, and I know exactly when I have an advantage. One thing I don't know is what kind of odds this system confers as a whole; I'd probably have to use some kind of simulator in order to figure that out.

So yeah, that's about it! I guess I'm just curious as to whether I've made any significant mistakes or forgotten to account for some large variable. Thoughts?

## Quick and dirty card counting system

**Moderators:** gmalivuk, Moderators General, Prelates

- sevenperforce
**Posts:**658**Joined:**Wed Feb 04, 2015 8:01 am UTC

### Re: Quick and dirty card counting system

Upon reading this, I wanted to answer the question "How frequently will the count go negative?" Even if this count is reliable, it won't be useful unless it goes negative often enough (for example I could declare that if all 24 aces were dealt off the top of the deck, this would be an extremely reliable indication that this deck will favor the house. It would also basically never happen, so though it is informative when it happens, it won't happen often enough to be useful).

To get a quick answer, I wrote a simplified sim to implement the count.

Here's what I do:

- deal two cards

- If there is a blackjack, add 6

- elseif there is an ace, add 3 (keeping in mind that they might both be aces)

- otherwise subtract 1

Discard the 2 cards and deal 2 more, until you get through the whole deck.

Since any play that doesn't involve aces and blackjacks doesn't affect the count, you don't need to model any of it.

What I find is that the count ALWAYS goes negative about halfway through the shoe (typically between 50 and 150 hands in). This means the count predicts that the back end of the shoe always has an advantage for the player. This doesn't seem to be a correct conclusion.

I can improve my sim to be more complete. Have you done any investigation to how frequently the count goes negative?

EDIT: I figured out my problem. By explicitly treating hands as 2 cards, I was over weighting the decrementer. I fixed that up and I find that the count will go negative around 18% of the time.

To get a quick answer, I wrote a simplified sim to implement the count.

Here's what I do:

- deal two cards

- If there is a blackjack, add 6

- elseif there is an ace, add 3 (keeping in mind that they might both be aces)

- otherwise subtract 1

Discard the 2 cards and deal 2 more, until you get through the whole deck.

Since any play that doesn't involve aces and blackjacks doesn't affect the count, you don't need to model any of it.

What I find is that the count ALWAYS goes negative about halfway through the shoe (typically between 50 and 150 hands in). This means the count predicts that the back end of the shoe always has an advantage for the player. This doesn't seem to be a correct conclusion.

I can improve my sim to be more complete. Have you done any investigation to how frequently the count goes negative?

EDIT: I figured out my problem. By explicitly treating hands as 2 cards, I was over weighting the decrementer. I fixed that up and I find that the count will go negative around 18% of the time.

- sevenperforce
**Posts:**658**Joined:**Wed Feb 04, 2015 8:01 am UTC

### Re: Quick and dirty card counting system

18% doesn't sound too bad, though in practice I imagine that you wouldn't be able to use a negative count close to the end of the shoe, and that's probably where the count would be most likely to go negative. I counted through a dozen or so shoes and only hit a negative count three times, and only for one or two hands.

I wonder if a better approach would be to use the total number of cards dealt divided by the number of hands? The average hand has 2.7 cards, so if you assume most people are playing something close to basic strategy, a greater number of cards per hand means that more low cards are being dealt, meaning there are a greater number of high cards left in the deck.

Something like "subtract 1 for each hand played; add 1 for each hit that doesn't result in a bust" might give a simple but useful count.

I wonder if a better approach would be to use the total number of cards dealt divided by the number of hands? The average hand has 2.7 cards, so if you assume most people are playing something close to basic strategy, a greater number of cards per hand means that more low cards are being dealt, meaning there are a greater number of high cards left in the deck.

Something like "subtract 1 for each hand played; add 1 for each hit that doesn't result in a bust" might give a simple but useful count.

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