Is a double deck game with good rules but only 50% penetration beatable?

Is a standard vegas double deck game with good rules but only 50% penetration beatable? If so what kind of minimum bet spreads need to be used? Any tricks to get the cut card lowered to around 75%?
Thanks.

Remember, it's not just the pen. The MAIN thing is the number of betting opportunities per shuffle. Ergo, you want to play with as few other players as possible -- ideally zero other players.

"The MAIN thing is the number of betting opportunities per shuffle."

There's a concept I've not heard before. :)

I've always felt penetration was the main thing. Considering your statement above, heads-up 50% 8D would give you even more 'betting opportunities per shuffle' and I'm thinking, generally, I want no part of that.

While 50% pen DD may be beatable heads-up, I think it is a tough beat.

In DD, playing head-up or with an absolutle max of 2 other players, you're going to see TCs of +2, +3, +4, +5 etc. much more often than you will with, for example, 8D even with good pen.

To illustrate, the same 26 cards come off the top of the shoe giving a hi-lo RC of +6. In 8D, you've got to divide that +6 by 7.5 to get the TC of a little less than +1. In DD, you divide by 1.5 to get a TC of +4.

With too many players at the DD table, that +4 TC will probably last only for that round, especially at a full table where you will only get 2 or 3 rounds per shuffle. But playing head up, using an average of 5.5 cards per round, you get 8 or 9 rounds/shuffle (or at least 6 if they're using RO6).

Is it hard to beat? Yes. But I find it easier to beat than 6D & 8D even with good pens, because the count seems to run much more even (far fewer extreme counts) the more decks there are in the shoe.

Proportionately more favorable betting opportunities per shuffle is the key, and pen is only one of the factors governing that number.

...you're going to see TCs of +2, +3, +4, +5 etc. much more often than you will with, for example, 8D...

Not sure I agree. For example: Let's say we compare 8D with 1.5 decks cut off and 2D with 1 deck cut off. We know that when an 8D pack gets to +5TC, on average it will stay there until the cut card/reshuffle comes. That also holds true for 2D.

Therefore, 8D X 10 shoes would equal 65 decks actually played. Whereas, 2D X 40 packs of 2D would only yeild 40 decks actually played.

My contention in the above scenerio is that a player will yeild more hands played at +5 on the 8D rather than the 2D. Therefore, pen is certainly an important factor.

> We know that when an 8D pack gets to +5TC, on average it will stay there until the cut card/reshuffle comes. <

If that were literally true, then that +5 would give you no advantage. It's not enough that the high cards are in there -- they've also got to come out to do you any good.

> That also holds true for 2D.<

Not really -- or should I say not nearly as much. A +5 TC in DD almost always means you are going to see many high cards THIS round -- which is much to your benefit, but which also drive the count down substantially.

Also, I repeat my purely emperical observation from over 30 years of play that the more decks in the shoe, the less often extreme counts tend to occur. Don't get me wrong -- I have seen VERY extreme counts in multi-deck shoe games -- the one that comes to mind was the 4-deck game at the old MGM Grand when the RC was -23 and there were only 23 cards left undealt [what's that -- a TC of -50?]. But my experience is that there are proportionately MANY more betting opportunities in DD than in 6+ deck games.

If that were literally true, then that +5 would give you no advantage. It's not enough that the high cards are in there -- they've also got to come out to do you any good.

Why do you say it would give no advantage? +5TC is +5TC. It doesn't have to dump back to 0 for you to be getting on average your +5 share of the big cards. If it dumps back to 0, all players and the D are getting more big cards than the average.

Once the count gets to what ever it happens to get to, it will on average stay at that TC to the end. If you pull a card from anywhere in the remainder of the 6D pack at the time of a +5TC, it will be respective of what typically comes from a +5TC pack, regardless of whether it's 6D, 2D 1D or 44D.

We have to assume an even distribution because we cannot know differently. And, on average, it is even. But in actuality, it is SELDOM even. All those UNeven distributions added up and meaned is what gives the even average.

Ergo, ON AVERAGE, the TC will stay at +5 as you have stated.

BUT, the mathematics also tell us that the more extreme the count (i.e., the more extreme the deviation from the mean), the more likely it becomes that cards will come out which move us back closer to the mean.

This effect is much more readily observed in SD and DD than in 6D & 8D play because the count is so much less diluted (i.e., there are so many fewer cards available which can cause the return to the mean).

Thus, a TC of +5 in SD or DD will have a faster return to the mean (i.e., high cards coming out quickly) than it will 6D & 8D games. Thus, the TC of +5 won't last as long, but the benefit is greater and more immediate.

BUT, the mathematics also tell us that the more extreme the count (i.e., the more extreme the deviation from the mean), the more likely it becomes that cards will come out which move us back closer to the mean.

I used to think the same way until Eliot Jacobson taught me different. I even played 1st base because I thought the count was more accurate there and that it would give me better odds of getting the big cards. I was wrong. On average, the RC will return back toward 0. Whereas, the TC, on average will stay the same to the end. At +5TC all players including the D will get more than average amount of big cards all respective of a +5TC all the way to the last few cards.

Thus, a TC of +5 in SD or DD will have a faster return to the mean (i.e., high cards coming out quickly) than it will 6D & 8D games.

This not true. Why do you think that a more-than-average-amount of high cards are more apt to fall soon after the count rises to a high point? If the TC gets to +5, there is just as much chance of the TC getting to +7 as it does to +3. Remember, a +5TC is randomly scatered throughout the remaining pack.

For instance; Let's say the TC reaches +5TC after about the first 52 cards into a 6D pack. The cut card has be placed about 52 cards in from the rear. Therefore, we now have approximately 4 decks worth of cards left to play with. Subject to how many people are on the table, you are now going to play with 4 decks worth of cards having an average count of +5TC until the end. Therefore, I personally would rather play 6D with good pen rather than 1 or 2D with 50% pen. I believe this makes for good math. However, I would like to here from some other veteran players and especially the math wizards.

the number of cards dealt/in-play. In a DD game, dealt 75%, whatever happens is going to happen over 78 cards. As opposed to a shoe dealt to 1 deck left which is going to be played for over 3X as long.

In a DD game, the count goes up and down much more frequently, which I think was his point. Clearly the DD game is a better game in terms of hourly win rate assuming penetration is decent.

"For instance; Let's say the TC reaches +5TC after about the first 52 cards into a 6D pack."

You've chosen an extreme scenario for your example. You're talking about an RC of +25 after only 52 cards have been played. You're certainly correct to believe that a TC of +5 for the duration of the four remaining decks is a fantastic scenario, but I'm not sure it's fair or accurate to use it in comparison to the 2-deck game. A +5 TC at such an early point in the 6-deck shoe is incredibly uncommon.

On a more positive note, what little I understand of the True Count Theorem leads me to believe that your assertion concerning "enduring" TCs is correct.

A simple comparison between games can be made by visiting the site bjstats.com

I'd suggest you go to this site and plug in a wide variety of combinations of rules & pen & bankroll etc. to see the impact of varying the various parameters.

Here are some examples of scenarios you might want to explore to answer questions of the type you have raised. Assume optimal betting ramps, Hi-Lo, I 18 and Fab 4 Surr for all:

6D, mediocre rules (H17, DAS), pen 1.5 decks cut off, spread 1:16 to a 10k bankroll ($5 to $80): SCORE 15, N0 54K hands, Win Rate $15.44

The importance of rules changes can be seen if you add Late Surr: SCORE 35, N0 28K, WR $36.64

For comparison, a 2D game H17 DAS with lousy pen (1.0 decks cut off), spread 1:8 to the same 10K Bankroll ($10 to $80): SCORE 20, N0 50k, WR $19.87

Boost the pen of the 2D game just a little, to more typical conditions (0.8 decks cut off): SCORE 48, N0 21K, WR $34.04

A simple analysis such as this will let you compare games quite easily. You will also want to explore changes in Standard Deviation with each of these games. Also, try tinkering with the bankroll - you'll be shocked at the impact. Note that in some venues the "Bankroll" you'll want to plug to the web site's calculators may not necessarily be your own bankroll, but may be limited by the max bet you can push out on your betting ramp without being thrown out. (Read as "Gold Spike" in LV :-)

Good luck

:: ::

> Why do you think that a more-than-average-amount of high cards are more apt to fall soon after the count rises to a high point? If the TC gets to +5, there is just as much chance of the TC getting to +7 as it does to +3. Remember, a +5TC is randomly scatered throughout the remaining pack.<

Consider the situation where in DD there are 1-1/4 decks remaining and a TC of +5. Assuming an even distribution of the uncounted cards, there are 15 of those remaining, leaving 50 counted cards. A TC of +5 means an RC of 6.25 (remember, this is theoretical and only leading up to my point). This means there's an average of 25 + 3.125 high cards remaining and 25 - 3.125 low cards. We can use those numerators to create all the fractions for the probs governing the next card, but remember the DENOMINATOR of each fraction is 65! In a comparable situation for 6D or 8D, the denominator would be several times as large, meaning each card which falls has a significantly smaller effect on the subsequent fractions than is the case in DD.

IOW you would probably have to draw an average of 4, 5, or 6 cards from a rich portion of an 8D shoe to have the same effect of a single high card coming out in DD.

Also, because the denominator is DD is so much smaller, it is MUCH less likely the count will rise to +7 than it is to fall to +3. In 6D and 8D, because of the number of total cards available it is only somewhat less likely to rise than to fall.

But let us regress to the main point I made earlier: DD with only 50% pen IS beatable provided there are very few players at the table, because pen is only one factor governing the average number of favorable betting opportunities per shuffle.

that if there is just as high a probability of the TC going from +5 to +7 is there is for it to drop from +5 to something lower, then our card-counting systems can't possibly be working.

A TC of +5 says that there is a _higher_ than normal probability of 10's falling out of the deck than usual. This means there is a _higher_ probability that the TC is going to fall than that it will rise...

On average the RC will drop; the TC should remain about the same.

A TC of +5 says that there is a _higher_ than normal probability of 10's falling out of the deck than usual.

Very true.

This means there is a _higher_ probability that the TC is going to fall than that it will rise...

False. If you had said "RC", that would be true. However, the TC averages the same throughout the entire pack regardless of the number of decks. My point is that the RC will always go toward 0 but the TC will remain the same, on average.

I will try and find some real proof, then post it.

See Below. This touches on the subject.

CardCounter.Com Main Board
6D - high C - full table?
Posted By: Stealth Bomber
Date: 1/2/04 8:41:53 p.m.
Is the following statement correct?
The RC will become lower on average to the player on 3rd base and the D, while the TC will remain the same on average for 1st base all the way through to 3rd base and the D, during a round of cards. Thus, even though the RC has more chance of becoming lower as each card falls, all positions at the table have the same chance of getting the big cards.
Stealth

CardCounter.Com Main Board
Re: 6D - high C - full table?
Posted By: The Mayor
Date: 1/3/04 6:39:54 p.m.
In Response To: 6D - high C - full table? (Stealth Bomber)
Yes, that statement is correct, it is just the "True Count Theorem" -- the RC must end up at 0, but the TC stays the same ... such is life.
--Mayor

Clearly the DD game is a better game in terms of hourly win rate assuming penetration is decent.

Maybe so. All depends on the exact amount of pen. However, I personally can do better on the shoe game because I wong, wong, wong. Hard to wong 2D.