FYIQuote:LostWages...Now, if I could only satisfy my curiosity about the max TC count in a 6 deck game. If all the 2s, 3s, 4s, 5s, and 6s somehow miraculously got dealt out, my bare bones math brain still says that's:

5 pts (1 for each 2, 3, 4, 5, 6) X 4 suits = TC +20 X 6 decks = 120?

Maybe I missed the boat on this one . . . I woudn't have a BR big enough to satisfy a TC + 120 anyway . . . :-)

Max RC in 6 deck (using Hi/Low): 2, 3, 4, 5, 6... 5*4*6 = +120, HOWEVER you must take in to account what the "neutral" cards do to effect the true count. They are worth 0, but they keep the deck diluted with not only "big" cards. THUS, to get the MAXIMUM TC possible you must ALSO remove the 7, 8, and 9. So the RC stays the same (RC +120), but now the number of cards remaining goes down from 192 to 120. With 120 cards remaining that's about 2.3 decks, thus the maximum true count one could possibly reach in a 6D game is 120/2.3 = 52.174. You will literally never see this in your lifetime, and if you do I would be that the deck was gaffed by human error rather than you seeing this "randomly" occur. Either way you should literally bet the table max and watch your 20 push every time and just try to trade blackjacks back and forth with the dealer (because your 3:2 payout will make you win that battle).

Thinking more about this, if there's only aces and 10's left in the deck, the TC division by 2.3 decks seems to be a moot point? Alas, it's all a moot point because this will literally never happen lol.

And this brings me back to a strict, British-run grade school in the heart of Djakarta, Indonesia. Wha? Well, I'm immediately reminded of Sister Nurberge (very feisty, 4' 5" French nun, with hands of steel and laser eyes) who taught us arithmetic. "You learn the basics, don't forget to APPLY . . . you APPLY and put into action!!! "Quote:Romes. . . thus the maximum true count one could possibly reach in a 6D game is 120/2.3 = 52.174!

So I learned about TC and RC. I learned about card-counting. I knew that TC = RC/remaining deck estimate.

I did not APPLY . . . forgot to divide RC by the estimated remaining decks, which you have once more eloquently explained.

I can sleep better tonight, knowing I'll never see a TC = +120 (my dreams for the last 2 weeks), but maybe a high 20 . . . that'll work. I'll even DOUBLE my humongous BR session ($100 + $50 emergency for double-down)!!!!

To trip #3!

Indeed, IBYA, you are correct, that the maximum TC is a 6-deck BJ game is 52 for Hi/lo. This assumes Romes's calculations are correct:Quote:Ibeatyouraces1) assuming hi-lo...(someone correct me if I'm wrong) +52. That would be nothing but 10's/A's.

Max RC in 6 deck (using Hi/Low): 2, 3, 4, 5, 6... 5*4*6 = +120, HOWEVER you must take in to account what the "neutral" cards do to effect the true count. They are worth 0, but they keep the deck diluted with not only "big" cards. THUS, to get the MAXIMUM TC possible you must ALSO remove the 7, 8, and 9. So the RC stays the same (RC +120), but now the number of cards remaining goes down from 192 to 120. With 120 cards remaining that's about 2.3 decks, thus the maximum true count one could possibly reach in a 6D game is 120/2.3 = 52.174.

The player would split his aces, and the dealer would be required to hit his. So you'd be doing more than just trading blackjacks.

A dealer A-A would remove more aces more quickly, and the dealer would need, at a minimum, two more cards (A-A-10-10 = 22 bust) and could possibly draw additional aces, perhaps making a non-busted total (A-A-10-A-A-A-A-A).

So. Assuming six decks with all the cards 2-9 removed, betting "1 unit" each time, what's the minimum and the maximum units that could be won or lost?

One scenario:

Player gets nothing but 10-10, dealer gets nothing other than A-10. Player loses 24 rounds. Then there are no more aces. Last 78 rounds are 10-10 v 10-10.

Player loses 24 units.

Another scenario:

Player gets all the blackjacks, wins 24 hands, pushes the rest. Dealer wins 36 units.

As far as how to get to the point where there are only 10s and Aces left in the shoe? Back counting. Simple as that.

How interesting! Something to at least give a passing thought of reflection. Your scenarios somewhat match the expected BJ outcomes of 43% win, 48% lose to the dealer, and 9% push, huh?Quote:racquetWith only tens and aces left in the deck, either player or dealer could possibly get A-A as well as 10-10 or A-10.

Thanks for your comment!

1) Dealer ace showing - you take insurance. Either blackjack or dealer will bust unless he draws 5 more aces.

2) Dealer 10 showing. Either blackjack or then you know the dealer has 20. You can only lose if you split aces and get another ace. Otherwise, split your 10's as much as possible to get aces or other 20's which you know will push.

In this "not real and will never happen scenario" your win rate should be rather huge. Hell, pretty sure you're supposed to be doubling down on your blackjacks here too =). The double, split, and 150% pay for blackjack would be amplified on steroids here. Aka TC +52 the player would absolutely decimate the game, as expected.

This is now my favorite (and recurring!) dream: TC+ is now in the 50s? What to do . . . ? (Heh! Heh!)Quote:RomesAka TC +52 the player would absolutely decimate the game, as expected.

Now that I'm trying to be more math-conscious, I'm remembering the calculations that produced 52.174 as the max TC+ in a 6-deck game.

Just for fun, I applied the concept to "What is the max TC + for an 8-deck game?" and made an interesting discovery.

My calculations (please correct as necessary):

8 decks (416 cards) x 5 (2s, 3s, 4s, 5s, 6s) x 4 suits/deck = 160 RC

416 cards less (8 decks x 3 (7s, 8s, 9s) x 4 suits/deck) 96 = 320 cards left in deck

320/52 = 6.15 decks left

TC = 160/6.15 = 26.016

Huh? For an 8-deck game the max TC+ is LESS than a 6-deck game (52.174)? Almost TWICE as less? Doesn't this signal that -- given a choice -- a 6-deck game is better to play than an 8-deck game? Or am I missing something, like an 8-deck game provides MORE ROUNDS to continue with ramped up bets?

Take out all the little cards (2, 3, 4, 5, 6).. 4 each, 8 decks... 5*4*8 = 160

Take out the middle cards (7, 8, 9)... 4 each, 8 decks... 3*4*8 = 96

Orignal 8D shoe has 416 cards... so 416 - 160 - 96 = 160 card remaining.

160 cards remain, and they're all aces and faces.... RC = +160, 3.0769 decks remaining (160 cards)...

TC = 160/3.0769 = 52.0004.

2D...

104 cards - (little & middle) = 104 - 64 = 40... +40 RC.... and 40 cards remaining (.8077 decks)... TC = 40/.8077 = 49.5233 ...less than 6D and 8D.

Also... What game has a bigger max true count IS A TERRIBLE WAY to decide what game to play. You'll never see TC +25, let alone TC +50. I've been counting for over 10 years and the biggest TC I've ever seen (maybe twice in 10 years) was like +18, if I remember correctly (after TC +10 I'm just max betting and maxing my deviations so they're all the same to me at that point).

And now I know:Quote:RomesAlso... What game has a bigger max true count IS A TERRIBLE WAY to decide what game to play.

- how to correct my 'rithmetic! 8D TC+ max is just like 6D, 52!

- don't decide on game to play based on max true count!

I think it's high time for me to decompress a week or two, and just practice card-counting.

Counting apps --> here I come!

BTW, BIG Thanks for taking time to explain!

LW