I believe your figure suggests a 75% pen. An event unlikely to occur.
A brand new 52 card deck would have 50% pen because 50% is automatically discarded, therefore removed from play. Each subsequent shuffle is only the cards in discard, which we assume to be all the cards that are not on the table in action. In order to have 75% pen, as you suggest it is, the dealer can only discard 13 cards to the tray. Since these cards are not going to be playable until the next shuffle, they are in effect, cut off. To cut off 13 cards, the dealer must have just shuffled 2X that. E.G., shuffle 26 cards, discard half of those.
In order to only be able to shuffle 26 cards, the other 26 must be in play. With 6 players, 1 dealer, that is 3.7 cards per hand unbusted AND 1 hand must be unresolved or the dealer would not be forced to shuffle.
If it were to occur, Even a flat bettor would be smart to have a good size bet in play until the next shuffle. At this point, you would be halfway thru the deck, with a very high count. It would have to be high, or how else could 7 people draw 3-4 cards each without breaking?
If the dealer didnt have blackjack I would split tens to try for the ten bonus or to draw aces because the dealer is most likely going to have a 20. If he had the Ace up, I would be fully insured.
Other things would have occurred to get a 75% pen at a full table. Two hands before the shuffle had a lot of 2 card 20s which tanked the count, the following hand has the dealer with a 7 or better as the upcard, but most likely an 8 or 9 and most of the players having 17 to 19 after drawing on stiff hands.
No Pen of 75% would ever happen if an intelligent counter were playing heads up. The best I could see it get is 68% IF the counter was taking a big chance. For a pen of 68%, the player and the dealer would have to eat up 18 cards and still be unfinished. To do this, the player would most likely have split to 4 hands against a dealer breaker, with the two card totals being less than what is worthy of doubling (remember the count is very low to start this hand, and the RC would have only gone up +6 before the first probable double). I don't think this situation would occur because the high cards would all be gone on the previous hand, so a counter should bail beforehand. Even if he did stick around, I doubt it would be worthwhile splitting when the dealer bust probability is ultra-low.
So, I think 75% is very unlikely cutting off halves and dealing down to 0.
> I've just begun working on the case of the 3:2 suited split tens, and I think this is where it's going to get just a *little bit* tricky.
The original post said "suited and connected" which I think means 10/J, J/Q or Q/K only - not 10/Q, 10/K or J/K. Don't forget to divide by 2!
> An additional note, I've done some work on the 5:2 insurance case, and initial calculations suggest an increase in Expectation of approximately .12% over 2:1 insurance.
I'm surprised it's so little. Doesn't 5:2 make insurance +EV with basic strategy?
Antichrist: you're not just having a little joke with us about these rules, are you? This has to be the most bizarre game I've ever heard of. Wouldn't there be frequent dealer errors with the different payoffs for front and back bets, and different options for front bet, en prise bet and back bet?
and the bonus after splitting 10s is only for suited AND connected 10's. 10-J; J-Q; Q-K.
Also, in response to someone below who asked if there were a lot of dealer errors - YES, even though backliners bets are separated by a hideous looking, fat, orange and silver slug. Almost like a "Look at me, I'm a dick" sign
Another thing to consider is that the table is almost always full and the game is VERY slow. Hands per hour go waaaay down due to the additional settlements of the side bets, special payouts on BJ, insurance, and 10 splits, en-prise settlements, so you need to factor this in. I would guestimate that hands-per hour get cut by at l/3, perhaps more with the constant arguing and bitching that goes along with all the backline action. Players slow way down when making their playing decisions as well
Last comment to whoever mentioned throwing purples on top of a teammates bet. Remember: you lose the ability to take insurance on the top bet, a very big money making play in high counts. No way that 10 splitting and BJ bonus offsets this.
"I believe your figure suggests a 75% pen."
No, merely the dealing procedure. Which makes for a deeper (and different) penetration than straight 50%. There is the small issue of knowing all the cards that go into the deal to complete the round.
"In order to have 75% pen, as you suggest it is, the dealer can only discard 13 cards to the tray. Since these cards are not going to be playable until the next shuffle, they are in effect, cut off. To cut off 13 cards, the dealer must have just shuffled 2X that. E.G., shuffle 26 cards, discard half of those."
You are confusing matters.
The dealer deals normally until the 26th card is reached. If the round is completed with that card, fine, the whole deck is shuffled and we start anew. If the round is not completed, the dealer picks up the 26 discards, shuffles them, they're cut in half, and the dealer will try to complete the round with the 13 cards in his hand.
"In order to only be able to shuffle 26 cards, the other 26 must be in play. With 6 players, 1 dealer, that is 3.7 cards per hand unbusted AND 1 hand must be unresolved or the dealer would not be forced to shuffle."
Why are you assuming that the 26th card must be reached at the first round? What if 3 players are playing? 1st round, 11 cards including the dealer's. 2nd round, 13 cards, including the dealer's. 3rd round, the dealer runs out of cards as she deals the second player's first card.
Quiz: How many boxes must be open in that third round of play?
"Two hands before the shuffle had a lot of 2 card 20s which tanked the count, the following hand has the dealer with a 7 or better as the upcard, but most likely an 8 or 9 and most of the players having 17 to 19 after drawing on stiff hands."
The second part of the above phrase is mystifying. Where do you get that "most likely" about the dealer having 7, 8 or whatever? Or about the players' hands? Those claims are not based on substance. But you're onto something about many aces & faces being dealt out in the first half of the deck!
Quiz: Would you want to play again those cards?
"No Pen of 75% would ever happen if an intelligent counter were playing heads up."
An intelligent player would destroy the house that offers that game. Two intelligent players, one back-betting the other, would fare even better. Locking up the table would be nirvana.
(The term "would" is used rhetorically.)
"The rest dealt to the bottom, if last card dealt in mid-hand, all discards shuffled, cut in half again, and the game proceeds form that point on,..."
Notice the phrase, ALL DISCARDS, that means discards from previous rounds, including the 11 and 13 cards from the two rounds you mentioned. So the dealer would be shuffling 26 + 13 + 11 (the total of all cards in the discard tray) for a total shuffle of 50 cards, of which 25 should be placed afterwards into the discard tray. You are assuming that the dealer doesnt include cards from previous rounds in the shuffle. If yours was the case, what would the dealer do with your 24 cards from rounds 1 and 2.Your example actually puts the pen one card off from exactly 50%.
As far as the composition of the players hands, I assume BS was used in the game. To give each player 3.7 cards, they have to be of low value. A player most likely wont have high value cards in his hand if he's holding 3.7 cards. They would also have to be hitting stiff hands, which suggests the dealer has a 7 or better. The likelihood of a dealer 10 is lower simply due to an extremely low count before the hand was started. I say most likely an 8 or 9 for the dealer because this would further reduce the player bust probability by increasing the number of 7s remaining, which is crucial when there are only a dozen or so cards left for the players to draw on low stiffs without busting. Try it yourself and deal 26 cards out between 6 players and 1 dealer without breaking the players.
I agree that counters might find a good game of this, however, to cause such a deep pen under these rules requires an extremely low count that a smart counter would avoid, which would leave the table open after a heads up game and causing a full deck shuffle.
"75% is still off."
I didn't mean to imply "75%" but merely that this is not the usual 50% penetration.
"Notice the phrase, ALL DISCARDS, that means discards from previous rounds, including the 11 and 13 cards from the two rounds you mentioned. So the dealer would be shuffling 26 + 13 + 11 (the total of all cards in the discard tray) for a total shuffle of 50 cards, of which 25 should be placed afterwards into the discard tray."
Those 26 cards you have the dealer shuffling again are the cut-offs, not the discards. The original poster wrote "SD shuffled, 1/2 cut and put in discard tray" but don't let that confuse you - those cards placed at the discard tray before the deal starts are not discards!
Nevertheless you are right about the dealing procedure : The dealer shuffles 1 deck, gets it cut in half (approximately, but let's say at the 26th card), and deals out. If she manages to complete the last round of play with that half-deck (meaning she completes the last round of play with the 26th card which is highly unlikely), fine, she starts all over again, by shuffling all the cards, cutting them in half, etc.
If the last round needs more cards, she doesn't continue into the cut-offs, as she would have done in normal non-pitch games, but instead takes the discards and the cut-offs (this is where I was wrong), which makes for [26 cards cut-off]+[X cards dealt in previous rounds], where X = 26 - [cards dealt at the last round which awaits completion].
So, if before the last round, 20 cards have been dealt out, the dealer deals out her last 6 cards and then shuffles [26 cards] + [ (26-6)] = 46 cards. Sorry for the confusion.
"As far as the composition of the players hands, I assume BS was used in the game. To give each player 3.7 cards, they have to be of low value."
Your figure of 3.7 cards per hand is calculated on a wrong assumption : You are assuming that the cut-card will be reached at the first round of play. So you are calculating for one round, 6 players + 1 dealer = 7 hands, 26/7=3.7 . However, the last round can be reached after playing any number of rounds, without limitation. The rest of your assumptions, abt the composition of those hands, is therefore also wrong.
The hands I speculated was purposely composed of 26 cards. The reason is to try to create a scenario where there would be a 75% pen. Since the pen is determined by the number of discards, I had to reduce the discard stack to 26 cards to get a pen of 75%. That being the case, there had to be 26 cards in play on the current hand or else the pen would have been lower because the discard stack would have been higher. Last round, first round, it doesn't matter. The round itself though, must be made up of 26 cards to reduce the discard stack sufficiently.
So, I guess we can agree that the pen will range between 50% and 65% the great majority of the time. Here's a Quiz for you: Under these rules, can the pen ever be lower than 50%?
"..to create a scenario where there would be a 75% pen. Since the pen is determined by the number of discards, I had to reduce the discard stack to 26 cards to get a pen of 75%."
The pen is determined by the number of cards dealt in the last round. If there are 7 boxes open in the last round and the 26th card is the first card at first base at the last round, then you have 51 cards re-shuffled to complete that last round.
There is no question of a 75% penetration. Not even of a 55% or 65%. This is because we'd have to use different terms than "penetration" for this game. Penetration is how deep into a deck (or shoe) you will go. Penetration gives you an idea of kind of information you will have about the remaining cards. Here, we re-shuffle both discards and cut-offs. It's a kind of end-play.
In my above example, with the 26th card starting the last round, we would have the least possible knowledge of the composition of the cards remaining to be played. In another scenario, with the dealer starting the last round with 20 cards, she deals out 14 cards to the 7 boxes and 2 to herself. Then she deals 4 cards to the first two boxes --and runs out of cards. She has to re-shuffle 32 cards in all, to complete the round. Wouldn't you agree that the players at boxes 3,4,5,6 and 7 will have quite useful infomation to consider before playing their hands?
"Here's a Quiz for you: Under these rules, can the pen ever be lower than 50%?"
No.
As I said, the term "pen" is not accurate here. But you will always see at least half of the cards when playing that game.
(Notice I didn't at all mind not getting an answer to my quizzes...)
--Cyrus
You didn't understand what Duane Vick was saying. He didn't say anything about the theoretical average number of cards per hand, ie 2.7, but is talking instead about the average number of cards needed to run out of cards in the deal when 6 boxes are open.
6 players' hands + 1 dealer's hand = 7 hands.
1 deck cut in half = 26 cards.
26 cards / 7 hands = 3.7 , a number which is irrelevant as I pointed out (the half-deck can be depleted by dealing more than 1 round). Read the "75% never implied" post again.
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