Studies on Deck Composition

[aslan]

Is there any body of knowledge available on the effect of various deck configurations on EV? For example, what is the effect of every other card being a high card? --every third card? What would be the effect of the first 160 cards in 8 deck being tens and aces? --being 2 through 6? What effect do various size slugs of tens or small cards have on EV? If three or four aces lump together?

 

[rrwoods]

I can't answer most of that. But for a portion of a deck you know to be composed entirely of tens and aces, the advantages is enormous. Correct strategy is to always split when you can and stand when you can't.

 

[SleightOfHand]

The only study that I can think of is one by Snyder (I think) where he looked at the effect of playing a shoe with fresh decks stacked on each other with no shuffling. If I remember right, where there were multiple players playing BS, first base would have a large advantage and the following players would have steadily decreasing advantages.

What is the point of knowing this kind of information?

 

[pit15]

The only study that I can think of is one by Snyder (I think) where he looked at the effect of playing a shoe with fresh decks stacked on each other with no shuffling. If I remember right, where there were multiple players playing BS, first base would have a large advantage and the following players would have steadily decreasing advantages.

What is the point of knowing this kind of information?

In case a casino tries to screw you by shuffling unwashed cards. If you can pick up on this and know what to do you can turn it around

 

[aslan]

The only study that I can think of is one by Snyder (I think) where he looked at the effect of playing a shoe with fresh decks stacked on each other with no shuffling. If I remember right, where there were multiple players playing BS, first base would have a large advantage and the following players would have steadily decreasing advantages.

What is the point of knowing this kind of information?

My intuition is that cards that are very well shuffled will usually result in small cards, medium cards and aces/tens being even dispersed throughout the shoe, with maybe a few small clumps. What effect do cards so configured have on EV?

Occasionally, you'll play a shoe where clumps of high cards and clumps of low cards are prevalent. Intuitively, this occurs less often than an even distribution does. What effect does this have on EV?

If the cards are evenly distributed, I would guess that they are more likely to stay that way after being shuffled than to suddenly become badly clumped. If you knew the effect on EV for evenly distributed cards, you would know something about you chances for such a shoe. For one thing, the count would tend to hover around zero.

If the cards are badly clumped into large cards and small cards, it would seem that a shuffle might easily return them back to an even distribution, but I don't know the likelihood of that. I have often noted that when three aces are in consecutive order, they sometimes continue reappearing in consecutive order in subsequent shoes. So I wonder what the chances are of a badly clumped shoe remaining badly clumped in subsequent shoes, and what the effect of such a shoe is on EV.

It just seems to me that such knowledge could influence decision making at blackjack.

It may be voodoo, but I suspect there may be an underlying science to it--maybe useless, maybe not.

 

[tensplitter]

The only study that I can think of is one by Snyder (I think) where he looked at the effect of playing a shoe with fresh decks stacked on each other with no shuffling. If I remember right, where there were multiple players playing BS, first base would have a large advantage and the following players would have steadily decreasing advantages.

What is the point of knowing this kind of information?

First base would always get the ace on the first hand.

Heads up
Player has soft 14 and hits to 19. Dealer shows a deuce with a 4 in the hole, hits to 19 for a push.

With 2 players
1st player gets a soft 15, hits and gets a 7, hits the 12 to get an 8 for a total of 20. 2nd player has 7 and hits to 16. Dealer shows a 3 with 6 in the hole and draws a ten. Player 1 wins.

With 3 players
1st base gets soft 16, doubles and gets an 9. 2nd player has 8 and hits to 18. 3rd player had 10, doubles and gets a jack. Dealer shows a 4, has 8 in the hole, draws a queen to bust. Everyone wins.

With 4 players
1st base gets soft 17, doubles and gets a jack. 2nd player has a 9 and doubles and gets a queen. 3rd player had 11, doubles and gets a king. 4th player stays on 13. Dealer shows a 5, has a ten in the hole, draws an ace and a deuce for 18. Players 2 and 3 win

With 5 players
1st base would get soft 18 and double to a hard 18, 2nd player would get 10 and double to get an ace, 3-5 would have stiffs and stay. Dealer would show a six and have a queen underneath, then draw a deuce. 2nd player would be the only winner.

With 6 players
First base gets soft 19 and stays, 2nd base gets 11 and doubles and gets a deuce, 3rd player would get 13 and hit to 16 and then 20, 4th player would have 14 and hit to 19, 5th player would have 15 and hit to 21, 6th player would have 16 and bust. Dealer would show a 7 and have an ace. Players 1, 3,4,5 win

With 7 players
First base gets soft 20 and stays, 2nd base gets 12, hits and gets a 4 then a 5 for 21, 3rd player would get 13 draw a 6 for 19, 4th player would have 14 draw a 7 for 21, 5th player would have 15 and bust, 6th player would have 16 and bust. 7th player would have soft 17, hit and get a 9 for a hard 16, then bust with a 10. Dealer would show a 8 and have a 3 in the hole, then draw a jack for 21. Players 2 and 4 push and everyone else loses

With 8 players
First base gets blackjack, 2nd base gets 12, hits and gets a 6 for 18, 3rd player would get 13 draw a 7 for 20, 4th player would have 14 draw a 8 to bust, 5th player would have soft 16 and draw a 9 and ten to bust, 6th player would have 8 and draw a jack for 18. 7th player would have 10, hit and get a 20. 8th player would have 12 and draw a king to bust. Dealer would show a 9 and have a 5 in the hole, then draw an ace and deuce for 17. Players 1, 2, 3, 6 and 7 would win.


Knowing this info and with 8 spots, a single player should bet the table max on spots 1, 2, 3, 6 and 7 and the table min on 4, 5, and 8.

Actually it would be better if the player bet table max on all 8 spots (or better yet, ask for the max to be raised and then take out all of your life savings). Double down the blackjack on first base, double down the hard 12 in spot 2, double down the hard 13 in spot 3, stay on the 14 in spot 4, double down the soft 16 in spot 5, double down the 8 in spot 6 and the 10 in spot 7 and stay on the 12 in spot 8. The dealer will bust. Then the dealer will be fired.
Alternatively, if allowed to bet anything, bet your life savings - 7 times the minimum bet on spot 1, and bet the minimum on the rest.

 

[pit15]

First base would always get the ace on the first hand.

Heads up
Player has soft 14 and hits to 19. Dealer shows a deuce with a 4 in the hole, hits to 19 for a push.

With 2 players
1st player gets a soft 15, hits and gets a 7, hits the 12 to get an 8 for a total of 20. 2nd player has 7 and hits to 16. Dealer shows a 3 with 6 in the hole and draws a ten. Player 2 wins.

With 3 players
1st base gets soft 16, doubles and gets an 9. 2nd player has 8 and hits to 18. 3rd player had 10, doubles and gets a jack. Dealer shows a 4, has 8 in the hole, draws a queen to bust. Everyone wins.

With 4 players
1st base gets soft 17, doubles and gets a jack. 2nd player has a 9 and doubles and gets a queen. 3rd player had 11, doubles and gets a king. 4th player stays on 13. Dealer shows a 5, has a ten in the hole, draws an ace and a deuce for 18. Players 2 and 3 win

With 5 players
1st base would get soft 18 and double to a hard 18, 2nd player would get 10 and double to get an ace, 3-5 would have stiffs and stay. Dealer would show a six and have a queen underneath, then draw a deuce. 2nd player would be the only winner.

With 6 players
First base gets soft 19 and stays, 2nd base gets 11 and doubles and gets a deuce, 3rd player would get 13 and hit to 16 and then 20, 4th player would have 14 and hit to 19, 5th player would have 15 and hit to 21, 6th player would have 16 and bust. Dealer would show a 7 and have an ace. Players 1, 3,4,5 win

With 7 players
First base gets soft 20 and stays, 2nd base gets 12, hits and gets a 4 then a 5 for 21, 3rd player would get 13 draw a 6 for 19, 4th player would have 14 draw a 7 for 21, 5th player would have 15 and bust, 6th player would have 16 and bust. 7th player would have soft 17, hit and get a 9 for a hard 16, then bust with a 10. Dealer would show a 8 and have a 3 in the hole, then draw a jack for 21. Players 2 and 4 push and everyone else loses

With 8 players
First base gets blackjack, 2nd base gets 12, hits and gets a 6 for 18, 3rd player would get 13 draw a 7 for 20, 4th player would have 14 draw a 8 to bust, 5th player would have soft 16 and draw a 9 and ten to bust, 6th player would have 8 and draw a jack for 18. 7th player would have 10, hit and get a 20. 8th player would have 12 and draw a king to bust. Dealer would show a 9 and have a 5 in the hole, then draw an ace and deuce for 17. Players 1, 2, 3, 6 and 7 would win.


Knowing this info and with 8 spots, a single player should bet the table max on spots 1, 2, 3, 6 and 7 and the table min on 4, 5, and 8.

Actually it would be better if the player bet table max on all 8 spots (or better yet, ask for the max to be raised and then take out all of your life savings). Double down the blackjack on first base, double down the hard 12 in spot 2, double down the hard 13 in spot 3, stay on the 14 in spot 4, double down the soft 16 in spot 5, double down the 8 in spot 6 and the 10 in spot 7 and stay on the 12 in spot 8. The dealer will bust. Then the dealer will be fired.
Alternatively, if allowed to bet anything, bet your life savings - 7 times the minimum bet on spot 1, and bet the minimum on the rest.

heh.. not only do most tables not have 8 spots (most have 6), good luck getting dealt from an unshuffled deck.

I was thinking more along the lines of an unwashed, but riffled deck. That changes the composition of hands that you get (stiff vs pat), and changes the strategy that you should play

 

[Automatic Monkey]

Is there any body of knowledge available on the effect of various deck configurations on EV? For example, what is the effect of every other card being a high card? --every third card? What would be the effect of the first 160 cards in 8 deck being tens and aces? --being 2 through 6? What effect do various size slugs of tens or small cards have on EV? If three or four aces lump together?

It's very advanced stuff, but it has some potential. The best application I see for that kind of theory is for use with certain sidebets.

 

[bigplayer]

My intuition is that cards that are very well shuffled will usually result in small cards, medium cards and aces/tens being even dispersed throughout the shoe, with maybe a few small clumps. What effect do cards so configured have on EV?

Occasionally, you'll play a shoe where clumps of high cards and clumps of low cards are prevalent. Intuitively, this occurs less often than an even distribution does. What effect does this have on EV?

If the cards are evenly distributed, I would guess that they are more likely to stay that way after being shuffled than to suddenly become badly clumped. If you knew the effect on EV for evenly distributed cards, you would know something about you chances for such a shoe. For one thing, the count would tend to hover around zero.

If the cards are badly clumped into large cards and small cards, it would seem that a shuffle might easily return them back to an even distribution, but I don't know the likelihood of that. I have often noted that when three aces are in consecutive order, they sometimes continue reappearing in consecutive order in subsequent shoes. So I wonder what the chances are of a badly clumped shoe remaining badly clumped in subsequent shoes, and what the effect of such a shoe is on EV.

It just seems to me that such knowledge could influence decision making at blackjack.

It may be voodoo, but I suspect there may be an underlying science to it--maybe useless, maybe not.

Your intuition would be wrong. Clumps of small and big cards are all part of randomness. Random does not mean evenly distributed...that would, in fact, be very non-random. Clumps of small and big cards are what create positive and negative counts...without these counting cards and shuffle tracking would not be possible. A badly clumped shoe once shuffled may result in another clumped shoe but often if is a different clump in a different place with completely different cards. If a clump is easily maintained from shoe to shoe chances are the shuffle is inadequate and you can exploit such shuffles by just keeping your eyes open.

 

[aslan]

Your intuition would be wrong. Clumps of small and big cards are all part of randomness. Random does not mean evenly distributed...that would, in fact, be very non-random. Clumps of small and big cards are what create positive and negative counts...without these counting cards and shuffle tracking would not be possible. A badly clumped shoe once shuffled may result in another clumped shoe but often if is a different clump in a different place with completely different cards. If a clump is easily maintained from shoe to shoe chances are the shuffle is inadequate and you can exploit such shuffles by just keeping your eyes open.

I hear what you're saying, and I agree even before my initial statement. Of course some clumps would be part of any random shuffle. That's not what I'm saying. There is probably a way to figure the probability of four card clumps, five card clumps, six card clumps, etc. in any random shuffle. I'll bet intuitively that there are more clumps of no clumps and two card clumps than anything else. When they go beyond the normal, has anyone studied the effect on EV? I am not satisfied to say without it having been studied that there is nothing possible to be gained from understanding shoe composition and shuffle impact on various shoe configurations. As you said, badly clumped shoes when shuffled may result in anything--but is one thing or the other more likely? It seems you don't know and I don't know either. Would it be a different result with a hand shuffle as opposed to an ASM shuffle? Also, I am not willing to concede that a hand shuffle, or even an ASM shuffle, is a random shuffle, that being because I have never studied these two types of shuffle--could be, I just don't know. I am keeping an open mind until someone shows me, or cites, the data that explains it all. But thanks for you take on it. It's about where I was before I asked the question.

 

[aslan]

First base would always get the ace on the first hand.

Heads up
Player has soft 14 and hits to 19. Dealer shows a deuce with a 4 in the hole, hits to 19 for a push.

With 2 players
1st player gets a soft 15, hits and gets a 7, hits the 12 to get an 8 for a total of 20. 2nd player has 7 and hits to 16. Dealer shows a 3 with 6 in the hole and draws a ten. Player 1 wins.

With 3 players
1st base gets soft 16, doubles and gets an 9. 2nd player has 8 and hits to 18. 3rd player had 10, doubles and gets a jack. Dealer shows a 4, has 8 in the hole, draws a queen to bust. Everyone wins.

With 4 players
1st base gets soft 17, doubles and gets a jack. 2nd player has a 9 and doubles and gets a queen. 3rd player had 11, doubles and gets a king. 4th player stays on 13. Dealer shows a 5, has a ten in the hole, draws an ace and a deuce for 18. Players 2 and 3 win

With 5 players
1st base would get soft 18 and double to a hard 18, 2nd player would get 10 and double to get an ace, 3-5 would have stiffs and stay. Dealer would show a six and have a queen underneath, then draw a deuce. 2nd player would be the only winner.

With 6 players
First base gets soft 19 and stays, 2nd base gets 11 and doubles and gets a deuce, 3rd player would get 13 and hit to 16 and then 20, 4th player would have 14 and hit to 19, 5th player would have 15 and hit to 21, 6th player would have 16 and bust. Dealer would show a 7 and have an ace. Players 1, 3,4,5 win

With 7 players
First base gets soft 20 and stays, 2nd base gets 12, hits and gets a 4 then a 5 for 21, 3rd player would get 13 draw a 6 for 19, 4th player would have 14 draw a 7 for 21, 5th player would have 15 and bust, 6th player would have 16 and bust. 7th player would have soft 17, hit and get a 9 for a hard 16, then bust with a 10. Dealer would show a 8 and have a 3 in the hole, then draw a jack for 21. Players 2 and 4 push and everyone else loses

With 8 players
First base gets blackjack, 2nd base gets 12, hits and gets a 6 for 18, 3rd player would get 13 draw a 7 for 20, 4th player would have 14 draw a 8 to bust, 5th player would have soft 16 and draw a 9 and ten to bust, 6th player would have 8 and draw a jack for 18. 7th player would have 10, hit and get a 20. 8th player would have 12 and draw a king to bust. Dealer would show a 9 and have a 5 in the hole, then draw an ace and deuce for 17. Players 1, 2, 3, 6 and 7 would win.


Knowing this info and with 8 spots, a single player should bet the table max on spots 1, 2, 3, 6 and 7 and the table min on 4, 5, and 8.

Actually it would be better if the player bet table max on all 8 spots (or better yet, ask for the max to be raised and then take out all of your life savings). Double down the blackjack on first base, double down the hard 12 in spot 2, double down the hard 13 in spot 3, stay on the 14 in spot 4, double down the soft 16 in spot 5, double down the 8 in spot 6 and the 10 in spot 7 and stay on the 12 in spot 8. The dealer will bust. Then the dealer will be fired.
Alternatively, if allowed to bet anything, bet your life savings - 7 times the minimum bet on spot 1, and bet the minimum on the rest.

I guess you missed my drift entirely. Anyone could figure out the result of dealing an unshuffled shoe, but what I was driving at was not that, but the effect of different configurations of cards in general on EV. For example, what would be the effect of every other card being a 9 or above, and every other card being a 2 thru 6, with 7s being evenly distributed between the two. In understanding the extreme cases, it might shed some light on the more usual cases, like when there are no "large clumps" of cards.

I have no idea how many clumps there are in a randomly shuffled deck of cards, but I am guessing that clumps of two outnumber clumps of three, which outnumber clumps of four, and so forth. When there are no clumps above three, could that have a predictable bearing (percentage-wise) on EV?

 

[bigplayer]

I guess you missed my drift entirely. Anyone could figure out the result of dealing an unshuffled shoe, but what I was driving at was not that, but the effect of different configurations of cards in general on EV. For example, what would be the effect of every other card being a 9 or above, and every other card being a 2 thru 6, with 7s being evenly distributed between the two. In understanding the extreme cases, it might shed some light on the more usual cases, like when there are no "large clumps" of cards.

I have no idea how many clumps there are in a randomly shuffled deck of cards, but I am guessing that clumps of two outnumber clumps of three, which outnumber clumps of four, and so forth. When there are no clumps above three, could that have a predictable bearing (percentage-wise) on EV?

I suggest you check out Wong's Professional Blackjack. There is an entire section devoted to shuffles starting on page 222. It may provide some answers to some of your questions.

 

[aslan]

I suggest you check out Wong's Professional Blackjack. There is an entire section devoted to shuffles starting on page 222. It may provide some answers to some of your questions.

Thanks. I have not read PB.

 

[k_c]

One composition

An interesting theoretical composition is to remove all of the odd cards (ace, 3, 5, 7, 9) from a shoe. The playing strategy for this composition is pretty simple and would yield a nice expected value.

Hit-stand
Always stand on 12 or more

Doubles
Always double on 10
Double on 8 versus 2,4,6
Double on 6 versus 2,4,6

Splits
Split T-T versus 2,4,6
Split 8-8 versus 2,4,6,8
Split 6-6 versus 2,4,6
Split 4-4 versus 2,4,6 if DAS, otherwise double versus 2,4,6 if NDAS
Split 2,2 versus 2,4,6

Ev(all odd cards removed) for above strategy (1 split allowed)
Single deck NDAS: +12.68%, DAS: +13.06%
8 decks NDAS: +12.23%, DAS: +12.67%