Stanford Wong writes in one of his books: "Basic strategy is the best way to play a blackjack hand on the first round
after a shuffle, assuming you see no cards other than your own and the deal-
er’s upcard".
During, let's say a six deck game with cut card, dealt from a shoe, as the shoe is depleted, all kind off different subsets (remaining depleted deck) are appearing. For each subset, there is a specific "best strategy" of Course which may deviate more or less from the generic basic strategy. Now, over all possible subsets that may appear in the long run (playing thousands of shoes), which is the best strategy ON AVERAGE, meaning which strategy played as a "composite" strategy will give the player the greatest advantage? If it is Basis Strategy as defined by Wong above, why so? Why is the best overall strategy the one valid for the first hand played from a full deck? i.e. why is it best representing all subsets? It could be well any other strategy, even though it may deviate only slightly from the best startegy for the first hand.
The only thing clear to me is that there is just one best/basic strategy for an infinite deck, as this deck never changes.
I have been dealing with BJ only since three months, so I apologize for any unclear or imprecise questions.
Thank you so much for your answer in advance.
Simba from Austria
For the average player, who can't track cards or knows the AP moves( called Indices, by most), Basic Strategy is your best way to play.
Most counters know the most popular deviations, called the Magnificent 18. Some know a hundred deviations, but I think most of those can also give you the value of Pi to eighteen digits. For most of us, 3.14 is sufficient. Learning to count and knowing the Mag. 18 is more than enough for most players.
These types of plays apply because the effect of your cards change the probabilities sufficiently to make the abnormal play better. In this case your chances of getting 21 are reduced as you already have two 7's.Quote: BleedingChipsSlowly...play only differs for...7,7v8....
In essence whenever the deck is depleted enough you should split 7's vs 8.
However unless you're tracking what's gone, after four decks in a six deck shoe, even knowing you have two (of the 24) 7's does not change the strategy. Knowing you have two of the 4 7's (single-deck) does change things.
btw similar logic applies to counting other ranks and adopting single-deck logic as applicable (e.g. lots of 6s gone and standing 15vs10), but for practical purposes it isn't worth the bother.
Let's further consider being dealt a pair of 7's and the dealer up card is 8. For the first hand out of a 6-deck shoe, There is a 7.12% (22/309) chance of drawing a remaining 7 for a 21 total. Basic strategy is hit.Quote: charliepatrick...
However unless you're tracking what's gone, after four decks in a six deck shoe, even knowing you have two (of the 24) 7's does not change the strategy. Knowing you have two of the 4 7's (single-deck) does change things.
...
Later on with 2 decks remaining in the shoe a hand is dealt. Absent counting, how many 7's should we guess are left in that shoe? 24? Not bloody likely. 2? Also not very likely. How about 8? Not likely, but I think it's the best guess. So, you are dealt 7's and the dealer has 8 up. Going with that best guess of now 6 7's left, there is a 5.94% (6/101) chance of drawing one of the remaining 7's for a 21 total. Perhaps that reduced probability is why basic strategy for 2-decks is to split, not hit.
I don't see a compelling reason to stick with 6-deck basic strategy when you are at the point of 2-decks remaining in the shoe. Other than you would have to know 2-deck basic strategy as well in order to shift. I realize that is a difficult reach for a lot of players, based on what I see at the tables. I'm not trying to be snarky, it's just a fact.
Approaching a table with 104 cards remaining out of a 6-deck or 8-deck shoe and with no knowledge of which cards have already been removed from the shoe, what is the best strategy for hit/stand/double/fold decisions?
I think the answer depends upon whether the influence of removing cards is symmetric between positive and negative "counts" for the close-call decisions.
Off the top of my head, I am guessing that the integral over all possible 104 card combinations would roughly correspond to a slightly negative count for most deciisons.
Initially the ratio of 7's to cards in a shoe is 1 in 13 (7.69%). Removing 2 7's and and 8 in six decks reduces this to 7.12%; doing the same in two decks reduces it to 5.94%. Therefore somewhere between these it becomes better to split rather than hit, presumably because your chances of 21 are reduced.Quote: BleedingChipsSlowlyI don't see a compelling reason to stick with 6-deck basic strategy when you are at the point of 2-decks remaining in the shoe.
Now consider six decks and the situation where you were dealt 2 7's and the dealer has an upcard of 8 but where there are only two decks left to deal. Your assertion is that
- on average there would have been 8 7's in the remaining shoe
- you've just seen two of them
- so on average there are now six left!
This logic is wrong.
(1)
Consider an extreme case where there was a bonus for having the seven of Spades and where you have just been dealt two. What are the chances you get a third?
With two decks you've obviously seen the only two so cannot get another one. In a six deck shoe there is clearly a possibility of you getting a third one but using your logic your chances would be zero (2-2=0). The answer should be there are now four left and, rather than 1 per deck, on average there will be about 2/3 per deck.
(2)
Compare various situations.
(i) You enter mid-shoe, see four decks in the discard tray - you know nothing about what's in it - and the dealer deals the three cards.
(ii) As above except the dealer shuffles the discards and then places the contents of the discard tray at the back of the shoe.
(iii) The dealer deals the three cards from a new six deck shoe and then, for some strange reason, takes the last four decks and throws them away.
In theory your third card is unaffected whether the discarded cards are at the back of the shoe or in the discard tray, or whether the later decks are in the shoe or not.
Saves a lot of time.
Yes, my logic was flawed in comparing two remaining decks of a six-deck shoe to a two-deck shoe for the problem presented. Your point was well presented that if I were to join a table with two decks remaining in a shoe that originally had six I would have no insight as to the remaining shoe composition. However, I think your logic was off in comparing that situation to facing a newly shuffled shoe. Four decks are discarded. Without knowing anything about what cards were discarded the remaining shoe composition is subject to the laws of probability. To explore that, I wrote a program to simulate random removal of cards from six decks down to two, then count the 7's remaining.Quote: charliepatrick... Now consider six decks and the situation where you were dealt 2 7's and the dealer has an upcard of 8 but where there are only two decks left to deal. Your assertion is that
- on average there would have been 8 7's in the remaining shoe
- you've just seen two of them
- so on average there are now six left!
This logic is wrong.
...
Compare various situations.
(i) You enter mid-shoe, see four decks in the discard tray - you know nothing about what's in it - and the dealer deals the three cards.
(ii) As above except the dealer shuffles the discards and then places the contents of the discard tray at the back of the shoe.
(iii) The dealer deals the three cards from a new six deck shoe and then, for some strange reason, takes the last four decks and throws them away.
In theory your third card is unaffected whether the discarded cards are at the back of the shoe or in the discard tray, or whether the later decks are in the shoe or not.
"use strict";
//Intialize a zero-filled array to tally 0-to-24 count occurances
let counts = [];
counts.length = 25;
counts.fill(0);
//Tally results for simulations and display.
for (let i=0, simulations = 1000000; i<simulations; i++) counts[sevenValuesLeft()]++;
console.log(counts);
function sevenValuesLeft() {
const ranks = 13, suits = 4, decks = 6, sevenRank = 6;
let shoe = [];
//Populate the shoe with cards
for (let h=0;h<decks;h++) {
for (let i=0; i<suits; i++) {
for (let j=0; j<ranks; j++) {
shoe.push(j);
}
}
}
//Remove random cards until two decks remain.
for (let i = (decks-2)*suits*ranks; i>0; i--) shoe.splice(Math.floor(shoe.length*Math.random()),1);
//Return the count of 7's left
return shoe.reduce( (sevens, card) => { return (card==sevenRank) ? ++sevens : sevens;}, 0);
}
I set the program to execute 1,000,000 simulations and report the occurrence of 7's remaining, reporting from 0 to 24 for each execution. I ran it five times and combined the data. For each number of 7's remaining I calculated the probability of that number occurring. Also for each number of 7's remaining I calculated the probability of drawing a 7 for a dealt 7,7 hand. I summed the product of those probabilities for each number of 7's and arrived at an overall probability of 5.94%, almost exactly what is calculated for a two-deck shoe. (I don't think a larger sampling would move that result much.) Here is a download link to a spreadsheet of the data and calculations an OpenDocument file format, which is compatible with most spreadsheet flavors.
Analysis spreadsheet download
[The link will be active for 7 days after this post. I will take it down sometime after that.]
As always, criticism and comments are welcome. Thanks, charliepatrick, for pointing out the error of my logic. Without that I never would have looked further. I would have arrived at the same conclusion for the wrong reason. So, I'm going to stick with switching to two-deck basic strategy when a six-deck shoe reaches that level of depletion. At least for 7's vs. 8.
And yes, that means that much of basic strategy for two decks will be correct when applied to 104 remaining cards. But that does not mean that all of basic strategy will be correct. I can't explain it without using integral signs and there is no easy way to use mathematical signs (such as integrals) in this forum.
Quote: QFITBS for the last deck of a show is the same as for the first deck of a shoe.
Nuff Said!
You begin with a shoe of 8 well-shuffled decks and begin discarding cards until there are only 52 left. You then are dealt a blackjack hand and get a pair of 2s. You hit and get a third 2. You hit again and get a fourth 2.
At that point, what is the probability that your next card is another 2? Obviously, if this had been just a single-deck game played with a regulation deck, there would be no chance at all of there being a fifth 2. But if we are considering the last 52 cards of what started as 8 decks, there is some chance that there were more than four 2s in those last 52 cards.
I ask this because I'm wondering about a different problem. Suppose in your hand dealt from those last 52 cards, you get a pair of 7s. Is the probability of a third 7 at that point really the same as it would have been for a single-deck game? Alternatively, if you are playing with the last 104 cards of an 8-deck shoe and are dealt that pair of 7s, is the probability of getting another 7 really the same as with a double-deck game?
For both of those questions, I was thinking "yes" until I started thinking about the case of being dealt quadruple 2s out of the last 52 cards. Now I'm now quite so confident.
There is a better chance than your hunch tells you according to my reconing. I ran a simulation for 10-million cases using this script:Quote: Doc... You begin with a shoe of 8 well-shuffled decks and begin discarding cards until there are only 52 left. You then are dealt a blackjack hand and get a pair of 2s. You hit and get a third 2. You hit again and get a fourth 2.
At that point, what is the probability that your next card is another 2? ...
"use strict";
//Intialize a zero-filled array to tally 0-to-32 count occurances
let counts = [];
counts.length = 33;
counts.fill(0);
//Tally results for simulations and display.
for (let i=0, simulations = 10000000; i<simulations; i++) counts[twoRanksLeft()]++;
console.log(counts);
function twoRanksLeft() {
const ranks = 13, suits = 4, decks = 8, twoRank = 1;
let shoe = [];
let singleDeck = [];
//Populate the shoe with cards
for (let h=0;h<decks;h++) {
for (let i=0; i<suits; i++) {
for (let j=0; j<ranks; j++) {
shoe.push(j);
}
}
}
//Remove random cards to create a single deck
for (let i = suits*ranks; i>0; i--)
singleDeck.push((shoe.splice(Math.floor(shoe.length*Math.random()),1))[0]);
//Return the count of 2's left
return singleDeck.reduce( (twos, card) => { return (card==twoRank) ? ++twos : twos;}, 0);
}
10-million might be a little light for a sample size, but here's the results of the run:
#: count
-----------
0: 116155
1: 580289
2: 1376746
3: 2057328
4: 2173015
5: 1731615
6: 1084976
7: 547305
8: 225702
9: 77450
10: 22660
11: 5355
12: 1193
13: 171
14: 33
15: 7
16: 0
17: 0
18: 0
19: 0
20: 0
21: 0
22: 0
23: 0
24: 0
25: 0
26: 0
27: 0
28: 0
29: 0
30: 0
31: 0
32: 0
Based on the data I recon there is a 37% chance the last deck will have 5 or more 2's, and that if you have been dealt 4 2's and there are one or more remaining in the shoe, you have a 4% chance of drawing that 2. (Assuming the dealer's up card is not a 2.)
Thanks!Quote: gordonm888Nice work. ...
Yes, a slightly negative bias for count would make a big difference. That's why I chose 7's initially, a rank that doesn't influence most counting systems. I took a look at the the possibility of a count bias. My method using simulation certainly isn't as elegant or accurate as an integral solution, but that's what's in the scope of my ability. So, I ran the following code to simulate what the running count would be when a six-deck shoe has two-decks left:Quote: gordonm888... I think the answer depends upon whether the influence of removing cards is symmetric between positive and negative "counts" for the close-call decisions.
Off the top of my head, I am guessing that the integral over all possible 104 card combinations would roughly correspond to a slightly negative count for most deciisons.
"use strict";
//Intialize a zero-filled array to tally running count occurances
let counts = [];
counts.length = 241; // -120..120
counts.fill(0);
//Tally results for simulations and display.
for (let i=0, simulations = 10000000; i<simulations; i++) counts[twoDeckRunningCount()+120]++;
console.log(counts.toString());
function twoDeckRunningCount() {
const ranks = 13, suits = 4, decks = 6;
let shoe = [];
//Populate the shoe with cards
for (let h=0;h<decks;h++) {
for (let i=0; i<suits; i++) {
for (let j=0; j<ranks; j++) {
shoe.push(j);
}
}
}
//Remove random cards until four decks remain that represent the discards
for (let i = (decks-4)*suits*ranks; i>0; i--) shoe.splice(Math.floor(shoe.length*Math.random()),1);
//Return the running count based on the discards
let runningCount =
shoe.reduce(
(runningCount, card) => {
const ace=0, ten=9, jack=10, queen=11, king=12;
const two=1, three=2, four=3, five=4, six=5;
switch (card) {
case ace:
case ten:
case jack:
case queen:
case king:
runningCount--;
break;
case two:
case three:
case four:
case five:
case six:
runningCount++;
break;
}
return runningCount;
}
, 0
);
return runningCount;
}
Here are the results for the run:
count : #
------------
-120 : 0
-119 : 0
-118 : 0
-117 : 0
-116 : 0
-115 : 0
-114 : 0
-113 : 0
-112 : 0
-111 : 0
-110 : 0
-109 : 0
-108 : 0
-107 : 0
-106 : 0
-105 : 0
-104 : 0
-103 : 0
-102 : 0
-101 : 0
-100 : 0
-99 : 0
-98 : 0
-97 : 0
-96 : 0
-95 : 0
-94 : 0
-93 : 0
-92 : 0
-91 : 0
-90 : 0
-89 : 0
-88 : 0
-87 : 0
-86 : 0
-85 : 0
-84 : 0
-83 : 0
-82 : 0
-81 : 0
-80 : 0
-79 : 0
-78 : 0
-77 : 0
-76 : 0
-75 : 0
-74 : 0
-73 : 0
-72 : 0
-71 : 0
-70 : 0
-69 : 0
-68 : 0
-67 : 0
-66 : 0
-65 : 0
-64 : 0
-63 : 0
-62 : 0
-61 : 0
-60 : 0
-59 : 0
-58 : 0
-57 : 0
-56 : 0
-55 : 0
-54 : 0
-53 : 0
-52 : 0
-51 : 0
-50 : 0
-49 : 0
-48 : 0
-47 : 0
-46 : 0
-45 : 0
-44 : 0
-43 : 0
-42 : 0
-41 : 0
-40 : 1
-39 : 0
-38 : 2
-37 : 0
-36 : 2
-35 : 2
-34 : 7
-33 : 21
-32 : 35
-31 : 57
-30 : 102
-29 : 195
-28 : 339
-27 : 596
-26 : 974
-25 : 1608
-24 : 2384
-23 : 3840
-22 : 5749
-21 : 8725
-20 : 12820
-19 : 18617
-18 : 26562
-17 : 36756
-16 : 50190
-15 : 66465
-14 : 87728
-13 : 112751
-12 : 142224
-11 : 176999
-10 : 214824
-9 : 256781
-8 : 300415
-7 : 345288
-6 : 389922
-5 : 431700
-4 : 469384
-3 : 501135
-2 : 525847
-1 : 539487
0 : 545418
1 : 538272
2 : 524310
3 : 499907
4 : 469383
5 : 432050
6 : 389253
7 : 345262
8 : 299969
9 : 256054
10 : 214101
11 : 176627
12 : 142433
13 : 113473
14 : 87937
15 : 66340
16 : 49828
17 : 36283
18 : 26485
19 : 18609
20 : 12914
21 : 8800
22 : 5810
23 : 3807
24 : 2371
25 : 1523
26 : 882
27 : 560
28 : 371
29 : 190
30 : 100
31 : 55
32 : 40
33 : 20
34 : 16
35 : 5
36 : 6
37 : 1
38 : 1
39 : 0
40 : 0
41 : 0
42 : 0
43 : 0
44 : 0
45 : 0
46 : 0
47 : 0
48 : 0
49 : 0
50 : 0
51 : 0
52 : 0
53 : 0
54 : 0
55 : 0
56 : 0
57 : 0
58 : 0
59 : 0
60 : 0
61 : 0
62 : 0
63 : 0
64 : 0
65 : 0
66 : 0
67 : 0
68 : 0
69 : 0
70 : 0
71 : 0
72 : 0
73 : 0
74 : 0
75 : 0
76 : 0
77 : 0
78 : 0
79 : 0
80 : 0
81 : 0
82 : 0
83 : 0
84 : 0
85 : 0
86 : 0
87 : 0
88 : 0
89 : 0
90 : 0
91 : 0
92 : 0
93 : 0
94 : 0
95 : 0
96 : 0
97 : 0
98 : 0
99 : 0
100 : 0
101 : 0
102 : 0
103 : 0
104 : 0
105 : 0
106 : 0
107 : 0
108 : 0
109 : 0
110 : 0
111 : 0
112 : 0
113 : 0
114 : 0
115 : 0
116 : 0
117 : 0
118 : 0
119 : 0
120 : 0
The negative occurrences do outnumber the positive ones 4730534:4724048, but I don't know if that's statistically significant given the sample size. Even if it is, it would not be much of a bias. So, I'm sticking with the idea that two-deck basic strategy is better than six-deck basic strategy when a six-deck shoe has two left without counting. Other opinions have been voiced, but without any reasoning to back up the assertions.
But in the end, we're only considering four marginal differences anyway.
I'm familiar with the work you've posted here. You have the knowledge, skill and tools needed to provide a much better answer than I can. Thanks for doing the work and posting!Quote: QFIT... I generated basic strategy charts for the last two decks in a six-deck shoe. It turned out to be exactly the same as the full six-deck strategy, ...
I'll believe you. My world doesn't include the deep end of the pool.Quote: QFIT... as logic would dictate
Therefore: say you have only 26 cards remaining in a shoe from which to deal a last BJ hand. Given that suits are irrelevant in BJ, is the probability for the composition(ranks) of these 26 cards any different if the 26 cards are the remnant of a 7-deck shoe vs being the remnant of a 10 deck shoe? (I think the answer is No.)
Therefore, if there were basic strategy differences between a 7-deck shoe and a 10-deck shoe would you not play a hand dealt from the remaining 26 cards as if they had originated from a 7-deck shoe even if they had originated from a 10 deck shoe?
Yes, the same set of combinations apply, but with different probabilities. For instance, with 8 decks the probability the 2 cards are aces is 0.0057. For two decks, the probability is 0.0052. I'm not sure, but I think that's related to QFIT's conclusion that basic strategy for a shoe doesn't change based on discards.Quote: gordonm888Reducto ad absurdo: you have 2 cards left from a shoe. The possible combinations of cards for those two cards is identical whether they originated from a 2-deck shoe or an 8 deck shoe.
Therefore: say you have only 26 cards remaining in a shoe from which to deal a last BJ hand. Given that suits are irrelevant in BJ, is the probability for the composition(ranks) of these 26 cards any different if the 26 cards are the remnant of a 7-deck shoe vs being the remnant of a 10 deck shoe? (I think the answer is No.)
Therefore, if there were basic strategy differences between a 7-deck shoe and a 10-deck shoe would you not play a hand dealt from the remaining 26 cards as if they had originated from a 7-deck shoe even if they had originated from a 10 deck shoe?