Tavish Srivastava, co-founder and Chief Strategy Officer of Analytics Vidhya, is an IIT Madras graduate and a passionate data-science professional with 8+ years of diverse experience in markets including the US, India and Singapore, domains including Digital Acquisitions, Customer Servicing and Customer Management, and industry including Retail Banking, Credit Cards and Insurance.
| Great work Wiz! | 7 votes (58.33%) |
| Meh. | 1 vote (8.33%) |
| I don't understand a word you're saying. | 1 vote (8.33%) |
| How does this match up with Griffin? | 2 votes (16.66%) |
| I think you should do a Benford test. | No votes (0%) |
| You should assume different rules. | No votes (0%) |
| I disagree with the 0.563%. | No votes (0%) |
| I prefer to combinatorial analysis. | No votes (0%) |
| I play 6-5 blackjack and make all the side bets. | No votes (0%) |
| The Queen's Gambit. | 3 votes (25%) |
12 members have voted
WizardAdministrator
This post shall address my initial analysis, which I open up for peer review.
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- Dealer hits soft 17
- Blackjack pays 3 to 2
- Dealer peeks for blackjack
- Player may double after a split
- Player may NOT surrender
- Player may re-split up to four hands, including aces
- Continuous shuffler used
- Total-dependent basic strategy followed
My house edge calculator shows the house edge as follows:
- One deck = 0.014%
- Eight decks = 0.577%
This makes the effect of the difference in number of decks 0.563%.
Before I go on, all figures in this analysis are the result of billions of hands played by random simulation.
Let's work on this like an onion, starting with a simplistic balanced version of blackjack and then add the effects of the rules that are not equal both ways.
Imagine a blackjack game where the player follows a 'mimic the dealer' strategy of hitting to 17 and also hitting a soft 17. Also let a winning blackjack pays 1-1. Finally, add a rule that if both the player and dealer bust, that the result is a push. This game is perfectly balanced for a zero house edge.
Next, let's remove the rule about a push if both player and dealer bust. Instead, we use the real blackjack rule that if both the player and dealer bust, then the dealer wins. Here is the player expected value adding that rule for one and eight decks:
- One deck = -8.237%
- Eight decks = -8.157%
- Difference = -0.079%.
Why does the player lose more with a single deck? It's because there is more busting on both sides in a single-deck game, resulting in more double busts. The absolute value of the percentages above, to be exact. Why is there more busting with a single deck? I figure it's because both sides are hitting hands with more small cards than large, resulting in hitting into a deck/shoe that is high-card rich. This effect of removal is simply stronger with a single deck.
Next, let's consider the value of a blackjack paying 3-2. It is easy to calculate the effect of this mathematically, as follows:
| Decks | Prob player BJ | Prob dealer BJ | Prob winning BJ | Value |
|---|---|---|---|---|
| 1 | 4.827% | 3.673% | 4.649% | 2.325% |
| 8 | 4.745% | 4.605% | 4.527% | 2.263% |
| Diff | 0.081% | -0.932% | 0.123% | 0.061% |
Here is the effect of the two rule changes thus far. So, the eight-deck game is still a little better.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Total = -0.018%
Next, let's improve the player's strategy by using basic strategy with a hard total of 12 to 16, which is:
Player has 13 to 16 vs. dealer 2-6 = Stand
Player has 12 vs. dealer 4-6 = Stand
Deck Estimator Blackjack Free
Otherwise, hit with hard 12 to 16.
Here is the effect of that strategy improvement.
- One deck = 3.703%
- Eight decks = 3.270%
- Difference = 0.433%
So this change in strategy is more helpful with a single deck. I assume this is because hitting with 12 to 16 is more dangerous in a single-deck game, because the 12 to 16 are likely composed of more small cards than large, resulting in hitting into a deck/shoe rich in high cards, thus more busting. In other words, the effect of removal again.
Here is where we are at now of the reason a single-deck game is better for the player.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Total = 0.415%
Next, let's add proper doubling strategy. We will use the appropriate basic strategy for the given number of decks. Here is the value of adding doubling the player's game.
- One deck = 1.653%
- Eight decks = 1.380%
- Difference = 0.273%
Doubling is thus much more valuable in a single-deck game than a shoe. I figure it's because the player likely has two small cards when doubling, like a 6 and 5, resulting on doubling into a shoe/deck that is ten-card rich. Again, this effect of removal is stronger with one deck than eight.
Let's update our effect list:
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Strategic doubling = 0.273%
- Total = 0.689%
The only thing left to change (remember, we are not allowing surrender) is the splitting rule. Here is the effect of adding the appropriate splitting strategy:
- One deck = 0.544%
- Eight decks = 0.669%
- Difference = -0.125%
Interesting! Splitting is more effective in an 8-deck shoe. This should no be surprising as it is harder to form a pair in a single deck game than a shoe. Be be specific, the probabilitiy of getting any pair in two cards is 1/17 = 5.882%. In an eight deck shoe it is 7.470%. Not to get off topic, but it interesting how little splitting actually helps the player. So, there is more splitting and re-splitting going on in the shoe game.
Let's update our table again.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Strategic doubling = 0.273%
- Strategic splitting = -0.125%
- Total = 0.563%
As a reminder, the difference from my blackjack house edge calculator matches the 0.563%. Yay!
I love graphs, so here is one of the various effects I looked at.
Again, I plan to write this all up in a more formal way and will indclude more details. For now, I open it up to comments and questions. I've been working on this for about a week, so hope to get some feedback.
I would like to thank Don Schlesinger for his help and advice.
Again, I plan to write this all up in a more formal way and will indclude more details. For now, I open it up to comments and questions. I've been working on this for about a week, so hope to get some feedback.
Feedback, well lets use a story. Way back, like 2010, I was on a crew with a guy was no BJ dummy. His dad was a mill guy had to take a late in life dealer job. He knew all the basics. and was a good craps student the night I later taught him. The subject of continuous shufflers came up probably over breakfast or looking at oil leases in the courthouse.
Anyhow, I tried to explain how a CSM gives a slight player edge, having spent more time studying WoV and WoO than on any course I took in college. I kept telling him to read the article on it. A couple weeks of this and I sent him the link and he promised to read it.
At breakfast the next morning he said, 'Bro, that was way deeper than I imagined.' I had warned him it was, but he still just kind of took it on faith.
'Bro, that was way deeper than I imagined!'
I read this a few times but am still mostly just taking the math on faith. The only thing I could easily explain to a table at a party night is more chance to split because there are more matched cards with more decks. If you explained more of it like that it would be easier to understand.
Picture how I would have to explain it. At a table of people who have probably half of them played before, and most who are not 'math people.'
The first question I would get would be what is the difference with more decks since the ratio of cards is the same just more. A clump of small cards is as possible as a clump of large cards. At least that is how it appears.
Maybe show how things differ going form 1 to 2 decks, then build on that. Then look at some random hands. Say player 10, 12, and 16. What are my probabilities of hitting 12 on 1 deck vs 2. Ditto 16. Think how many ways to bust a 16 with one deck then 2. How many ways to improve on 1 deck then 2. If I had something that simple, even if for just say 12 and 16 for 1 vs 2 decks I could explain it to people at a table.
IOW, can you show it with cards and not decimals?
Just a thought.
Administrator
Second, it's always been my policy to teach to near the top. If you put everybody on the bell curve, I try to aim at the person about one standard deviation above average (meaning 84% are less intelligent and 16% are more). There are already plenty of advanced books for those 2 or 3 standard deviations above the norm and plenty of sources for those who need things simplified as much as possible. I aim for those in between.
The topic at hand is complicated and difficult to dumb down. Other sources on this question tend to emphasize that blackjacks are more likely the fewer the decks and the player gets that extra half unit on a winning blackjack. That is pretty easy to explain, but it accounts for about 11% of the total effect of the number of decks only.
Nevertheless, I appreciate the suggestion.
I will read Wizards post several more times in the coming days and weeks, and it is possible some things might click and make sense as time goes on, but the majority of it will remain a foreign language to me. :/
Same thing happened with Don's BJA3. First time I read it....foreign language. Understood almost none of the math, But I applied many of the conclusions. BJA3 happens to be my Bible. I have re-read that 100 times over the years and frequently reference different things, and every once in a while something 'clicks' and I say oh yeah that makes sense, but even still most of the math is foreign to me. I am probably the stupidest, most math challenged person on this forum or in the blackjack community. I'm ok with that.
Kind of like a car engine. I drive a vehicle, but if you tell me that the spark tube (The Office mafia episode reference) is bad and needs to be replaced, you are talking a foreign language.
so in another thread Wizard suggested I take a look at this thread. And so I have. I have to say none of the conclusions surprised me. I knew most of that. But the math to get there....that is another matter. Despite that I play blackjack for a living, I am not a math guy nor understand many of the math and formulas behind things. I take what other, much smarter people than me conclude and apply it. People like Wizard, Don Schlesinger, Doghand and many others.
I will read Wizards post several more times in the coming days and weeks, and it is possible some things might click and make sense as time goes on, but the majority of it will remain a foreign language to me. :/
Same thing happened with Don's BJA3. First time I read it....foreign language. Understood almost none of the math, But I applied many of the conclusions. BJA3 happens to be my Bible. I have re-read that 100 times over the years and frequently reference different things, and every once in a while something 'clicks' and I say oh yeah that makes sense, but even still most of the math is foreign to me. I am probably the stupidest, most math challenged person on this forum or in the blackjack community. I'm ok with that.
Kind of like a car engine. I drive a vehicle, but if you tell me that the spark tube (The Office mafia episode reference) is bad and needs to be replaced, you are talking a foreign language.
What about Griffin's Theory of Blackjack? Fuggedaboudit!
What about Griffin's Theory of Blackjack? Fuggedaboudit!
I'm still waiting on the English language version to be published.
Busting rule? Not a hint that is a problem!
In Craps, when playing the darkside, the idea that a 12 rolled on the comeout is a push instead of a win is the real killer is about as hidden an effect as you can possibly experience. I in fact have experienced some awful sessions where a 12 was rolled on the comeout maybe only once the entire time. It is so counter-intuitive to grasp the real reasons the house has an edge in these things!
Administrator
It is so counter-intuitive to grasp the real reasons the house has an edge in these things!
That is one of the essences of a well-designed casino game -- the player doesn't see where the house has the advantage. In blackjack and most poker games it is a player positional disadvantage. The idea to remove the 10's in Spanish 21 was also a good one.
I get asked from time to time by the other side about how to significantly increase the house edge, like by 1%, in blackjack without changing the 3-2 payoff, mandating a side bet, charging a commission on wins, nor change the deck composition in a way the average player won't notice. Nothing good has ever come to mind.
That is one of the essences of a well-designed casino game -- the player doesn't see where the house has the advantage. In blackjack and most poker games it is a player positional disadvantage. The idea to remove the 10's in Spanish 21 was also a good one.
I get asked from time to time by the other side about how to significantly increase the house edge, like by 1%, in blackjack without changing the 3-2 payoff, mandating a side bet, charging a commission on wins, nor change the deck composition in a way the average player won't notice. Nothing good has ever come to mind.
By Ion Saliu, Founder of Blackjack Mathematics
I. Probability, Odds for a Blackjack or Natural 21
II. House Edge on Insurance Bet at Blackjack
III. Calculate Double-Down Hands
IV. Calculate Blackjack Pairs: Strict or Mixed Ten-Cards
V. Free Blackjack Resources, Basic Strategy, Casino Gambling Systems
1.1. Calculate Probability (Odds) for a Blackjack or Natural 21
First capture by the WayBack Machine
I have seen lots of search strings in the statistics of my Web site related to the probability to get a blackjack (natural 21). This time (November 15, 2012), the request (repeated 5 times) was personal and targeted directly at yours truly:
- 'In the game of blackjack determine the probability of dealing yourself a blackjack (ace face-card or ten) from a single deck. Show how you arrived at your answer. If you are not sure post an idea to get us started!'
Oh, yes, I am very sure! As specified in this eBook, the blackjack hands can be viewed as combinations or arrangements (the order of the elements counts; like in horse racing trifectas).
1) Let's take first the combinations. There are 52 cards in one deck of cards. There are 4 Aces and 16 face-cards and 10s. The blackjack (or natural) can occur only in the first 2 cards. We calculate first all combinations of 52 elements taken 2 at a time: C(52, 2) = (52 * 51) / 2 = 1326.
We combine now each of the 4 Aces with each of the 16 ten-valued cards: 4 * 16 = 64.
The probability to get a blackjack (natural): 64 / 1326 = .0483 = 4.83%.
2) Let's do now the calculations for arrangements. (The combinations are also considered boxed arrangements; i.e. the order of the elements does not count).
We calculate total arrangements for 52 cards taken 2 at a time: A(52, 2) = 52 * 51 = 2652.
In arrangements, the order of the cards is essential. Thus, King + Ace is distinct from Ace + King. Thus, total arrangements of 4 Aces and 16 ten-valued cards: 4 * 16 * 2 = 128.
The odds to get a blackjack (natural) as arrangement: 128 / 2652 = .0483 = 4.83%.
4.83% is equivalent to about 1 in 21 blackjack hands. (No wonder the game is called Twenty-one!)
Calculations for the Number of Cards Left in the Deck, Number of Decks
There were questions regarding the number of cards left in the deck, number of decks, number of players, even the position at the table.1) The previous probability calculations were based on one deck of cards, at the beginning of the deck (no cards burnt). But we can easily calculate the blackjack (natural) odds for partial decks, provided that we know the number of remaining cards (total), Aces and Ten-Value cards.
Let's take the situation heads-up: One player against the dealer. Suppose that 12 cards were played, including 2 Tens; no Aces out. What is the new probability to get a natural blackjack?
Total cards remaining (R) = 52 - 12 = 40
Aces remaining in the deck (A): 4 - 0 = 4
Ten-Valued cards remaining (T): 16 - 2 = 14
Odds of a natural: (4 * 14) / C(40, 2) = 56 / 780 = 7.2%
(C represents the combination formula; e.g. combinations of 40 taken 2 at a time.)
The probability for a blackjack is higher than at the beginning of a full deck of cards. The odds are exactly the same for both Player and Dealer. But - the advantage goes to the Player! If the Player has the BJ and the Dealer doesn't, the Player is paid 150%. If the Dealer has the blackjack and the Player doesn't, the Player loses 100% of his initial bet!
This situation is valid only for one Player against casino. Also, this situation allows for a higher bet before the round starts. For multiple players, the situation becomes uncontrollable. Everybody at the table receives one card in succession, and then the second card. The bet cannot be increased during the dealing of the cards. Hint: try as much as you can to play heads-up against the Dealer!
The generalized formula is:
Probability of a blackjack: (A * T) / C(R, 2)
2) How about multiple decks of cards? The calculations are not exactly linear because of the combination formula. For example, 2 decks, (104 cards):
~ the 2-deck case:
C(52, 2) = 1326
C(104, 2) = 5356 (4.04 times larger than total combinations for one deck.)
8 (Aces) * 32 (Tens) = 256
Odds of BJ for 2 decks = 256 / 5356 = 4.78% (a little lower than the one-deck case of 4.83%).
~ the 8-deck case, 416 total cards:
C(52, 2) = 1326
C(416, 2) = 86320 (65.1 times larger than total combinations for one deck.)
32 (Aces) * 128 (Tens) = 4096
Odds of BJ for 8 decks = 4096 / 86320 = 4.75% (a little lower than the two-deck situation and even lower than the one-deck case of 4.83%).
There are NO significant differences regarding the number of decks. If we round the figures, the general odds to get a natural blackjack can be expressed as 4.8%.
The advantage to the blackjack player after cards were played: Not nearly as significant as the one-deck situation.
3) The position at the table is inconsequential for the blackjack player. Only heads-up and one deck of cards make a difference as far the improved odds for a natural are concerned.
- Axiomatic one, let's cover all the bases, as it were. The original question was, exactly, as this: 'Dealing yourself a blackjack (Ace AND Face-card or Ten) from a single deck'. The calculations above are accurate for this unique situation: ONE player dealing cards to himself/herself. The odds of getting a natural blackjack are, undoubtedly, 1 in 21 hands (a hand consisting of exactly 2 cards).
- Such a case is non-existent in real-life gambling, however. There are at least TWO participants in a blackjack game: Dealer and one player. Is the probability for a natural blackjack the same – regardless of number of participants? NOT! The 21 hands (as in probability p = 1 / 21) are equally distributed among multiple game agents (or elements in probability theory). Mathematics — and software — to the rescue! We apply the formula known as exactly M successes in N trials. The best software for the task is known as SuperFormula (also component of the integrated Scientia software package).
- Undoubtedly, your chance to get a natural BJ is higher when playing heads-up against the dealer. The degree of certainty DC decreases with an increase in the number of players at the blackjack table. I did a few calculations: Heads-up (2 elements), 4 players and dealer (5 elements), 7 players and dealer (8 elements).
- The degree of certainty DC for 2 elements (one player and dealer), one success in 2 trials (2-card hands) is 9.1%; divided by 2 elements: the chance of a natural is 9.1% / 2 = 4.6% = the closest to the 'Dealing yourself a blackjack (Ace AND Face-card or Ten) from a single deck' situation.
- The chance for 5 elements (4 players and dealer), one success in 5 trials (2-card hands) is 19.6%; distributed among 5 elements, the degree of certainty DC for a blackjack natural is 19.6% / 5 = 3.9%.
- The probability for 8 elements (7 players and dealer), one success in 8 trials (2-card hands) is 27.1%; equally distributed among 8 elements, the degree of certainty DC of a blackjack natural is 27.1% / 8 = 3.4%.
- That's mathematics and nobody can manufacture extra BJ natural 21 hands... not even the staunchest and thickest card-counting system vendors! The PI... er, pie is small to begin with; the slices get smaller with more mouths at the table. Ever wondered why the casinos only offer alcohol for free — but no pizza?
1.2. Probability, Odds for a Blackjack Playing through a Deck of Cards
The probabilities in the first chapter were calculated for one trial. That is, the odds to get a blackjack in the first two cards. But what are the probabilities to get a natural 21 dealing an entire deck?
1.2.A. Dealing 2-card hands until the deck is dealt entirely
There are 52 cards in the deck. Total number of trials (2-card hands) is 52 / 2 = 26. SuperFormula probability software does the following calculation:- The probability of at least one success in 26 trials for an event of individual probability p=0.0483 is 72.39%.
1.2.B. Dealing 2-card hands in heads-up play until the deck is dealt entirely
There are 52 cards in the deck. We are now in the simplest real-life situation: heads-up play. There is one player and the dealer in the game. We suppose an average of 6 cards dealt in one round. Total number of trials in this case is equivalent to the number of rounds played. 52 / 6 makes approximately 9 rounds per deck. SuperFormula does the following calculation:- The probability of at least one success in 9 trials for an event of individual probability p=0.0483 is 35.95%.
You, the player, can expect one blackjack every 3 decks in heads-up play.
2. House Edge on the Insurance Bet at Blackjack
'Insurance, anyone?' you can hear the dealer when her face card is an Ace. Players can choose to insure their hands against a potential dealer's natural. The player is allowed to bet half of his initial bet. Is insurance a good side bet in blackjack? What are the odds? What is the house edge for insurance? As in the case of calculating the odds for a natural blackjack, the situation is fluid. The
This post shall address my initial analysis, which I open up for peer review.
- Start your casino journey with an amazing 200% up to £400, plus 100 spins Blackjack Deck Estimator at Betfair! Play Casino, Live casino, Sportsbook and Poker, all under one roof.
- Start your casino journey with an amazing 200% up to £400, plus 100 spins Blackjack Deck Estimator at Betfair! Play Casino, Live casino, Sportsbook and Poker, all under one roof. Play Casino, Live casino, Sportsbook and Poker, all under one roof.
- Dealer hits soft 17
- Blackjack pays 3 to 2
- Dealer peeks for blackjack
- Player may double after a split
- Player may NOT surrender
- Player may re-split up to four hands, including aces
- Continuous shuffler used
- Total-dependent basic strategy followed
My house edge calculator shows the house edge as follows:
- One deck = 0.014%
- Eight decks = 0.577%
This makes the effect of the difference in number of decks 0.563%.
Before I go on, all figures in this analysis are the result of billions of hands played by random simulation.
Let's work on this like an onion, starting with a simplistic balanced version of blackjack and then add the effects of the rules that are not equal both ways.
Imagine a blackjack game where the player follows a 'mimic the dealer' strategy of hitting to 17 and also hitting a soft 17. Also let a winning blackjack pays 1-1. Finally, add a rule that if both the player and dealer bust, that the result is a push. This game is perfectly balanced for a zero house edge.
Next, let's remove the rule about a push if both player and dealer bust. Instead, we use the real blackjack rule that if both the player and dealer bust, then the dealer wins. Here is the player expected value adding that rule for one and eight decks:
- One deck = -8.237%
- Eight decks = -8.157%
- Difference = -0.079%.
Why does the player lose more with a single deck? It's because there is more busting on both sides in a single-deck game, resulting in more double busts. The absolute value of the percentages above, to be exact. Why is there more busting with a single deck? I figure it's because both sides are hitting hands with more small cards than large, resulting in hitting into a deck/shoe that is high-card rich. This effect of removal is simply stronger with a single deck.
Next, let's consider the value of a blackjack paying 3-2. It is easy to calculate the effect of this mathematically, as follows:
| Decks | Prob player BJ | Prob dealer BJ | Prob winning BJ | Value |
|---|---|---|---|---|
| 1 | 4.827% | 3.673% | 4.649% | 2.325% |
| 8 | 4.745% | 4.605% | 4.527% | 2.263% |
| Diff | 0.081% | -0.932% | 0.123% | 0.061% |
Here is the effect of the two rule changes thus far. So, the eight-deck game is still a little better.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Total = -0.018%
Next, let's improve the player's strategy by using basic strategy with a hard total of 12 to 16, which is:
Player has 13 to 16 vs. dealer 2-6 = Stand
Player has 12 vs. dealer 4-6 = Stand
Deck Estimator Blackjack Free
Otherwise, hit with hard 12 to 16.
Here is the effect of that strategy improvement.
- One deck = 3.703%
- Eight decks = 3.270%
- Difference = 0.433%
So this change in strategy is more helpful with a single deck. I assume this is because hitting with 12 to 16 is more dangerous in a single-deck game, because the 12 to 16 are likely composed of more small cards than large, resulting in hitting into a deck/shoe rich in high cards, thus more busting. In other words, the effect of removal again.
Here is where we are at now of the reason a single-deck game is better for the player.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Total = 0.415%
Next, let's add proper doubling strategy. We will use the appropriate basic strategy for the given number of decks. Here is the value of adding doubling the player's game.
- One deck = 1.653%
- Eight decks = 1.380%
- Difference = 0.273%
Doubling is thus much more valuable in a single-deck game than a shoe. I figure it's because the player likely has two small cards when doubling, like a 6 and 5, resulting on doubling into a shoe/deck that is ten-card rich. Again, this effect of removal is stronger with one deck than eight.
Let's update our effect list:
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Strategic doubling = 0.273%
- Total = 0.689%
The only thing left to change (remember, we are not allowing surrender) is the splitting rule. Here is the effect of adding the appropriate splitting strategy:
- One deck = 0.544%
- Eight decks = 0.669%
- Difference = -0.125%
Interesting! Splitting is more effective in an 8-deck shoe. This should no be surprising as it is harder to form a pair in a single deck game than a shoe. Be be specific, the probabilitiy of getting any pair in two cards is 1/17 = 5.882%. In an eight deck shoe it is 7.470%. Not to get off topic, but it interesting how little splitting actually helps the player. So, there is more splitting and re-splitting going on in the shoe game.
Let's update our table again.
- Player loses if both bust = -0.079%
- Blackjack pays 3-2 = 0.061%
- Strategic standing on 12 to 16 = 0.433%
- Strategic doubling = 0.273%
- Strategic splitting = -0.125%
- Total = 0.563%
As a reminder, the difference from my blackjack house edge calculator matches the 0.563%. Yay!
I love graphs, so here is one of the various effects I looked at.
Again, I plan to write this all up in a more formal way and will indclude more details. For now, I open it up to comments and questions. I've been working on this for about a week, so hope to get some feedback.
I would like to thank Don Schlesinger for his help and advice.
Again, I plan to write this all up in a more formal way and will indclude more details. For now, I open it up to comments and questions. I've been working on this for about a week, so hope to get some feedback.
Feedback, well lets use a story. Way back, like 2010, I was on a crew with a guy was no BJ dummy. His dad was a mill guy had to take a late in life dealer job. He knew all the basics. and was a good craps student the night I later taught him. The subject of continuous shufflers came up probably over breakfast or looking at oil leases in the courthouse.
Anyhow, I tried to explain how a CSM gives a slight player edge, having spent more time studying WoV and WoO than on any course I took in college. I kept telling him to read the article on it. A couple weeks of this and I sent him the link and he promised to read it.
At breakfast the next morning he said, 'Bro, that was way deeper than I imagined.' I had warned him it was, but he still just kind of took it on faith.
'Bro, that was way deeper than I imagined!'
I read this a few times but am still mostly just taking the math on faith. The only thing I could easily explain to a table at a party night is more chance to split because there are more matched cards with more decks. If you explained more of it like that it would be easier to understand.
Picture how I would have to explain it. At a table of people who have probably half of them played before, and most who are not 'math people.'
The first question I would get would be what is the difference with more decks since the ratio of cards is the same just more. A clump of small cards is as possible as a clump of large cards. At least that is how it appears.
Maybe show how things differ going form 1 to 2 decks, then build on that. Then look at some random hands. Say player 10, 12, and 16. What are my probabilities of hitting 12 on 1 deck vs 2. Ditto 16. Think how many ways to bust a 16 with one deck then 2. How many ways to improve on 1 deck then 2. If I had something that simple, even if for just say 12 and 16 for 1 vs 2 decks I could explain it to people at a table.
IOW, can you show it with cards and not decimals?
Just a thought.
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Second, it's always been my policy to teach to near the top. If you put everybody on the bell curve, I try to aim at the person about one standard deviation above average (meaning 84% are less intelligent and 16% are more). There are already plenty of advanced books for those 2 or 3 standard deviations above the norm and plenty of sources for those who need things simplified as much as possible. I aim for those in between.
The topic at hand is complicated and difficult to dumb down. Other sources on this question tend to emphasize that blackjacks are more likely the fewer the decks and the player gets that extra half unit on a winning blackjack. That is pretty easy to explain, but it accounts for about 11% of the total effect of the number of decks only.
Nevertheless, I appreciate the suggestion.
I will read Wizards post several more times in the coming days and weeks, and it is possible some things might click and make sense as time goes on, but the majority of it will remain a foreign language to me. :/
Same thing happened with Don's BJA3. First time I read it....foreign language. Understood almost none of the math, But I applied many of the conclusions. BJA3 happens to be my Bible. I have re-read that 100 times over the years and frequently reference different things, and every once in a while something 'clicks' and I say oh yeah that makes sense, but even still most of the math is foreign to me. I am probably the stupidest, most math challenged person on this forum or in the blackjack community. I'm ok with that.
Kind of like a car engine. I drive a vehicle, but if you tell me that the spark tube (The Office mafia episode reference) is bad and needs to be replaced, you are talking a foreign language.
so in another thread Wizard suggested I take a look at this thread. And so I have. I have to say none of the conclusions surprised me. I knew most of that. But the math to get there....that is another matter. Despite that I play blackjack for a living, I am not a math guy nor understand many of the math and formulas behind things. I take what other, much smarter people than me conclude and apply it. People like Wizard, Don Schlesinger, Doghand and many others.
I will read Wizards post several more times in the coming days and weeks, and it is possible some things might click and make sense as time goes on, but the majority of it will remain a foreign language to me. :/
Same thing happened with Don's BJA3. First time I read it....foreign language. Understood almost none of the math, But I applied many of the conclusions. BJA3 happens to be my Bible. I have re-read that 100 times over the years and frequently reference different things, and every once in a while something 'clicks' and I say oh yeah that makes sense, but even still most of the math is foreign to me. I am probably the stupidest, most math challenged person on this forum or in the blackjack community. I'm ok with that.
Kind of like a car engine. I drive a vehicle, but if you tell me that the spark tube (The Office mafia episode reference) is bad and needs to be replaced, you are talking a foreign language.
What about Griffin's Theory of Blackjack? Fuggedaboudit!
What about Griffin's Theory of Blackjack? Fuggedaboudit!
I'm still waiting on the English language version to be published.
Maybe it's just me, but I never quite get over how when you are actually playing you get no sense that any of this is in play. For me in fact I get no sense whatsoever while playing BJ that the real advantage the dealer has is the rule about who busts first. Instead, I might get irritated by how many BJs he gets, how he can draw a bunch of small cards when he needs them, or how often he deals me a 12-16 hand while he so often gets great starting cards. Yeah, 'seemingly' should have been inserted in all the above. The sense I get, though, it doesn't seem like 'seemingly'
Busting rule? Not a hint that is a problem!
In Craps, when playing the darkside, the idea that a 12 rolled on the comeout is a push instead of a win is the real killer is about as hidden an effect as you can possibly experience. I in fact have experienced some awful sessions where a 12 was rolled on the comeout maybe only once the entire time. It is so counter-intuitive to grasp the real reasons the house has an edge in these things!
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It is so counter-intuitive to grasp the real reasons the house has an edge in these things!
That is one of the essences of a well-designed casino game -- the player doesn't see where the house has the advantage. In blackjack and most poker games it is a player positional disadvantage. The idea to remove the 10's in Spanish 21 was also a good one.
I get asked from time to time by the other side about how to significantly increase the house edge, like by 1%, in blackjack without changing the 3-2 payoff, mandating a side bet, charging a commission on wins, nor change the deck composition in a way the average player won't notice. Nothing good has ever come to mind.
That is one of the essences of a well-designed casino game -- the player doesn't see where the house has the advantage. In blackjack and most poker games it is a player positional disadvantage. The idea to remove the 10's in Spanish 21 was also a good one.
I get asked from time to time by the other side about how to significantly increase the house edge, like by 1%, in blackjack without changing the 3-2 payoff, mandating a side bet, charging a commission on wins, nor change the deck composition in a way the average player won't notice. Nothing good has ever come to mind.
By Ion Saliu, Founder of Blackjack Mathematics
I. Probability, Odds for a Blackjack or Natural 21
II. House Edge on Insurance Bet at Blackjack
III. Calculate Double-Down Hands
IV. Calculate Blackjack Pairs: Strict or Mixed Ten-Cards
V. Free Blackjack Resources, Basic Strategy, Casino Gambling Systems
1.1. Calculate Probability (Odds) for a Blackjack or Natural 21
First capture by the WayBack Machine (web.archive.org) Sectember (Sect Month) 1, 2015.I have seen lots of search strings in the statistics of my Web site related to the probability to get a blackjack (natural 21). This time (November 15, 2012), the request (repeated 5 times) was personal and targeted directly at yours truly:
- 'In the game of blackjack determine the probability of dealing yourself a blackjack (ace face-card or ten) from a single deck. Show how you arrived at your answer. If you are not sure post an idea to get us started!'
Oh, yes, I am very sure! As specified in this eBook, the blackjack hands can be viewed as combinations or arrangements (the order of the elements counts; like in horse racing trifectas).
1) Let's take first the combinations. There are 52 cards in one deck of cards. There are 4 Aces and 16 face-cards and 10s. The blackjack (or natural) can occur only in the first 2 cards. We calculate first all combinations of 52 elements taken 2 at a time: C(52, 2) = (52 * 51) / 2 = 1326.
We combine now each of the 4 Aces with each of the 16 ten-valued cards: 4 * 16 = 64.
The probability to get a blackjack (natural): 64 / 1326 = .0483 = 4.83%.
2) Let's do now the calculations for arrangements. (The combinations are also considered boxed arrangements; i.e. the order of the elements does not count).
We calculate total arrangements for 52 cards taken 2 at a time: A(52, 2) = 52 * 51 = 2652.
In arrangements, the order of the cards is essential. Thus, King + Ace is distinct from Ace + King. Thus, total arrangements of 4 Aces and 16 ten-valued cards: 4 * 16 * 2 = 128.
The odds to get a blackjack (natural) as arrangement: 128 / 2652 = .0483 = 4.83%.
4.83% is equivalent to about 1 in 21 blackjack hands. (No wonder the game is called Twenty-one!)
Calculations for the Number of Cards Left in the Deck, Number of Decks
There were questions regarding the number of cards left in the deck, number of decks, number of players, even the position at the table.1) The previous probability calculations were based on one deck of cards, at the beginning of the deck (no cards burnt). But we can easily calculate the blackjack (natural) odds for partial decks, provided that we know the number of remaining cards (total), Aces and Ten-Value cards.
Let's take the situation heads-up: One player against the dealer. Suppose that 12 cards were played, including 2 Tens; no Aces out. What is the new probability to get a natural blackjack?
Total cards remaining (R) = 52 - 12 = 40
Aces remaining in the deck (A): 4 - 0 = 4
Ten-Valued cards remaining (T): 16 - 2 = 14
Odds of a natural: (4 * 14) / C(40, 2) = 56 / 780 = 7.2%
(C represents the combination formula; e.g. combinations of 40 taken 2 at a time.)
The probability for a blackjack is higher than at the beginning of a full deck of cards. The odds are exactly the same for both Player and Dealer. But - the advantage goes to the Player! If the Player has the BJ and the Dealer doesn't, the Player is paid 150%. If the Dealer has the blackjack and the Player doesn't, the Player loses 100% of his initial bet!
This situation is valid only for one Player against casino. Also, this situation allows for a higher bet before the round starts. For multiple players, the situation becomes uncontrollable. Everybody at the table receives one card in succession, and then the second card. The bet cannot be increased during the dealing of the cards. Hint: try as much as you can to play heads-up against the Dealer!
The generalized formula is:
Probability of a blackjack: (A * T) / C(R, 2)
2) How about multiple decks of cards? The calculations are not exactly linear because of the combination formula. For example, 2 decks, (104 cards):
~ the 2-deck case:
C(52, 2) = 1326
C(104, 2) = 5356 (4.04 times larger than total combinations for one deck.)
8 (Aces) * 32 (Tens) = 256
Odds of BJ for 2 decks = 256 / 5356 = 4.78% (a little lower than the one-deck case of 4.83%).
~ the 8-deck case, 416 total cards:
C(52, 2) = 1326
C(416, 2) = 86320 (65.1 times larger than total combinations for one deck.)
32 (Aces) * 128 (Tens) = 4096
Odds of BJ for 8 decks = 4096 / 86320 = 4.75% (a little lower than the two-deck situation and even lower than the one-deck case of 4.83%).
There are NO significant differences regarding the number of decks. If we round the figures, the general odds to get a natural blackjack can be expressed as 4.8%.
The advantage to the blackjack player after cards were played: Not nearly as significant as the one-deck situation.
3) The position at the table is inconsequential for the blackjack player. Only heads-up and one deck of cards make a difference as far the improved odds for a natural are concerned.
- Axiomatic one, let's cover all the bases, as it were. The original question was, exactly, as this: 'Dealing yourself a blackjack (Ace AND Face-card or Ten) from a single deck'. The calculations above are accurate for this unique situation: ONE player dealing cards to himself/herself. The odds of getting a natural blackjack are, undoubtedly, 1 in 21 hands (a hand consisting of exactly 2 cards).
- Such a case is non-existent in real-life gambling, however. There are at least TWO participants in a blackjack game: Dealer and one player. Is the probability for a natural blackjack the same – regardless of number of participants? NOT! The 21 hands (as in probability p = 1 / 21) are equally distributed among multiple game agents (or elements in probability theory). Mathematics — and software — to the rescue! We apply the formula known as exactly M successes in N trials. The best software for the task is known as SuperFormula (also component of the integrated Scientia software package).
- Undoubtedly, your chance to get a natural BJ is higher when playing heads-up against the dealer. The degree of certainty DC decreases with an increase in the number of players at the blackjack table. I did a few calculations: Heads-up (2 elements), 4 players and dealer (5 elements), 7 players and dealer (8 elements).
- The degree of certainty DC for 2 elements (one player and dealer), one success in 2 trials (2-card hands) is 9.1%; divided by 2 elements: the chance of a natural is 9.1% / 2 = 4.6% = the closest to the 'Dealing yourself a blackjack (Ace AND Face-card or Ten) from a single deck' situation.
- The chance for 5 elements (4 players and dealer), one success in 5 trials (2-card hands) is 19.6%; distributed among 5 elements, the degree of certainty DC for a blackjack natural is 19.6% / 5 = 3.9%.
- The probability for 8 elements (7 players and dealer), one success in 8 trials (2-card hands) is 27.1%; equally distributed among 8 elements, the degree of certainty DC of a blackjack natural is 27.1% / 8 = 3.4%.
- That's mathematics and nobody can manufacture extra BJ natural 21 hands... not even the staunchest and thickest card-counting system vendors! The PI... er, pie is small to begin with; the slices get smaller with more mouths at the table. Ever wondered why the casinos only offer alcohol for free — but no pizza?
1.2. Probability, Odds for a Blackjack Playing through a Deck of Cards
The probabilities in the first chapter were calculated for one trial. That is, the odds to get a blackjack in the first two cards. But what are the probabilities to get a natural 21 dealing an entire deck?
1.2.A. Dealing 2-card hands until the deck is dealt entirely
There are 52 cards in the deck. Total number of trials (2-card hands) is 52 / 2 = 26. SuperFormula probability software does the following calculation:- The probability of at least one success in 26 trials for an event of individual probability p=0.0483 is 72.39%.
1.2.B. Dealing 2-card hands in heads-up play until the deck is dealt entirely
There are 52 cards in the deck. We are now in the simplest real-life situation: heads-up play. There is one player and the dealer in the game. We suppose an average of 6 cards dealt in one round. Total number of trials in this case is equivalent to the number of rounds played. 52 / 6 makes approximately 9 rounds per deck. SuperFormula does the following calculation:- The probability of at least one success in 9 trials for an event of individual probability p=0.0483 is 35.95%.
You, the player, can expect one blackjack every 3 decks in heads-up play.
2. House Edge on the Insurance Bet at Blackjack
'Insurance, anyone?' you can hear the dealer when her face card is an Ace. Players can choose to insure their hands against a potential dealer's natural. The player is allowed to bet half of his initial bet. Is insurance a good side bet in blackjack? What are the odds? What is the house edge for insurance? As in the case of calculating the odds for a natural blackjack, the situation is fluid. The odds and therefore the house edge are proportionately dependent on the amount of 10-valued cards and total remaining cards in the deck.We can devise precise mathematical formulas based on the Tens remaining in the deck. We know for sure that the casino pays 2 to 1 for a successful insurance (i.e. the dealer does have Ten as her hole card).
We start with the most easily manageable case: One deck of cards, one player, the very beginning of the game. There is a total of 16 Teens in the deck (10, J, Q, K). The dealer has dealt 2 cards to the player and one card to herself that we can see exactly — the face card being an Ace. Therefore, 52 – 3 = 49 cards remaining in the deck. There are 3 possible situations, axiomatic one:
- 1) The player has 2 non-ten cards; there are 16 Teens in the deck = the favorable situations to the player if taking insurance. There are 49 – 16 = 33 unfavorable cards to insurance. However, the 16 favorable cards amount to 32, as the insurance pays 2 to 1. The balance is 33 – 32 = +1 unfavorable situation to the player but favorable to the casino (the + sign indicates a casino edge). In this case, there is a house advantage of 1/49 = 2%.
- 2) The player has 1 Ten and 1 non-ten card; there are 15 Teens remaining in the deck = the favorable situations to the player if taking insurance. There are 49 – 15 = 34 unfavorable cards to insurance. However, the 15 favorable cards amount to 30, as the insurance pays 2 to 1. The balance is 34 – 30 = +4 unfavorable situations to the player but favorable to the casino. In this case, there is a house advantage of 4/49 = 8%.
- This can be also the case of insuring one's blackjack natural: an 8% disadvantage for the player.
- This figure of 8% represents the average house edge regarding the insurance bet. I did calculations for various situations — number of decks and number of players.
- 3) The player has 2 Ten-count cards; there are 14 Teens in the deck = the favorable situations to the player if taking insurance. There are 49 – 14 = 35 unfavorable cards to insurance. However, the 14 favorable cards amount to 28, as the insurance pays 2 to 1. The balance is 35 – 28 = +7 unfavorable situations to the player but favorable to the casino. In this case, there is a house advantage of 7/49 = 14%. This is the worst-case scenario: The player should never — ever — even think about insurance with that strong hand of 2 Tens!
Believe it or not, the insurance can be a really sweet deal if there are multiple players at the blackjack table! Let's say, 5 players, the very beginning of the game. There is a total of 16 Teens in the deck (10, J, Q, K). The dealer has dealt 10 cards to the players and one card to herself that we can see exactly — the face card being an Ace. Therefore, 52 – (10 + 1) = 41 cards remaining in the deck. There are many more possible situations, some very different from the previous scenario:
- 1) The players have 10 non-ten cards; there are still 16 Tens in the deck = the favorable situations to the player if taking insurance. There are 41 – 16 = 25 unfavorable cards to insurance. However, the 16 favorable cards amount to 32, as the insurance pays 2 to 1. The balance is 25 – 32 = –7 favorable situation to the player but unfavorable to the casino (the – sign indicates a player advantage now). In this case, there is a house advantage of 7/41 = –17%. The Player has a whopping 17% advantage if taking insurance in a case like this one!
- 2) The players have 10 Ten-count cards; there are 6 Teens in the deck = the favorable situations to the player if taking insurance. There are 41 – 6 = 35 unfavorable cards to insurance. However, the 6 favorable cards amount to 12, as the insurance pays 2 to 1. The balance is 35 – 12 = +23 unfavorable situations to the player but favorable to the casino. In this case, there is a house advantage of 23/41 = 56%. This is the worst-case scenario: None of the players should ever even think about insurance with those strong hands of 2 Tens per capita!
- 3) Applying the wise aurea mediocritas adagio, there should be an average of 3 or 4 Teens coming out in 11 cards; thus, 12 or 13 Tens remaining in the deck. There are 41 – 13 = 28 unfavorable cards to insurance. However, the 12.5 favorable cards amount to an average of 25, as the insurance pays 2 to 1. The balance is 30 – 25 = +5 unfavorable situations to the player but favorable to the casino. In this case, there is a house advantage of 5/41 = 12%. Unfortunately, even if we consider averages, taking insurance is a repelling bet for the player.
- A formula? It would look complicated symbolically, but it is very easy to follow.
- HA = house advantage
- R = cards remaining in the deck
- T = Tens remaining in the deck.
HA = {(R – T) – T*2} / R
where —
• Axiomatic one, buying (taking) insurance can be a favorable bet for all blackjack players, indeed. Of course, under special circumstances — if you see certain amounts of ten-valued cards on the table. The favorable situations are calculated by the formula above.
But, then again, a dealer natural 21 occurs about 5%- of the time — the insurance alone won't turn the blackjack game entirely in your favor.
3. Calculate Blackjack Double-Down Hands
Strictly-axiomatic colleague of mine, writing software leads me into new-ideas territory far more often than not. I discovered something new and intriguing while programming software to calculate the blackjack odds totally mathematically. By that I mean generating all possible elements and distinguishing the favorable elements. After all, the formula for probability is the rapport of favorable cases, F, over total possible cases, N: p = F/N.Up until yours truly wrote such software, total elements in blackjack (i.e. hands) were obtained via simulation. Problem with simulation is incomplete generation. According to by-now famed Ion Saliu's Probability Paradox, only some 63% of possible elements are generated in a simulation of N random cases.
I tested my software a variable number of card decks and various deck compositions. I noticed that decks with lower proportions of ten-valued cards provided higher percentages of potential double-down hands. It is natural, of course, as Tens are the only cards that cannot contribute to a hand to possibly double down. However, the double-down hands provide the most advantageous situations for blackjack player. Indeed, it sounds like 'heresy' to all followers of the cult or voodoo ritual of card counting!
I rolled up my sleeves and performed comprehensive calculations of blackjack double-downs (2-card hands). The single deck is mostly covered, but the calculations can be extended to any number of decks.
At the beginning of the deck (shoe): Total combinations of 52 cards taken 2 at a time is C(52, 2) = 1326 hands. Possible 2-card combinations that can be double-down hands:
- 9-value cards AND 2-value cards: 4 9s * 4 2s = 16 two-card possibilities
- 8-value cards AND 2-value cards: 4 8s * 4 2s = 16 two-card configurations
- 8-value cards AND 3-value cards: 4 8s * 4 3s = 16 two-card possibilities
- 7-value cards AND 2-value cards: 4 7s * 4 2s = 16 two-card configurations
- 7-value cards AND 3-value cards: 4 7s * 4 3s = 16 two-card possibilities
- 7-value cards AND 4-value cards: 4 7s * 4 4s = 16 two-card configurations
- 6-value cards AND 3-value cards: 4 6s * 4 3s = 16 two-card configurations
- 6-value cards AND 4-value cards: 4 6s * 4 4s = 16 two-card combinations
- 6-value cards AND 5-value cards: 4 6s * 4 5s = 16 two-card possibilities
- 5-value cards AND 4-value cards: 4 5s * 4 4s = 16 two-card combinations
- 5-value cards AND 5-value cards: C(4, 2) = 6 two-card hands (5 + 5 can appear 6 ways).
- Ace AND 2-value cards: 4 As * 4 2s = 16 two-card combinations
- Ace AND 3-value cards: 4 As * 4 3s = 16 two-card possibilities
- Ace AND 4-value cards: 4 As * 4 4s = 16 two-card hands
- Ace AND 5-value cards: 4 As * 4 5s = 16 two-card possibilities
- Ace AND 6-value cards: 4 As * 4 6s = 16 two-card hands
- Ace AND 7-value cards: 4 As * 4 7s = 16 two-card combinations.
- Total possible 2-card hands in doubling down configuration: 262. Not every configuration can be doubled down (e.g. 4+5 against Dealer's 9 or A+2 against 7).
- We look at a double down blackjack basic strategy chart. Some 42% of the hands ought to be doubled-down (strongly recommended): 262 * 0.42 = 110. That figure represents 8% of total possible 2-hand combinations (1362), or a chance equal to once in 12 hands.
- The chance for double-down situations increases with an increase in tens out over the one third cutoff count. The probability for a natural blackjack decreases also — one reason the traditional plus-count systems anathema the negative counts. But what's lost in naturals is gained in double downs — and then some.
- A sui generisblackjack card-counting strategy was devised by yours truly and it beats the traditionalist plus count systems hands down, as it were.
- Be mindful, however, that nothing beats the The Best Casino Gambling Systems: Blackjack, Roulette, Limited Martingale Betting, Progressions. That's the only way to go, the tao of gambling.
4. Calculate Blackjack Pairs: Strict or Mixed Ten-Cards
The odds-calculating software I mentioned above (section III) also counts all possible blackjack pairs. The software, however, considers pairs to be two cards of the same value. In other words, 10, J, Q, K are treated as the same rank (value). My software reports data as this fragment (single deck of cards):Mixed Pairs: All 10-Valued Cards Taken 2 at a Time
Deck Estimator Blackjack Classic
Evidently, there are 13 ranks. Nine ranks (2 to 9 and Ace) consist of 4 cards each (in a single deck). Four ranks (the Tenners) consist of 16 cards. Total of mixed pairs is calculated by the combination formula for every rank. C(4, 2) = 6; 6 * 9 = 54 (for the non-10 cards). The Ten-ranks contribute: C(16, 2) = 120. Total mixed pairs: 54 + 120 = 174. Probability to get a mixed pair: 174 / 1326 = 13%.
Strict Pairs: Only 10+10, J+J, Q+Q, K+K
But for the purpose of splitting pairs, most casinos don't legitimize 10+J, or Q+K, or 10+Q, for example, as pairs. Only 10+10, J+J, Q+Q, K+K are accepted as pairs. Allow me to call them strict pairs, as opposed to the above mixed pairs.There are 13 ranks of 4 cards each. Each rank contributes C(4, 2) = 6 pairs. Total strict pairs: 13 * 6 = 78. Probability to get a mixed pair: 78 / 1326 = 5.9%.Total strict pairs = 78 2-card hands (5.9%, but...).
However, not all blackjack pairs should be split; e.g. 10+10 or 5+5 should not be split, but stood on or doubled down. Only around 3% of strict pairsDeck Estimator Blackjack Game
should be legitimately split. See the optimal split pairsblack jack strategy card.5. Free Blackjack Resources, Basic Strategy, Casino Gambling Systems
- Blackjack Mathematics Probability Odds Basic Strategy Tables Charts.
- The Best Blackjack Basic Strategy: Free Cards, Charts.
~ All playing decisions on one page — absolutely the best method of learning Blackjack Basic Strategy (BBS) quickly (guaranteed and also free!) - Blackjack Gambling System Based on Mathematics of Streaks.
- Blackjack Card Counting Cult, Deception in Gambling Systems.
- The Best Blackjack Strategy, System Tested with the Best Blackjack Software.
- Reality Blackjack: Real, Fake Odds, House Advantage, Edge.
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