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Ask The Wizard #34
When utilizing a truly infinite deck in blackjack (or similar card games), is it futile to count cards?
Yes.
I must say, your website stands out as the best resource I've encountered for gambling insights. I appreciate how you bring clarity and honesty in a landscape overwhelmed with dubious 'winning strategies, tips & tricks.' I have a question: although I'm not a frequent player of slots, I understand that when a progressive jackpot hits a certain threshold, the odds may turn in favor of the player. Are there any organized groups or clubs that gather at casinos during such events to take advantage of this, monopolizing the machines and sharing the winnings? I've never come across any mention of such groups, but I imagine they must exist.
Thank you for the compliment. While I haven't come across any teams specifically focused on slot machines, it's a widespread practice among progressive video poker players. These professional teams regularly monitor the jackpot levels, and when they see a favorable opportunity, they call their members to dominate the machines until the jackpot is won.
The challenge with slot machines is that players are often unaware of the specific jackpot odds, making it hard to determine when the machine becomes favorable for play. Additionally, it’s uncommon for a jackpot meter to rise high enough to offset the house advantage.
Do you adjust your approach based on how others at your blackjack table are playing? For instance, if some players choose to hit when the dealer has a potential bust card, they may inadvertently affect the dealer’s chances of going bust themselves.
Unless you are proficient in card counting, the strategies employed by other players should not influence your gameplay. Adherents of basic strategy should remain true to their approach regardless of any perceived errors from others. Fellow players can impact your results either positively or negatively, but ultimately, their gameplay doesn’t change your optimal strategy.
Is it possible to gain an advantage over the house in blackjack under these specified conditions:
- The game is played with a shoe containing eight decks, and the cut card is placed after five decks have been dealt (three decks remain behind the cut).
- Dealer stands on soft 17.
- No surrender.
- Players can double down on any total consisting of two cards, even if they do not include an ace.
- Aces can be split only once, with one card placed on each ace.
- Any other pairs can be split to a maximum of three hands.
- Can double after split.
- The dealer maintains the original player's bets only on a blackjack.
- Players can opt for even money on a blackjack if the dealer reveals an ace.
- Table max is 50 times table min.
- Card counting is allowed as long as the counter plays the first hand of the shoe and participates in every subsequent hand. The counter may play multiple boxes and wager any amount, with the freedom to stop playing at any time. However, they cannot re-enter the shoe after missing a hand or join a game that is already in progress.
While I haven't performed any simulations, my educated guess is a strong yes—this game can indeed be beaten. The recommended strategy would be to place smaller bets when odds are against you and maximize your stakes when odds shift in your favor. Typically, a sudden jump in bet size would raise red flags, but in this scenario, the counter could execute this without repercussions. Initially, when Atlantic City opened, casinos were unable to evict card counters, resulting in tables filled with players making dramatic shifts from $5 to $300 bets. After facing large losses, casinos petitioned for regulatory changes, which were subsequently implemented. This situation not only presents opportunities but could also be a dream scenario for a skilled card counter.
Could you explain how to compute the likelihood of drawing a blackjack in a single deck? I manage well with other hands, but the dual possibilities make my calculations a bit challenging.
To calculate the probability of drawing an ace first, it's 4 out of 52 possibilities. Then, the probability that the second card drawn is a ten-point card is 16 out of the remaining 51. Thus, the chance of an ace followed by a ten for a blackjack is (4/52)*(16/51). Multiply that by 2 since the ten could also appear first, resulting in 2*(4/52)*(16/51) = 128/2652, or approximately 0.0482655, translating roughly to 1 in about 20.7 draws.
According to your calculations, the odds of achieving a royal flush is 4 out of 2,598,960 hands, which simplifies to 1 in 649,740. Therefore, if I engage in Caribbean Stud one-on-one with the dealer, the total hands played would indeed amount to 649,740*2, equal to 1,299,480, suggesting that there would be two royal flushes after this total number of hands. Please confirm if I've grasped the odds correctly.
You are correct that, on average, a royal flush will appear once every 649,740 hands, which means in 1,299,480 hands, we would statistically expect two occurrences. However, keep in mind that this is merely an average; with each hand dealt, you are not any closer to hitting a royal flush. Each round of activity functions independently, consequently, a royal flush is never 'due'.
The probability of getting zero royal flushes over the span of 1,299,480 hands sits at 13.53%.
Numerous online casinos boast payout rates around 98% or something similarly impressive, often claiming that big accounting firms verify these figures. Could you clarify how this percentage is calculated? Is there a way for me to determine my own payout ratio on a specific game over a given time frame?
Payout ratio is determined by comparing the total money won against the money wagered. For instance, if players collectively bet one million dollars and winnings amounted to $998,000, then the payout ratio would be 98%. It's essential to understand that as players continue to reinvest their winnings, the house edge ultimately reduces the average amount players retain to far less than 98%. To calculate your own ratio, tally the total amount bet and total winnings, then divide the two.
Are there basic strategies that you recommend for the following scenario? The dealer uses 17 as the stopping point, ties on 18 and 19, allows doubling after splitting, incorporates three resplits, utilizes a no-peek method, permits doubling on totals of 7-11 (both soft and hard), and the dealer stands on soft 17 using six decks.
Stanford Wong's Blackjack Count Analyzer is an excellent tool for answering such inquiries. Just input the relevant game rules, and it will furnish you with an immediate basic strategy along with a simulation option. Here are his suggested strategies based on these rules. My simulation of 31 million hands using the Blackjack Count Analyzer suggested a house edge of 4.13% under the given conditions. During my visit to Finland, I encountered single-zero roulette, which offers a significantly lower house edge compared to blackjack. I wonder why blackjack rules in Finland are so unfavorable.
PLR DEALER PLR DEALER 23456789XA 23456789XA 21 ---------- soft 21 ---------- 20 ---------- soft 20 ---------- 19 ---------- soft 19 ---------- 18 ---------- soft 18 +DDDd-++++ 17 ---------+ soft 17 +++DD+++++ 16 -----++--+ soft 16 +++DD+++++ 15 -----+++++ soft 15 +++DD+++++ 14 -----+++++ soft 14 ++++D+++++ 13 -----+++++ soft 13 ++++++++++ 12 +----+++++ 11 DDDDDDDD++ pair A /////////+ 10 DDDDDDDD++ pair 10 ---------- 9 ++DDD+++++ pair 9 /////-/--- 8 ++++++++++ pair 8 ///////--+ 7 ++++++++++ pair 7 -////+++++ 6 ++++++++++ pair 6 /////+++++ 5 ++++++++++ pair 5 DDDDDDDD++ 4 ++++++++++ pair 4 +++//+++++ pair 3 +////+++++ pair 2 +////+++++ Insurance: No + = hit - = stand D = Double if allowed otherwise hit d = Double if permitted, otherwise stand, / = split.
Hi, I'm an avid player of Pai Gow Poker, and I appreciate the wealth of information on your site about the game. The other day whilst playing with a friend, he was dealt a 9-high hand, which I believe is the lowest possible in the game. In all my experience, I've only seen it once before. Remarkably, just five hands later, he received the very same hand (2-3-4-5-7-8-9). We found it quite surprising and would like to know the odds of this occurrence, so we thought we would reach out to you. Thank you for your valuable time and insights.
A 9-high hand can be arranged in two different ways, as you've noted, and also as 2-3-4-6-7-8-9. The combinations of suits that do not result in a flush are calculated as 4 - 4*(combin(7,5)*3^2 + 6*3 + 1) = 15,552. Therefore, the probability of drawing a 9-high hand is 2*15,552/combin(53,7) = 31,104/154,143,080, or roughly 1 in 9,911. If you were to play five rounds, the chances of seeing two 9-high hands would drop to about 1 in 9,826,685. The fact that this occurred seems more like a coincidence rather than a glitch in the random number generator or the programming itself.7We provide mathematically sound strategies and information for a variety of casino games, including blackjack, craps, roulette, and hundreds more.