Ask The Wizard #34
Is counting cards in blackjack (or any other card game for that matter) pointless if one is using a truly infinite deck?
Yes.
Your site is definitely the best one I've ever seen regarding gambling, and I commend you for providing some light and truth in a seemingly endless sea of "winning gambling strategies, tips & tricks." My question is this. I'm no slot player, but obviously when a progressive jackpot reaches a certain point, the edge would shift from the house to the player. I was wondering if there are any "groups" or "clubs" that go out into the casinos when this happens, (virtually) monopolize the machines, capitalize on the opportunity, and split the winnings? I have never heard of any, but they've got to be out there.
Thanks for the kind words. I have barely heard of teams of slot players doing this. However, this is very common with progressive video poker players. There are teams of these professional players who routinely check the meters and when they find one high enough they call their teammates in an attempt to monopolize the machines until somebody hits the jackpot.
The problem with slots is that it is not clear to the player what the odds are of hitting the jackpot so it is not obvious what the jackpot size has to reach for the machine to become profitable. Plus, it probably rarely happens that a meter gets high enough to overcome the house edge.
Do you change your strategy based on the play of other players at your table? For example, you have players who hit when the dealer has a bust card face up taking the bust cards and therefore the dealer does not bust.
Unless you are a card counter, how other players play should not affect what you do. Basic strategy players should stick to the basic strategy no matter how badly the other players play. Other players are just as likely to help you as hurt you. In the end, it makes no difference how they play.
Can blackjack be beaten under the following casino conditions:
- The game is dealt face up from an 8 deck shoe, with the cut card appearing after 5 decks have been dealt (3 decks behind the cut).
- Dealer stands on soft 17.
- No surrender.
- Can double down on any 2 card total without an ace.
- Can split aces once only, one card on each.
- Can split any other pair to a maximum of 3 hands.
- Can double after split.
- Dealer takes original bets only on blackjack.
- Can take even money on blackjack when dealer’s upcard is an ace.
- Table max is 50 times table min.
- Card counting is permitted if the counter plays the first hand of the shoe, and plays every hand. Counter can play any number of boxes, and any bet amount. Counter can stop at any time, but cannot rejoin a shoe after missing a hand, or join a shoe that is partway through.
I haven't done any simulations, but my educated option is a definite yes, this game can be beaten. The strategy to use in this game would be to bet the minimum when the odds are against you and the maximum when they are in your favor. Normally a sudden 50 times increase in bet size would set off a huge red flag but it seems the counter could do this with impunity in your game. When Atlantic City first opened the casinos could not ask card counters to leave and entire tables were filled with people jumping suddenly from a $5 bet to $300, or whatever the minimums and maximums were. After taking a huge beating, the Atlantic City casinos begged the gaming authorities for a change in the rules, which they got. Not only could this be beaten, but I think it would be a card counters dream.
Please explain how to calculate the probability of a blackjack occurring in a single deck. I can easily work other hands but when a card can be either/or my brain cramps.
The probability that the first card is an ace is 4/52. The probability that the second card is a 10 point card is 16/51. So the probability of an ace first blackjack is (4/52)*(16/51). Multiply this by 2 because the ten could just as easily be the first card and the answer is 2*(4/52)*(16/51) = 128/2652 = 0.0482655, or about 1 in 20.7 .
The odds according to your formula for a royal flush is 4/2,598,960 = 1/649,740. So, if I was playing Caribbean Stud one-on-one with the dealer, then my hand and the dealers would equal 649,740*2=1,299,480. Therefore, according to the math, after 1,299,480 hands there should be two royal flushes. Please tell me if I understand the odds correctly.
You, are right that on average a royal flush will occur once in every 649,740 hands, and that in 1,299,480 hands the expected number of royal flushes is 2. However, this is only the average. With every hand that goes by you are no closer to getting a royal. Every game of independent trials has this memory-less property so a royal flush is never overdue.
The probability of zero royals in 1,299,480 hands is 13.53%.
Many on-line casinos advertise that they payout 98% or a number close to that. They also claim that this number is audited by one of the big-six accounting firms. How is this number calculated? Also, is there a way to calculate my own personal payout ratio for a certain game within certain period of time?
The payout is the ratio of money won to money bet. For example if players bet a total of one million dollars and the total amount paid to winning bets was $998,000 then the payout ratio would be 98%. Keep in mind that as players recirculate the same money the house edge grinds them down so the typical player will retain much less than 98% of their original purchase. The way to calculate your own ratio if to keep track of the total amount you bet and the total amount you win and divide.
Do you have basic strategy for the following rules? Dealer takes 17-17, 18-18 and 19-19 ties, doubling after split is allowed, 3 resplits, no-peek, player can double 7-11 totals (soft and hard), dealer stands on soft 17, six decks.
Stanford Wong's Blackjack Count Analyzer is perfect for questions like this. Just plug in the rules and it produces an immediately basic strategy and is ready to run a simulation. Following is his basis strategy under these rules. I did a 31-million hand simulation using Blackjack Count Analyzer, which shows a house edge of 4.13%, under these rules. When I was in Finland they had single zero roulette, which has a much lower house edge than this game. Why the blackjack rules are so stingy in Finland I would like to know.
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 allowed otherwise stand, / = split.
Hi, I am regular player of Pai Gow Poker, and I noticed your site has a lot of great information on the game. The other day when I was playing with a friend of mine he was dealt a 9-high hand, which I believe is the lowest hand possible. In all the time I had spent playing the game I had only seem it happen once before. Then five hands later he got the exact same hand(2-3-4-5-7-8-9). We couldn't believe it and were wondering what the odds of that happening were so we thought we would ask you. Thanks for your time and your great site.
There are two ways to arrange the ranks for form a 9 high hand, the one you mentioned and 2-3-4-6-7-8-9. The number of suit combinations without forming a flush is 47-4*(combin(7,5)*3^2+6*3+1) = 15,552. So the probability of a 9 high hand is 2*15,552/combin(53,7) = 31,104/154,143,080, or 1 in 9,911. If you were to play just five times the probability of getting 2 9-high hands would be 1 in 9,826,685. That this happened is a coincidence I believe as opposed to a fault in the random number generator or the coding of the program.