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Reason #4 why the Wizard likes Bodog:
One-stop shopping
Bodog offers the triple crown of gambling: casino, poker, and sports. Many other casinos have tacked on poker as an afterthought, and many poker rooms have tacked on a casino as an afterthought, and the lack of attention shows, sometimes painfully. And very few of these sites let you make sports wagers. But Bodog doesn't just offer all three, they do each one well, and everything's integrated. It's easy to play all three off one deposit, off just one account. Another nice thing about Bodog is that you don't need a separate account to play casino games with fake money. In fact you do not even need an account for that at all, you can just click over there and play. Finally, Bodog usernames are only six or seven characters long making them possible to remember. By contrast some competitors' usernames are extremely long and cumbersome. (Visit Bodog) One click and you're in:
No popups, no download, no registration, no B.S., just the game. See important note about Bodog payouts & deposits. |
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Thank you for all the great information on your webpage. I am currently on active duty in the Air Force and will be giving a seminar on responsible gambling. My history professor at NMSU told our class that the only way to win at Blackjack was to bet a little at a time and walk away will small profits..$25. This logic doesn't seem to work in my books...I know it's false. My question is...let's say I have $1,000,000 to gamble with in my lifetime. Do I have "better odds" by betting the whole million at one hand of Blackjack vs. small hands or are the odds always the same regardless? You have a great website and keep up the good work. Thanks a lot for your help! – Bryan from Alamogordo, NM
You're welcome. Your history professor is wrong. This "small win" strategy is nothing new. Usually it does result in a small win, but the occasional big losses more them wipe them out. To answer your question, it depends on what you mean by “better odds.” If you mean which way results in greatest average balance, it doesn’t make any difference. The expected loss is the same with one bet of $1,000,000 or one million bets of $1, assuming basic strategy, and you have reserve money to double or split. However, if you mean which has the greater probability of a net win, your chances are much better off with a single bet. If you make one million bets of $1 the expected loss is $2,850, with a standard deviation of $1,142. The probability of showing a profit is 0.6%. Betting one hand of $1,000,000, the probability of a win is 42.4%, with a push at 8.5%, and a net loss at 49.1%. January 2, 2008
As part of my desire to increase my game knowledge, as a casino dealer, I began reading your site years ago. I seem to remember reading that the dealer's average non-blackjack winning hand is 18.56. Today, as an instructor, I sought that fact to show my students and was unable to locate it. What is the correct answer and how is it calculated? Thank you. – Fred P. from New Orleans The table below shows the average dealer score, assuming the dealer does not bust, and the dealer already checked for blackjack, according to various rules. Note how the average score increases with the number of decks. More importantly, note that the average score is 0.0405 higher when the dealer hits soft 17. The probability that the dealer will bust is only 0.00403 higher when the dealer hits a soft 17. That is 10.05 extra dealer points per dealer bust. This, hopefully, goes to show why it is bad for the player if the dealer hits a soft 17.
To answer your second question, I used a brute force combinatorial program in C++ to cycle through all the possible combinations of dealer hands. May 2, 2007
I've played a lot of Blackjack over the years but have never struck anything like the situation I experienced on the weekend. Playing $25 a hand I lost 19 hands in a row with no pushes. One of the hands was a double down, so effectively I lost 20 x $25 bets in a row. I was playing strict Basic Strategy for New Zealand conditions (not counting, CSM in use). Have you ever heard of such a horror streak? I was ahead about $300 when the sky fell in but stuck to the strategy and eventually left the session $200 ahead and very relieved. My calculations estimate the probability of 19 straight losses as 1 chance in about 207,000; you may well correct me on this. I play to a betting progression system, purely for discipline/money management purposes which has me betting 1 unit after every loss. Had I done anything differently, I would have been cleaned out well before the 19 hands came up. – Ken from Auckland New Zealand
From my blackjack appendix 4 we see the following probabilities for each initial hand.
So the probability of going exactly 19 losses in a row is 0.4909^19*(1-0.4909) = 1 in 1,459,921. By way of comparison, the probability of being dealt a royal flush in video poker is 1 in 649,740, or 2.25 times as likely. Avid video poker players have been known to receive several dealt royals, so if you play a lot of blackjack you’ll likely hit such a losing streak eventually.February 14, 2007
According to standard BJ rules and perfect basic strategy, how many percent of my DOUBLED DOWN hands should I expect to win, push and lose? – Cameron from Melbourne, Australia Assuming liberal Vegas Strip rules (six decks, dealer stands on soft 17, double after split allowed, late surrender allowed, resplitting aces allowed) the following are the probabilities of each possible outcome when doubling on the initial two cards. This does not include doubling after splitting.
First off, my apologies if you consider this a basic math question. I'm a dealer at a Northern Ontario casino, and last night (for the dealer) drew a 12-card 17 (A-A-A-A-A-A-6-A-A-A-A-A). We use six decks. Neither my player or I had ever seen this before. What are the odds of this? - Timothy Rowland from Orillia
Wow! The probability of this is (combin(24,6)/combin(312,6)) * (24/306) * (combin(18,5)/combin(305,5)) = 1 in 287,209,346,813,617.
January 18, 2007
Fantastic site - A gambling bible no less. Which would be the better rule for the player in an otherwise normal blackjack game with six decks. The first rule is the insertion of two jokers in the shoe. If the player gets a joker in his first two cards it is an automatic winner, paying even money. If a joker comes out at any other time, including to the dealer, it is burned. The second rule is a Five-Card Charlie. – Gordon G. from Paramaribo
Thanks. The probability of getting at least one joker in two cards dealt from a 314-card shoe is 1-(combin(312,2)/combin(314,2)) = 1.27%. So the probability is 1.27% of turning an average hand into an automatic winner. If we assume an average hand has an expected value of –0.005 then the value of the first rule is 0.0127*(1-(-0.005)) = 1.28%. You can see from my blackjack section that the five-card Charlie rule is worth 1.46%. Assuming the cards are shuffled after every hand, or you are forced to flat bet, then given the choice as a player I would pick the five-card Charlie rule. However, the joker rule would be very easy to exploit further. The greater the ratio of jokers to cards in the shoe the more you should bet. With at least 50% deck penetration this should easily be enough to make it the better rule.
June 23, 2006
I lost a lot of money playing Cryptologic Blackjack today. While I don't think anything is fixed, one aspect of my play seemed well outside the range of probability. Within 35 hands, the dealer showed a 6 seven times and won each time. This was verified through the logs. If the probability of a dealer bust is 56% with a six, my calculation suggests the odds of this independent event happening six consecutive times is 0.23%. – Adam from Toronto
At Cryptologic they use 8 decks and the dealer stands on a soft 17. According to my blackjack appendix 2, the probability of the dealer busting with a 6 up is 0.422922. So the probability of not busing is 1 - 0.422922
= 0.577078. The probability of not busing 7 times out of 7 is (0.577078)7 = 2.13%.
June 9, 2006
Hi Wiz, as always, thanks for the great site and love your columns and especially the relationships questions. I have a question about my online blackjack play. I have been playing for quite a while and keep a record or my results. I play two $5 hands against the dealer, thus a total initial bet of $10. My total amount wagered, including doubles and splits, is just over $680,000 at this moment. I calculate my expected loss to have been $3,000 (after factoring the number of total bets into initial bets/hands) at a house edge of 0.5%. However, my actual loss is a far greater amount of $8,500. For such a reasonable sized sample, that seems to be a fairly large discrepancy. I don't know if that means the game is less than fair (I have no reason to believe so) or I am just experiencing an unfortunate result. What do you think? Are there any reasons to be suspicious based on these results? PS: fortunately, I am still making money due to the bonuses but not as much as I would have expected, and that is the disappointing part. Thanks and all the best. -- Mick from Port Kembla
Thanks for the kind words. So you have played 680,000/5 = 136,000 hands. According to my blackjack appendix 4 the standard deviation per hand, playing two hands at a time, is 1.91. So the standard deviation of 136,000 such hands would be 136,0000.5*1.91 = 704 hands. Your losses above expectations are $5,500, or 1,100 hands. So you are 1,100/704 = 1.56 standard deviations south of expectations. The probability of doing this badly or worse can be found in Excel as normsdist(-1.56) = 5.94%.
May 18, 2006
Great website! Would you please tell me how often the player and the dealer will both bust, on the same hand, when the player is using the basic strategy? I went to the math links that you have listed, but couldn't figure out how to set up the equation. I would appreciate any help you can provide. - Paul from Portland
The following table shows the probability in a single-player game. Remember, the dealer will not play out his hand if the player busts first. If you add players the dealer's probability of busting will go up because there will be a greater probability of at least one player not busting.
I have tried, quite unsuccessfully, to locate on your excellent site and elsewhere how to properly calculate a confidence interval based on 1, 2, or 3 standard deviations and number of hands played with Expected Value. For example, I'd like to learn how to calculate the +/- standard deviations for playing $10/hand at blackjack for 300 hands using the .50% house edge for basic strategy. Thanks for your GREAT support and advice. - Peter from Orlando The expected loss would be 300*$10*0.005 = $15. As I state in my blackjack appendix 4 the standard deviation is 1.17 (based on Atlantic City rules). The standard deviation on 300 hands at $10 each would be 3001/2 * $10 * 1.17 = $202.65. So here are the confidence intervals on the expected win for 1, 2, and 3 standard deviations: In blackjack what is the probability of the dealer making a stopping hand (17-21) drawing 8 cards? This happened to a friend of mine online and I think it's an extremely rare occurance (over 1 in a million). How about 7 cards? Thanks for the great site and keep up the awesome work! Thanks for the compliment. Assuming a 6-deck game, where the dealer stands on soft 17, and the player plays basic strategy here are the rounded results based on a 100 million hand simulation. Jan. 2, 2005 I am a blackjack dealer and last night I amazed my table on a single-deck blackjack game (the horrible 6 to 5). My hand consisted of an Ace up, Ace in the hole and then I drew the other 2 Aces and then a 7 for 21! What are the odds of this happening and I am especially interested in knowing the math. Thanks!- Brian The probability is (4/52)*(3/51)*(2/50)*(1/49)*(4/48) = 1 in 3,248,700. Dec. 27, 2004 What is the probability of a blackjack for n decks? 2*(4/13)*(4n/(52n-1)) Dec. 27, 2004 If playing Basic Strategy significantly helps the player not lose as much to the casinos, why do some casinos freely give BS cards out and allow players to consult BS cards while playing? I have seen this in Shreveport, Vicksburg and Black Hawk, CO. Friends have told me that they have done this in Vegas too. I have played blackjack at casinos all over the United States and have never seen a basic strategy card in blackjack given out for free. However most casino gift shops sell them and they are indeed allowed at the tables. I think the casinos aren't crazy about the cards the alternative of prohibiting them would be even worse. It would cause a lot of bad player relations to try to enforce a no strategy card rule. Furthermore, where would they draw the line? What if the player wrote the basic strategy on his hand, could the casino prevent a player from looking at his own hand? Dec. 2, 2003 At a single deck game what is the probability all three players and the dealer get a blackjack the first round after a shuffle? Following are the probabilities: First I wanted to tell you how much I look at and love your web site, and admire your math skills. I use 6 decks to deal blackjack, and added 3 jokers for reasons I won't waste your time with but, what are the odds of dealing all 3 jokers to a player right in a row. Thank you very much. You're welcome, thanks for you compliments. The probability of being dealt 3 jokers in a row from a six deck shoe (plus the 3 jokers) is 1/combin(315,3) = 1 in 5,159,805. Another solution is (3/315)*(2/314)*(1/313). Aug. 25, 2003 I just witnessed a friend get four blackjacks in a row starting with the first hand of a newly shuffled single deck playing head to head against the dealer. I looked at the FAQ's and saw the odds for getting one blackjack in single deck, but don't know how to calculate them for getting four in a row off the top. Instead of a decimal probability, could you tell me the odds of this? It must be astronomical. Hope to hear from you. I seem to get a variation of this question at least once a month. Let's assume for now the deck is shuffled after every hand, to make the math easier. If the probability of something happening is p then the probability of it happening n times in a row is pn. The probability of a blackjack in a single deck game is 4*16/combin(52,2) = 64/1326. So the probability of four in a row is (64/1326)4 = 16777216/ 3091534492176 = 1 in 184270. However the actual probability is much less, because as the player gets each blackjack the ratio of aces to cards left in the deck decreases. Without knowing what cards the dealer got I can't tell you the exact answer. Aug. 25, 2003 I recently went to Vegas and had an incredible hand of blackjack... received an ace as first card, split, received another ace, split, received a third ace, split, and got one last ace... Then was dealt blackjack on all 4 hands! No lie! 2 of my friends were witnesses, as was the entire Luxor gaming gods...What are the odds on this? It was a 6 card deck shoe, I was sitting in #3 seat of a 4 person game. Assume a fresh shuffle? Not too many places allow resplitting aces, so be glad you were playing somewhere that did. Your seat position does not matter. The probability of this is the probability that the first four cards out of the shoe are aces, and the next four are tens, or (combin(24,4)/combin(312,4))*(combin(96,4)/combin(308,4)) = 1 in 4,034,213. Aug. 2003 Dear Wiz, How do you calculate the probability of getting three sevens, three colored sevens, and three suited sevens in blackjack? Thanks, Geoff. Let's assume six decks of cards and the player always takes a third card (whether by hitting or splitting). The number of ways to draw 3 suited sevens is the number of suits (4) times the number of ways to choose 3 out of 6 sevens of that suit in the shoe. In other words 4*combin(6,3)=4*20=80. The number of ways to draw 3 colored sevens, including 3 suited sevens, is the number of colors times the number of ways to pick 3 out of the 12 sevens in the shoe of that color, or 2*combin(12,3)=2*220=440. The number of ways to draw any 3 sevens, including 3 colored and suited sevens, is the number of ways to pick 3 cards out of the 24 sevens in the shoe, or combin(24,3)=2024. The total number of combinations for any 3 cards out of 312 is combin(312,3)=5013320. So the probability of 3 suited sevens is 80/5013320=0.000015957. The probability of three colored, but not suited, sevens is (440-24)/5013320=0.0000830. The probability of three unsuited suited and uncolored sevens is (2024-440)/5013320= 0.00031596. For help with the combin function visit my poker probability or lottery probability sections. June 14, 2003 Dear Wiz, I am a blackjack dealer here in Vegas and the other night dealing, I had 4 out of the 6 ace of spades in my hand. I had A-A-K-A-A-10, so good think is I busted, but quick calculations on the game, we figured getting 4 out of the six aces on one had is around 7mil to 1. Is this number a little high? Thanks, Jason The probability of this occurring in which your other two cards are any two 10-point cards is 4*COMBIN(6,4)*COMBIN(6*16,2)*(4/6)*(3/5)*(1/2)/combin(312,6) 1 in 22,307,231. However there are other ways you get four aces in the same hand, for example the last card might be an 8 or 9. I would have to do a computer simulation to consider all the other combinations. However to make a rough guess I'd say the 7 million looks about right. What are the odds of a dealer getting 3 blackjacks in a row on a single deck table with two players? - J.A.S. from Las Vegas, USA It depends whether there is a shuffle between the blackjacks. Assuming there isn't the probability would be 8*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47) = 0.000044011058. The number of other players doesn't matter, except if they cause a shuffle. Mar. 24, 2002 In two handed blackjack using one deck, what is the probability of the dealer having a blackjack? - Steve Neill from Solva, United Kingdom The number of hands doesn't matter. The probability is 2*(4/13)*(8/103) = 0.0478. March 17, 2002 What is the probability that you play ten hands and never obtain a (two-card) 21? Assume the cards are reshuffled after each play? - Matt from Radford, USA If the probability of a blackjack is p then the probability of not getting any blackjacks in 10 hands is 1-(1-p)10. For example in a six deck game the answer would be 1- 0.95251110= 0.385251. Mar. 6, 2002 What are the odds of getting 3 blackjacks in a row with 1 deck 4 players and one dealer. - Joe P from Parma Heights, USA I'm going to assume there is never a shuffle between hands. The three other players don't matter. The answer would be 23*(16/52)*(4/51)*(15/50)*(3/49)*(14/48)*(2/47)= 0.00004401, or about 1 in 22722. If there were a shuffle between hands the probability would increase substantially. Mar. 6, 2002 Under the card game 21(A.K.A. Blackjack), you win if your two cards are an ace and either a 10, jack, queen, or king. What is the probability that you draw such a 21? It depends on the number of decks. If the number of decks is n then the probability is 2*pr(ace)*pr(10) = 2*(4/13)*(4*n/(52*n-1)), which is conveniently about 1 in 21. Here is the exact answer for various numbers of decks. I play 6 deck blackjack in Tunica, MS. The dealer hits on soft 17. I wonder what the odds are of standing on 16 when the dealer's upcard is 7. It seems only a 10 or face card can beat this and the odds would be in my favor if the dealer draws more than one card. Also, since most strategies are based on millions of calculations done on a computer, I wonder if those of us who will never play a million hands can rely on slight variations like this one. Is this a poor, fair or bad move to make? - Richard Sadler from Memphis, USA According to my blackjack appendix 9H the expected return of standing is -0.476476 and of hitting is -0.408624, assuming the 16 is composed of a 10 and 6. So my hitting you will save 6.79 cents for each dollar bet. This is not even a marginal play. There is no sound bite answer to explain why you should hit. These expected values consider all the numerous ways the hand can play out. The best play for a billion hands is the best play for one hand. If you want to deviate from the basic strategy here are some borderline plays: 12 against 3, 12 against 4, 13 against 2, 16 against 10. Deviating on these hands will cost you much less. Jan. 2, 2002 For blackjack, which is the probability to obtain three suited sevens in a 6-deck shoe. - Rodrigo Castro from Costa Rica I attempt to work this out in my blackjack appendix 8 but I'll work through it more slowly here. We'll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin(312,3)=5,013,320. There are 24 sevens in the shoe. The number of ways to arrange 3 sevens out of 24 is combin(24,3)=2024. The probability is the number of winning combinations divided by total combinations, or 2024/5013320=0.0004, or about 1 in 2477. June 29, 2001 What piece of information am I missing? If the odds of pulling a ten count card out of a deck is about 30.7% and the odds of pulling out an ace is 7.8% then it seems to me that the combined odds of this happening are about 2.4%. Why do blackjack simulators and blackjack authors state that the odds for a blackjack are 4.7% which happens to twice the calculated odds. What am I missing? - Jeffrey Cason from Loveland You are forgetting that there are two possible orders, either the ace or the ten can be first. Multiply by 2 and you'll have your answer. June 29, 2001 I was sitting at first base in an 8-deck blackjack game. The dealer finished shuffling, and as she put the cards in the shoe she fumbled them slightly, flashing the first two cards: a jack and an ace. Since I knew the jack would be the burn card, I also knew I would get the ace as my first card. This is obviously to my advantage, but how much, percentage-wise? I ended up betting $50 instead of my normal $5 on the first hand from the shoe. I wish it had a happy ending, but I was dealt a soft 18, and ended up busting since the dealer was showing a ten. Thanks for your time, and for the great site! - David Thanks for the compliment. I don't have numbers readily available for eight decks but in a four deck game where the dealer stands on soft 17 I get the player advantage given that the first card is an ace to be 51.66%. In Basic Blackjack by Stanford Wong he says the advantage in a six-deck game where the dealer stands on soft 17 to be 50.5%. Sometimes a Las Vegas fun book will have a coupon which can be used as an ace for the first card in blackjack. Wong also mentions the kind of situation that happened to you in his book. June 6, 2001 In blackjack, I've read that the dealer breaks about 25% of the time. If that figure is correct, is that 25% of all the hands dealt, or 25% of just the hands that he hits? Also, in blackjack the house edge varies according to the rules of that particular game. What I'd like to know is, what effect, if any does playing head to head with the dealer have. It would seem to me that both of you will get more blackjacks, thereby reducing the house edge somewhat. - Jim Malone from Rome, New York If the dealer stands on soft 17 he will bust about 29.1% of the time, and 29.6% of the time if he hits a soft 17. This does not count in the total the times when the dealer has blackjack. It does not matter how many players are at the table, everyone will get a blackjack about once every 21 hands. Feb. 10, 2001 If the first card dealt is an ace what is the probability the dealer will have a blackjack? Assume two decks. - T from Las Vegas, U.S. There are 103 cards remaining in the two decks and 32 are tens. So the probability of a blackjack is 32/103=31.07%. Jan. 20, 2001 What percentage of hands are suited blackjacks? Six-deck shoe, any suit. - RWR from Tucson, USA The probability of a suited blackjack in a six-deck game is 2*(4/13)*(6/311) = 0.0118723. Feb. 10, 2001 If I play 100 hands of blackjack at $5 a hand at an 0.5% house edge how much can I lose and still be above three standard deviations south of expectation? - Blair Gorman from Christchurch, New Zealand Your expected loss would be 100*$5*.005=$2.50. The standard deviation of one hand is 1.17, which can be found in my blackjack appendix 7. So one standard deviation in your example is $5*1.17*sqr(100)=$58.5. So the probability of losing $295 or more due to bad luck is .00135 (the Z statistic for -3). Jan. 20, 2001 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. ñ Mike Tobias of Bossier City, USA 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 Dec. 10, 2000 In a six deck shoe what is the percentage of times that a blackjack (ace face card or ten)will come up? ñ Ed of Lynnwood, USA Let n be the number of decks. The probability of a blackjack is 2*(4/13)*(4n/(52*n-1)). If n=6 the probability is 192/4043 = 4.75%. Nov. 19, 2000 What are the odds against winning seven hands of black jack in a row? How about six? - James Dal Bon of Palo Alto, California According to my FAQ the probability of an overall win in blackjack is 43.33%, a tie is 8.78%, and a loss is 47.89%. I'm going to assume you wish to ignore ties for purposes of the streak. The probability of winning n hands is a row is .475n. So the probability of winning six in a row is 1.1487% and seven in a row is 0.5456%. Nov. 4, 2000 Q: I have a few questions regarding blackjack: How often can one expect the dealer to bust and how often can a player expect to win 4 hands in a row? - John Brock of Westminster, USA A: When the dealer stands on a soft 17 the dealer will bust about 29.1% of the time. When the dealer hits on a soft 17 the dealer will bust about 29.6% of the time. Assuming you skip of ties the probability of the winning four hands in a row is about 5.1%. Aug. 13, 2000 Q: I'm a dealer at Casino Niagara and what to know what the odds of a dealer making a hand are when the up card is a 5. It seems to me and the other dealers that we all agree that we make a hand more times than not, and usually a good one at that. Also, what are the odds of a dealer having a blackjack with an ace as an up card? - John from Niagara Falls, Canada A: You're right, it is more likely the dealer will make a pat hand. From my blackjack appendix 2 the following are the probabilities of the dealer's final hand given a 5 as an up card. This assumes the dealer stands on a soft 17, which I believe is what you do.
Q: How can I determine the odds of flat betting ( no counting, no progressions , etc ) of being ahead in a negative game such as blackjack w/o counting with a .5% disadvantage after 45000 or so hands? Is it even possible? - Kevin A: This is a typical question one might encounter in an introductory statistics class. Because the sum of a large number of random variables always will approach a bell curve we can use the central limit theorem to get at the answer. From my section on the house edge we find the standard deviation in blackjack to be 1.17. You won't understand this if you haven't studied statistics but the probability of being at a loss in your example will be the Z statistic of 45000*0.005/(450001/2*1.17) =~ 0.91. Any basic statistics book should have a standard normal table which will give the Z statistic of 0.8186. So the probability of being ahead in your example is about 18%. June 18, 2000
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