Love your site! In your blackjack section, you discuss tipping, but you don't suggest an amount. I'm going to Vegas for the first time in about two weeks, and I never knew you were supposed to tip dealers AT ALL, let alone how much. Do you have any recommendations on tip amounts?
Sarah from Chicago, USA
There is no firm standard but I would recommend tipping about half your average bet per hour. More for good service, less for bad.
I am thinking of taking the following strategies to play mini baccarat. I only bet after either the Banker or Player has appeared four times in a roll. I double my money if I don't win the first time. However, if the second time I don't win, I stop betting for the time being until the next four continuous appearance coming again. Once I win, I also stop betting until the next 4 continuous appearance coming. Please evaluate my strategy. Thank you!
Mandy from Gold Coast, Australia
Waiting for streaks of four in a row is not going to help. The cards do not have a memory. Doubling after a loss is also not going to help. I would recommend betting on the banker every time. Skipping hands is fine, in fact not playing at all is the best possible strategy.
What is the best method for receiving comps from slot attendants when playing slot machines?
Donald from Golconda, USA
This is not my area of expertise. However Jean Scott "the Queen of Comps" says you should establish a relationship with a casino host anywhere you plan to play a lot. Then ask them for a comp after you have given them sufficient play.
I realize that card counting is pretty much a thing of the past and only if you can find a single-deck game. I also realize that if you are good at it you will probably be asked to leave which just about loses my interest in it. My question is when the deck is "rich" why do the odds favor the player and not the dealer as well since he/she is part of the hand?
Harry from Geneva, USA
Card counting is still alive and well. It is just as easy to count one deck as it is to count eight. Many card counters prefer the multi-deck games. There is lots that can be done to avoid being asked to leave, like short playing sessions and acting like a typical bad player. I try to explain why card counting works in my card counting page.
While Roulette clearly cannot be beaten by chance, I have heard that it can be beaten by physics two ways (in theory). Way one: a high tech device, which measures the velocity of the ball against the velocity of the wheel and predicts the outcome sector of the wheel with like a 40% accuracy. Way two: Wheel bias. Obviously a wheel would have to have a bias of at least 5.26% to get the player to an even keel. The question is, how many spins would you say, Wizard, does it take to determine wheel bias, if there is any?
JF from Providence, USA
I've heard of both these techniques being used. I don't know much about devices to clock wheels, except they are known to exist and be used from time to time. Here in Nevada such a device would be highly illegal.
Taking advantage of biased wheels I hear a lot more about. It has been done lots of times. I think casinos with old wheels are the most vulnerable target. I've been saying for years that I think Argentina is a ripe target for that.
Bank of America is offering to triple a selected deposit a day made at an ATM. The contest is for about two months. Are my odds better when depositing $300 ... to make three deposits of $100 or one of $300...or are my overall odds so low that the difference isn't worth the effort?
Daniel from Las Vegas, USA
Your expected win is the same regardless of how many times you divide your total deposits. A good strategy would be to deposit and withdraw the same money over and over as many times as possible. However, your odds may be so bad that it isn't worth the bother.
What percentage of hands are suited blackjacks? Six-deck shoe, any suit.
RWR from Tuscon, USA
The probability of a suited blackjack in a six-deck game is 2*(4/13)*(6/311) = 0.0118723.
I don't have a blackjack simulator personally, but wondered what the house edge would be on the video blackjack here in Rhode Island. It is single deck (reshuffled every round of course), split pairs only once (no splitting unlike tens), double on hard 10&11 only, pull six cards automatic winner regardless of dealers hand (i.e. If you have a soft 19 or 20 after 5 cards, you automatically hit because it is 100% impossible to bust on 6th card) double on splits on hard 10&11 only) split aces receive only one card each. BUT, here's the clincher, split Blackjacks pay 3 to 2 whereas traditionally they pay 1 to one) I know this can make a difference of maybe 0.5 % in the overall house edge. Also late surrender on any hand, and dealer hits soft 17. So, playing optimum basic strategy for these situations, what is the overall house edge. My guess is that it' in the 1.5% range because of the single deck and 3-2 pay out on split blackjacks.
JF from Providence, USA
According to my blackjack house edge calculator, the house edge before the 6-card Charlie rule and the 3-2 on blackjack after splitting, is 0.34% with perfect strategy.
My list of rule variations says that the 6-card Charlie rule is worth 0.16%, and a 3-2 paying blackjack after splitting aces is worth 0.19%. It still doesn't pay to split tens. So, the overall house edge is 0.34% - 0.16% - 0.19% = -0.01%. In other words, a player advantage of 0.01%.
Just a question about an Oriental dice game, where the players are supposed to guess which side of the die shows up. The players will first place their bets on 1,2,3,4,5,6 (like roulette) and then the "dealer" will roll 3 dice simultaneously. Payouts are 1:1 if the chosen numbers shows up once (on any of the 3 dice), 2:1 if the chosen no shows up twice, and 3:1 if the chosen number appears on all 3 dice. As the player can place any number of bets of the board, what will be the optimum number of bets to place? (assuming all my bets are equal in size)
Jansen from Toronto, Canada
The probability of three matching is 1/216. The probability of two matching is 3*5/216. The probability of one matching is 25*5/216. The probability of 0 matching is 5*5*5/216. So the expected return is 3*(1/216)+2*(15/216)+1*(75/216)-1*(125/216)=-17/216=-7.87%. There is no optimal number of bets, you will give up an expected 7.87% of total money bet no matter what you do.
These bets can be made in both sic bo and chuck a luck.
In you play-for-fun pai-gow poker, if you have nothing and drop your 2nd and 3rd highest cards to the front hand, the Advice will occasionally swap the 3rd highest for the 4th highest. I haven't been able to figure out the rule -- first I thought it always did that when your back hand was ace-high, but that didn't hold up. Is it because the house way requires the dealer to drop the 2nd and 3rd highest, so if you and the dealer have equally bad hands, the front hand will be decided by your 2nd highest card but keeping the 3rd highest in the back hand has a better chance of winning? If so, why doesn't it do that all the time? I'm confounded by this. I haven't been able to see the pattern. Please enlighten me!
Sarah from Chicago, USA
Sometimes the two-card hand is so bad that it is better to put the third highest card as a kicker in the five-card hand. There is no rule of thumb to advise exactly when to do so but I have noticed it tends to happen when the second highest card is rather low. The logic my program uses is a large array of probabilities that each hand will win and goes with the hand with the greatest possible sum of probabilities that won't spoil the hand.
I was visiting an Indian casino the other day where the blackjack minimum bet was $2, but for each bet, they asked for an additional 25 cents (they did this by giving you 4 special 25 cent chips for each dollar chip you had) so essentially, you bet $2.25 to win back $2. While I was still able to walk away with a bit of a profit, I was wondering if there was a quick and easy way to calculate just how much of a house edge this "commission" gave them? I know that compared to dealer hitting on soft 17, it's gotta be a killer.
The total house edge is [fee + (house edge)*(bet)]/[fee + bet]. Let's say the house edge is 0.8%. Then the house edge including the fee would be [$0.25 + $2.00*0.008][$0.25 + $2.00] = 11.82%.
However, a quick and easy estimate is to simply divide the fee by the bet as the increase in house edge due to the fee.
Hey Wizard...I have just got into online gambling and have referred to your site for pretty much everything...I appreciate all the information on the single-deck Unified Gaming
information, but what about the six deck game? Can you tell me the player (or dealer) edge in this game as well as a basic strategy? It would seem to me that this would be one of the best bets at which to count cards...what do you think? Thank you in advance!
Chris from Denton, TX
You should follow my multiple-deck strategy but surrender against aces in the same hands as the single-deck strategy. I haven't studied if counting this game is more profitable than the single-deck game, but I would be interested to hear from anyone who is doing as you suggest.
When playing the Martingale double-up system against the single-zero roulette wheel on any one of the even chances. I figured that you will lose one time in every 248 sessions. Meaning a session that runs to completion with either a win of one unit or a loss of 255 units. Am I figuring correctly, if not could you please give the correct odds?
Jack from Neenah, Wisconsin
If the maximum loss is 255 units then you can bet up to 8 times. The probability of losing eight bets in a row is (19/37)8=.004835. So, you have 99.52% of winning one unit, and 0.48% of losing 255 units.
I was recently looking at a football pool that was taking place. This was one of those where there is a grid of 100 boxes and the numbers 0-9 run along the X and Y axises and correspond to the last number of the score. I am not a football fan and did not bet on this pool but I am a gambler and don't think it is such a great bet.
I am taking for granted that you know the type of pool that I am referring to. Each box costs $5 and pay outs are each quarter. If your box wins you win $125 and it is possible to win all 4 quarters if the last numbers remain the same winning $500 for a payout of 100 to 1.
The person soliciting the bet was trying to tell me that the odds of winning the $500 are 100 to 1. I disagree. First the box that has 0 + 7 stands a much better chance of winning than the 2 + 9 box. However, the odds of 0 + 7 remaining the last 2 numbers throughout the game must be high. If the boxes are chosen for you through a random process, can you tell me the approximate odds of winning the $500 prize?
Sam from Phillipsburg, USA
Assuming the cells in the grid are chosen at random, then the odds of winning any one quarter would be 1/100. Assuming each quarter was an independent event, which they aren't, the odds of winning all four quarters would be (1/100)4 = 1 in 100 million.
Mr. Wizard, what is the probability of rolling two pair when rolling four dice?
Brian from St. Catherines, Canada
There are combin(6,2)=15 different sets of pairs possible. There are combin(4,2)=6 ways the dice can roll any specific two pair. There are 6^4=1296 ways to roll four dice. So the probability is 90/1296=6.9444%.
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 from Rome, New York
Assuming six decks, and the dealer hits on soft 17, if the dealer is forced to play out every hand them my blackjack appendix 2B says the probability of the dealer busting is 28.58%. However, in a head-to-head game where the dealer doesn't bother to draw cards if the player has a blackjack or busts first, then my blackjack appendix 4 says the probability of busting is 24.36%.