Ask The Wizard #140
If you have reason to believe that the seven is weighted and is coming up more than it should, does that favor the don’t or the pass side of craps?
The fewer the sevens the greater the odds favor the pass line bet. The following table shows the house edge according to the percentage of sevens, assuming the probability of all other numbers is proportional to the fair probability.
House Edge in Craps According to Seven Probability
| Seven Probability | Pass House Edge | Don’t Pass House Edge |
| 15.000% | -0.666% | 3.499% |
| 15.333% | -0.202% | 3.024% |
| 15.667% | 0.237% | 2.574% |
| 16.000% | 0.652% | 2.148% |
| 16.333% | 1.044% | 1.744% |
| 16.667% | 1.414% | 1.364% |
| 17.000% | 1.762% | 1.005% |
| 17.333% | 2.089% | 0.667% |
| 17.667% | 2.395% | 0.349% |
| 18.000% | 2.682% | 0.051% |
| 18.333% | 2.949% | -0.227% |
In a recent programming exercise myself and other students were asked to describe a six-sided die in code, and then use our dice to determine play simple game. The object of the game was to roll the dice until the sum of the tosses reached exactly 100. Any toss that put the total over 100 would not be added and merely added to statistics. Quickly it was determined that 17 throws would be the least amount of throws needed to reach 100. However calculating the odds of that occurring has proved elusive. Calculating the odds of a specific sequence of throws is rather straight forward, but how might one factor in both non-specific ordering of throws, and the different ways of reaching 100 in 17 throws (16*6 + 1*4 and 15*6 + 2*5)?
The two ways you mention are the only ways to throw a total of 100 in 17 throws. The probability of throwing 16 sixes and one four is 17*(1/6)17. There are 17 possible positions of the 4 and each sequence has a probability of (1/6)*(1/6)*...*(1/6) with 17 terms. The number of ways to get 15 sixes and 2 fives is combin(17,2) = 136. So the probability of 15 sixes and 2 fives is 136*(1/6)17. So the total probability is (17+136)*(1/6)17. = 1 in 110,631,761,077.
I’m walking through a casino and see a craps table with a shooter winning lots of money. I feel lucky and want to place a bet. What can/should I do? Do I have to first make a pass line bet? Can I make a come bet? Can I make either bet with odds? Or, do I wait until the next round of play?
First, it doesn’t make any difference that the shooter is making lots of money. Your odds are the same on an ice-cold table. The past does not matter. However if you are going to play then wait patiently for a come out roll. Never make a pass bet after a point has been established.
I’m getting married in October. Since my fiancée and I are inveterate gamblers, we’re going to honeymoon in Antigua at a resort with a casino. Are there any particular differences between gambling in the states and the Caribbean? Also, do you believe gambling to be an aphrodisiac?
Speaking only for the Bahamas and Curacao the gambling is similar to the U.S.. Don’t expect to find any great rules. The blackjack is played with six decks and the dealer hits a soft 17. Gambling is a good aphrodisiac if you win.
Why do tables have a maximum limit, (i.e. Roulette)? A friend told me that this is to prevent a player from continuing to double his losing bet when betting red or black. I disagree as the odds obviously stay the same and the strategy does not pay well risk to reward. Why are there max limits at tables?
Every casino has some kind of limits to protect itself from losing more than they are comfortable with. However, on most tables the maximum is much less than it is in the high-limit area. The reason for this has nothing to do with protecting the casino against is Martingale players. Any casino manager worth his weight in salt knows betting systems always lose in the long run. I asked an executive with a major Las Vegas casino, who wishes to remain anonymous, why a casino would refuse a $10,000 in the main casino when they would accept it in the high-limit room. He said a casino manager only has so many employees he truly trusts. The big action he prefers to be under the watch of those people.
I was at Casino On Net. I was playing Roulette. I was making safe bets, only betting on the 1st 12(L), 2nd 12(M) & 3rd 12(H). I spun the wheel 5 times without betting, waiting for a pattern of one of the sets to not come up so I could bet on it, hoping this would shift probability of it landing in my favor. 5 spins later L didn’t show. I kept betting in L, I figured the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 OR 12 would land within 12 Spins so that I could at least recoup my money...but they didn’t. The table went 17 spins in a row without a LOW number & I went from $258 to $0.00...it was Bonus Money anyway. This is a 3 part question:
- Did my waiting for 5 spins without a low number popping actually increase the chances of probability for a L to show up?
- What are the odds that a Low number won’t show up 16 times in a row?
- What are the odds that a Low number won’t show up 17 times in a row?
Thank you, I also want to thank you for that Blackjack Guide, I turned $5.00 into $100.00 using your method.
- No
- A single-zero wheel is used at Casino on Net. So the probability of going 16 times with a zero is (25/37)16 = 0.1887%.
- (25/37)17 = 0.1275%.
I looked in your baccarat appendices and did not find odds for a "Super 6" that paid 12-to-1 as found the Casino Filipino Online. In sum they state: In general, winning hands are paid even money. If the final count of the Banker is 6 while Player has less than 6, bets on the Banker and Super 6 win and the game is a Super 6. In this case, Banker bets are paid 0.5-to-1, while Super 6 bets are paid 12-to-1. The Super 6 bet is based on the proposition that a particular deal would result in a Super 6. Bets on the Super 6 lose when the deal results in a Draw, or any other outcome that is not a Super 6. I was wondering how bad this game could be. Thanks.
The same game without the side bet was once played in Atlantic City and is analyzed in my baccarat appendix 6. There it show that the probability of a Super 6 is 5.3864%. So the house edge at 12 to 1 would be 29.977% (ouch!).
In "Ken Warren Teaches Texas Hold ’Em" he states that against Pocket Aces (where the suits are the same color) that no other hand will win more often than 6/5 suited of the opposite color. I accept the fact that this hand will win 22.89% of the time in the above situation. However, there are twelve ways to arrange pocket aces, and only 2 of them include the situation above. My question is then: What is the best starting hand against 2 random aces?
The best hand against two random aces are the other two aces. Aside from that using my 2-player hold ’em calculator we can see the expected value (probability of winning less probability of losing) is greatest with a suited 5 and 6, whether or not one of the aces share the same suit as the 5 and 6. So against a random set of pocket aces the suited 5,6 is still better. To correct you, if you have a suited 5 and 6 and you know somebody else has pocket aces there is a 50% chance that one of them match your cards in suit.
Me and this guy have been togehter on and off a couple times well i finally ended it about 8 months ago and i moved on to somebody new... he is a really nice guy and i couldn’t be happier... well i’ve been having dreams about me and my ex getting back together... so my question for you is.... do you see us together again in the future? maybe married or something? There is no really good reason as to why i broke up with him in the first place so hows about helping me out and answering my damn question!!!! THANKS!!!
My advice is to stick with the nice guy you have now. You seem to have a fear of commitment, which is causing you to second guess yourself. The grass is not greener on the other side of the fence. Hold onto the good thing you have now.