Betting Systems - Specific Games and Systems
If one bet on two columns in roulette the probability of winning would be 24/38, or 63%. This seems like a winning strategy to me, what is your opinion?
In roulette any bet or combination of bets carries a high house edge. The more likely you are to win the more you will have to risk relative to the reward. If you do this 10 times the probability of showing a profit is 46.42%. At 100 times the probability drops to 24.6%.
How would you fare if you played roulette like this -- bet $5 on both 0 and 00, bet $15 on two of the columns. Wouldn't you have like a 70% of winning?
You would have a 2/38 chance of winning $140, a 24/38 of winning $5, and a 12/38 chance of losing $40. The overall expected return is [(2/38)*140 + (24/38)*5 + (12/38)*-40]/40 = -5.26%. The same house edge as on every bet in double-zero roulette (except the 0-00-1-2-3 combination, which is 7.89%).
I had the good fortune of hitting four of a kind at a local casino, and was subsequently invited to play in a Let it Ride tournament, where approximately 300 players will compete for quite substantial prize money. My question is, what do you think would be the optimal strategy? Each player is to be given $5,000 in play chips, and the minimum bet will be $25 a hand. There will be "heats", with the first round having all but 100 eliminated, second all but 25, third will have 6 left, and then the final round.
Table game tournament strategy is very complicated. However, briefly, I would bide by time in the early hands of each round. Sometimes your opponents will all burn themselves out and you can advance without any effort. When it gets down to about five hands to go you'll need to make your move on any players way ahead of you. This is the time when you want to pull into first or go bust trying. It is also good to wait to save your big bets for when you act AFTER your biggest competitors.
First of all, thanks for providing reliable gambling info. You are one of only about four or five sites on the net doing so. In your opinion, is it possible that a mathematically sound method (card counting etc.) could ever be devised to give a positive expectation in baccarat? There has recently been some speculation (and wild claims) on the bj21 and other gambling forums.
Thanks for the compliment. I address the vulnerability to card counting in my baccarat appendix 2. To make a long story short, no, baccarat is not countable unless you use a computer.
I liked your article about the cancellation betting system. What sort of a cancellation system would you follow if instead of betting on either black or red, you placed a bet (of the same amount) on two of the three blocks that cover 12 numbers (e.g: 1 to 12 & 25 to 36) on each roll?
Start the same way as I explained with even money bets. Each bet should be the sum of the left and right numbers. However, following your two column strategy you should add double the amount to the right if you lose.
Is there a way of combining bets in roulette to maximize one's odds? For example, a dozen bet pays out 2 to 1. If I place two dozen bets, say the first and second set of 12, I have a 63.16% chance of having it pay off. These are better odds then a simple red/black, even/odd, or high/low bet. Although I really only gain 1 to 1 rather than 2 to 1(if I win, since part of my bet has to lose since the winning number cannot be in both the first and second set of twelves), the odds have been slightly shifted in my favor by combining two bets. Have the odds on these sorts of combinations been determined? If they have been, where might I be able to find them?
As long as you stay away from the 0-00-1-2-3 combination, the house edge on any combination of bets is always exactly 1/19, or 5.26%. There are ways to increase your probability of winning, but at the cost of winning less relative to your total wager.
Isn't it an even worse roulette betting strategy to bet multiple numbers on the inside during one bet (as most players do) vs. making a sequence of independent bets on one number? For example, if one had $100, betting 10 bets of $10 dollars on the number "8" would lose less than betting $10 on 10 numbers on one spin? It seems to me that "hedging" just guarantees that certain (in the above case 9 bets) will ALWAYS lose? You don't address "hedging" on your page?
See my Ten Commandments of Gambling. The sixth commandment is "never hedge thy bets." About your roulette question, the probability of losing all ten bets by betting one at a time is (37/38)10=76.59%. The probability of losing all ten bets by betting them all at once on different numbers is (28/38)=73.68%. By hedging, or betting ten numbers at once, you lower your probability of a total loss but also limit your maximum win to $26. The player betting one at a time could win up to $350. Both these methods have the same total expected return of 94.74%.
I read your page on systems and I have been telling people this for years! I deal roulette in a casino and I have seen all of the systems at one time or another. I have seen one system that, even though on a computer simulation might not work (probably won’t), "Seems" to work in real life. That means that I have seen it win more than lose.
The way it works is a player will put $75 dollars on the 1 to 18 $50 dollars on the 3rd 12 and $10 dollars on the 0-00 split for a total of $135 dollars. This covers all but six numbers (19 through 22) and will yield a 15 dollar payout every time the ball misses those 6 numbers EXCEPT when 0 or 00 hits in which case it's 40 dollars. I know it sounds nuts!!! But trust me, I'm here to tell you I have seen it win more than lose. It also works in reverse (duh). I would love to know the true odds of this system, but it's hard to tell someone that it doesn't work when they are walking off my table 2 grand richer:-)
There are 30 ways to win $15, 6 ways to lose $135, and 2 way to win $45 (not $40). The expected return of this combination of bets is ((30/38)*15 + (6/38)*-135 + (2/38)*(45))/135 = -.0526, or 5.26%, the house edge on any one bet or combination of bets as long as the dreaded 0-00-1-2-3 combination is avoided. In your observations you have likely seen fewer than expected 19-24 occurrences, which accounts for the illusion that this method is winning.
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!
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.
I agree with you that there are no system that can beat a negative expectation game. Anyway, I take a look at the cancellation system and keep wondering ... what about using it in a bet like the banker in baccarat, where you have a POSITIVE expectation outcome? In which extension would the commission payed to the house erode your gains in the long run? I apologize for my shady English.
The Banker is baccarat is not a positive expectation bet. You're confusing the probability of winning the bet with having a positive expectation. Even without a betting system, you will probably win any banker bet but you will win less than what you bet, because of the 5% commission. This makes the banker bet a negative expectation bet.
Have you heard of the "Reverse Labouchere" method, described in detail in Norman Leigh's book "Thirteen Against the Bank"?
No, and I don't really care what it is either. All betting systems are equally worthless.
Have you ever heard of the Ken Fuchs progression. If so, would you please e-mail me or post the details on your site.
I’m not familiar with it. Ken Fuchs co-wrote Knock-Out Blackjack so he can’t be all bad. However I just hear the word progression and I’m immediately skeptical.
Hi. You say all betting systems will fail. If you play roulette and bet one unit on number 1-12 and 2 units on number 13-24 wouldn’t you then have 66.66% chance to break even or win.
Not quite. You would have a 12/38 chance of winning 3 units, 12/38 of breaking even, and 14/38 of losing 3 units. The expected value is [(12/38)*3 + (12/38)*0 + (14/38)*-3]/3 = (-6/38)/3 = -2/38 = -5.26%. This will be true of any combination of bets as long as you avoid the dreaded 5 number combo (0/00/1/2/3). If you only play for one spin and want to maximize your probability of winning then bet equally on 35 of the numbers. You’ll have a 92.11% chance of winning 1 unit and a 7.89% chance of losing 35 units.
Hello. I’ve been an avid roulette gambler for some years now and for the first time ever I’m thinking of trying out a roulette system...Now I know how you feel about these so called "systems" and the scammers behind them, and belive me, I feel the same way, but I’ve come across two systems which can’t be ignored...
The first one is the 3q/A-strategy found in R.D Ellison’s book "Gamble to win: Roulette", which has a verified win rate of 7.94% (7500 spins). The system was tested and developed in conjunction with" Spin roultte Gold" by Frank Scoblete and "Roulette system tester" by Eric St. Germain.
The second one is Don Young’s roulette system which is verified to beat the Roulette System Tester from Zumma Publishing(15000 spins).
Now, I must say I’m still a bit sceptical about spending money on these systems, but since they've proved themselves over the long haul, I can't really see no reason why I shouldn’t. I mean, beating these testbooks have to mean something...
What’s your opinion on these systems? And do you think I should try them out??
Thanks alot! Have nice day. Best wishes
7500 spins? Is that it? Anyone can show a profit of 7.94% of total money bet over 7500 spins if they bet aggressively. Same is true about 15000 spins. Most systems are designed to have a lot of small wins and small number of large losses. A system requiring a huge bankroll can easily go 15000 spins and show a profit. Eventually the losses will come in and it won't pass the test of time. The big losses might also come at the beginning. The true way to put a system to the test is to play it over billions of trials. My opinion about these systems is the same as all systems, they are worthless. I have no problem with you trying them out but I do have a problem with anyone putting one dime in the pockets of those selling them.
Note: See the follow up to this question in the next column.
In your last column you said anybody could create a roulette system that showed a 6.5% profit over 7500 spins. Well, I’m anybody and am challenging you to give me one.
You got it! Actually the system boasted an advantage of 7.94%. I'll up that and go to 8.00%. So here is "Wizards 8.0% advantage system." Here is how to play it.
- This system can be played on any even money game, including roulette, but craps is strongly suggested due to the lower house edge.
- Player makes only even money bets. In roulette any even money bet will do and the player may change the bet at will (as always the past does not matter).
- Player must be comfortable with a betting range of 1 to 1000 units.
- The first bet is 1 unit.
- After each bet the player will determine 8.1% (the extra 0.1% is a margin of safety) of his total past wagers. If his net win is less than this figure he will bet the lesser of the difference and 1000 units. If his net win is more then he will bet one unit.
- Repeat until 7500 bets are made.
In roulette I did a computer simulation of this experiment 10,000 times and the player made his 8.0% 4236 times and failed 5764 times. So the first time with live play it would not be unlikely that the player would report a success story. In craps betting on the pass line using the same system resulted in 6648 wins and 3352 losses, for a success rate of 66.48%. Going back to roulette, if the spread is 1 to 10,000 units the numbers of wins was 8,036 and 1,964 losses. In all cases when the system doesn't hold up over 7,500 spins the loss is big, more than 8.0% on average.
Of course this system is just as worthless as every other. The point I hope I have made is that it is easy to easy to design a system that usually wins. However when you do lose you lose big. Over the long run the losses will be more than the wins and the player will have a lot less money in his pocket.
Dear Sir, In a single zero roulette game, the PROBABILTY of winning increases if you place a portion of your money on fewer numbers for more spins versus covering more numbers per spin, an example: If you are willing to risk 500$ in order to win 250$ then you could: Option (A): place 250$ on any of two dozen and should you be a winner you will win 250$. The probabilty of that happening is 24/37=(.648648). Option (B): Place 125 on any one dozen and should you be a winner you will win 250$ and walk away. However, should you lose you can now bet187.5$ on the same dozen and should you be a winner you will win 375$ which will get you the 250$ and the 125$ you lost on the previous spin. Now should you lose on both spins you still have 187.5$ to play with and you can place 20.833333$ on any nine numbers and should you be a winner you will get 750$ which is equal to your 500 original capital plus 250$ in winning which was your goal. The probability of that happening meaning either hitting a dozen OR nine numbers at LEAST once in three spin is equal to[1-(25/37)x(25/37)x(28/37)]=0.65451. Hence, for the SAME capital and for the SAME payoff you are able to increase your PROBABILITY of success as in option (B) if you play fewer numbers with less money but for MAYBE more spins.(As you might win on the first spin) You can even improve your probability further if you play only six numberes at a time and try to win 250$. Any Explanation??!!!! Assuring you of my highest regards and awaiting the favor of your reply I remain.
You are correct that option B has the greater probability of success, although the goal and the capital are the same. The reason is the average amount bet in option B is less, thus your money is exposed to the house edge less, thus the probability of winning increases. The amount bet in option A is always $500. The average amount bet in option B is (12/37)*125 + (25/37)*(12/37)*(125+187.5)+ (25/37)*(25/37)*(125+187.5+187.5) = 337.29.
When I was on the Vegas Challenge, with a few minutes to go, I had about $8,000 and needed to get to at least $24,000. So I split my bankroll into four piles of $2000 each and bet each one on a 4-number combination, each of which would have paid $22,000. This way I was not necessarily exposing my entire stake to the house edge, which increased my probability of winning.
The Daniel Rainsong challenge made good reading. However after some thought I can only conclude that: 1. Daniel wanted a professional analysis of his system, to confirm for himself whether or not it had a positive expectation 2. Daniel concluded that risking $2000 for $20000 was clearly preferable to simply paying someone [like yourself] to analyze his ’procedure’. I don’t know what you charge to analyze a game, but I am confident it would be more than $2000 in this case. If so, using the Wizard’s Challenge was a very smart choice -- no matter how small the chance of success. Nevertheless, your decision to accept the challenge and decline the surrender was an even smarter choice! The final irony is that this story will probably result in MORE ill-conceived systems. What do you think?
I offered a straight no-bet analysis to Mr. Rainsong for $1000 but he declined. He felt very strongly that he would win. The less evidence there is for anything, the stronger the faith of the believers. I charge more than $2000 to analyze most games but this was just a simulation. All I had to do was modify my existing blackjack simulator to fit the rules of this procedure. Not surprisingly Mr. Rainsong has had lots of interested buyers in his system despite the fact that it failed my test. All I can do is tell the truth that systems don’t work over the long run. Whatever the masses do with my advice is up to them.
Dear Wizard - Mathematically, why doesn’t the cancellation system work? (The system goes by a lot of other names too. Just to be clear, I mean the system where you start with a series of number and bet the total of the outside numbers, canceling them out when you win, etc.) It seems that all you have to win is 1/3 plus two of you bets to come out ahead. In roulette you have about a 45% chance of winning. So you should win over the long haul, but you don’t. Why not?
As with most betting systems the cancellation system usually does result in a session win, at the cost of occasional huge losses. When the cancellation system does lose the results can be your worst nightmare. During those times when you just seem to lose almost all the time the bet sizes start adding up geometrically, which can quickly deplete a bankroll if the cards don’t run your way.
www.ccc-casino.com has no zero roulette which they call Super Chance Roulette. Are there any systems that would be effective since there is no zero? Without the zero could one effectively play both black and red at the same time since there is no fear of the zero?
I tried to find that game but the site was down when I checked. However, assuming such a game did exist, the answer is no. No system could be expected to beat it, nor lose to it, over the long-run. The expected value of every system would be exactly zero.
I submit to you that the advantage video poker player should sometimes deviate from optimal strategy, if following the Kelly Criterion. In borderline hands, I think Kelly may favor going for the less volatile play, even at a lower return, although I can’t think of a particular example. What are your thoughts?
I agree! As discussed in my section on the Kelly Criterion, there is an optimal bet size for any given bet with a player advantage, for purposes of balancing both risk and reward. Betting the exact Kelly amount will result in the greatest bankroll growth for the player with average luck.
For example, in full pay deuces wild, with a return of 100.76%, the optimal amount to bet every hand is 0.03419% of bankroll. These days if you can find full pay deuces wild it will probably be only at the 25-cent denomination, but if you could bet anything, 0.03419% of your total bankroll would be the optimal amount for long-term bankroll growth. For a player with a bankroll of $3,656, a quarter denomination game would be the perfect Kelly bet size.
As I discuss in my section on Kelly, the optimal bet amount is the one that maximizes the expected log of the bankroll after the bet, which I will call the Kelly Utility. Usually the Kelly Utility is maximized by making the optimal strategy play. However, one exception would be a five 3s to 9s, with three deuces. In particular, let’s look at 22277. The expected value of keeping the deuces only is 15.057354, and keeping the five of a kind is always worth 15 exactly.
The following table shows both the conventional expected value and the Kelly Utility holding the three deuces. The Kelly Utility for any given hand on the draw is p*log(1+0.0003419*w), where p is the probability, and w is the win.
Player Holds Three Deuces
| Hand | Pays | Combinations | Probability | Return | Kelly Utility |
| Four deuces | 200 | 46 | 0.042553 | 8.510638 | 0.001222 |
| Wild royal | 25 | 40 | 0.037003 | 0.925069 | 0.000137 |
| Five of a kind | 15 | 67 | 0.06198 | 0.929695 | 0.000138 |
| Straight flush | 9 | 108 | 0.099907 | 0.899167 | 0.000133 |
| Four of a kind | 5 | 820 | 0.758557 | 3.792784 | 0.000563 |
| Total | 1081 | 1 | 15.057354 | 0.002193 |
The next table shows the same figures for holding the five of a kind.
Player Holds Five of a Kind
| Hand | Pays | Combinations | Probability | Return | Kelly Utility |
| Four deuces | 200 | 0 | 0 | 0 | 0 |
| Wild royal | 25 | 0 | 0 | 0 | 0 |
| Five of a kind | 15 | 1 | 1 | 15 | 0.002222 |
| Straight flush | 9 | 0 | 0 | 0 | 0 |
| Four of a kind | 5 | 0 | 0 | 0 | 0 |
| Total | 1 | 1 | 15 | 0.002222 |
You can see that the Kelly Utility is higher keeping the pat five of a kind, at 0.002222 vs. 0.002193. For this particular hand, keeping the five of a kind will be the correct play under the Kelly Criterion for bankrolls up to 13,290 units, or for quarter players up to $16,613.
As I said, the optimal Kelly bet size for the optimal strategy player is 0.03419% of bankroll. The optimal bet size for the player playing optimal strategy, except keeping a dealt 22233 to 22299 is 0.03434% of bankroll. The bankroll growth of the optimal strategy player will be 0.0002605% per bet made. For the Kelly player it will be 0.0002615% per bet made. Every 40,000 hands the player following optimal straegy, and Kelly bet sizing, can expect bankroll growth of 10.98%. The conservative player keeping a dealt 22233 to 22299, and Kelly bet sizing based on that strategy, can expect growth of 11.03% per 40,000 hands.
So, I maintain that in some situations, indeed, you should go against optimal strategy and go for the more conservative play. I just hope Rob Singer doesn’t hear about this.A reader asked about a slot tournament at the Wynn. The cost to enter was $25,000, and the average prize was $30,000. You said that you need a bankroll of about three million to enter, according to the Kelly Criterion. I have two questions:
1. Does this take into account the unknown house edge on the slot machines?
2. What would be the playing strategy for the best overall return? Could you just sit back and not gamble, and hope that the other 49 players all end up behind, while you break even and take the grand prize of $1,000,000?
Slot tournaments are always held on dedicated tournament machines. Usually these machines don’t accept bets, so your balance will either stay even or go up, after each play. So it doesn’t make any difference what the return is; the more you play, the more you can expect your balance to go up. Even if you had to play conventional slot machines, I would still bet as fast as possible, stopping only if I got a jackpot large enough to likely win the tournament. The reason is that it is very unlikely that 49 out of 49 players would be negative.
Interestingly, there was once a slot tournament at Caesars Palace where they gave a prize to the person who finished last. However, they didn’t announce this rule until the award ceremony. If you somehow knew of such a rule, indeed, it might be best to not bet.
Some gambling books say that the correct Kelly bet is advantage/variance. However, you say that is just an approximation and the correct answer is to maximize the expected log of the bankroll after the bet. My question is, how much error is there in the variance approximation?
Advantage/variance is a pretty good approximation. Let’s look at full pay deuces wild, for example. The variance formula says to make a bet of 0.000295 times bankroll. Exact Kelly results in a bet of 0.000345 times bankroll.