Ask the Wizard: Betting Systems Explained

Last update: May 13, 2007

Betting Systems

Good question. I have run across numerous system believers and the one thing they all seem to have is a lot of conceit. Despite the fact that they never seem to have gone much past algebra, if they made it that far, they all think they know better than greatest names in mathematics. This inability to consider contradicting evidence or other points of view is certainly not confined to betting system chasers. The more ridiculous a belief is the more tenaciously it tends to be held, and there is no shortage of ridiculous things for weak-minded people to believe in. Oct. 3, 2005

Dear Wizard, I know from reading your website and from other sources that betting systems do not give you an advantage over the house. My question is do they decrease the house advantage? I have been playing the betting system outlined in Progression Blackjack by Donald Dahl for the last 8 years and it gives me the excitement of betting higher amounts then I normally would. I usually play the $10 tables and I often get up to $30 bets and on my last trip to Vegas I got up to the $100 level at Sam's Town which really got my heart pumping not to mention a $600 profit that I left the table with. Thanks for your help. Regards, Chris W.

No! Not only do betting systems not overcome the house edge but they can't even put a dent in it. Nor can they increase the house edge. All they can do is affect volatility. Since it sounds like you like a volatile exciting game than your system is fulfilling its purpose. Just don't expect to win. Sept. 30, 2004

I would like to encourage your readers to continue to utilize their favorite betting system. They all work. We've never lost long term on any of them. Of course when I say "we" I mean those of us who work in the casino. -Mark, A casino manager

I couldn't have said it better myself. Sept. 14, 2004

Do you mean to tell me that man has designed a way to put 3 million transistors on a single chip (microprocessor) the size of a finger nail, and we don't have a way to beat a 50/50 even money game bet. I find that to be unbelievable, besides I found that computer simulations are definitely not the same as live world action. Also why don't casinos introduce video blackjack to thwart the card counters and get rid of dealers? - Mark Grzeskowiak from Chicago, USA

I have said numerous times that there is no long-term way to beat a game with a house edge. If there were a true 50/50 game with no house edge it would be impossible to guarantee beating or losing to it under real world conditions. The results always approach the house edge in the long-term. It is not just computer simulations that back this up but the fundamental laws of probability. About video blackjack that may be the way of the future. I have seen fully electronic tables with video display at the World Gaming Expo. I have also seen tables that with cameras can track every bet and every play each player makes. This enables the house to accurately comp players and alert them to card counters. These tables look and feel like any other blackjack table, so you card counters may be out of business if these tables are successful. Jan. 14, 2001

In general, when betting on anything which pays even odds, is there any kind of "system" to help improve chances and or payoffs? - JJ from Boston, USA

No. Nov. 11, 2000

You take a lot of your time or space on your web site to explain that no system can beat the house edge. I'd like to know what drives you to continue to play Casino games if you know that you will lose in the end? - Yvan from Quebec, Canada

My own philosophy of gambling is that I bet big when the odds are in my favor and small when they are not. The reason I bet at all when the odds are against me is for the entertainment value. Considering the low house edge I can get in blackjack, video poker, and craps, gambling is less expensive than most any other form of entertainment on an expected-loss-per-hour basis. Nov. 4, 2000

Q: I have read a couple of articles about 'Parrando's Paradox'. Is there a way that you could explain what is going on as it is extremely counter-intuitive that two losing games played in a certain sequence could add up to a winner - I thought I understood the mathematics of gambling / probability! I can see that there is a subtle link between games A & B as game B is dependent on capital that is affected by game A; I am unable to carry the logic any further. Does Parrando's Paradox have any implications for negative expectation casino gamblers? I doubt it but would love to hear it from someone with a greater understanding. - Gavin Parnell from Burt St Edmunds, England

A: Parrando's Paradox states that two suboptimal games of chance can show a long term gain if played alternately. However the games can not be independent of each other, which eliminates any comparison to casino games. July 30, 2000

Q: Would it be that in general the longer you play the more likely you are to lose, with the exception of games where the player can overcome the dealer advantage? Do you recommend a strategy to overcome this, such as initially making large bets as a percentage of deposit. If you come out ahead on larger bets then you can reduce the size of your bets to try to stay in the game longer. - Chris Baker from Andover

A: Yes, in a negative expectation game (as most are) the longer you play the more likely you are to lose. If you want to maximize your chances of a net win, at the cost of a possible very large loss, then the best strategy is to bet more after losing, thus trying to recoup past losses. The Martingale is an example of a very aggressive form of this strategy. April 29, 2000

Q: Are you saying in the long run you will lose at every casino game no matter what you do? - Joe from Harrisburg, US

A: With the exception of rare positive expectation opportunites in blackjack and video poker, yes, that is what I'm saying. March 11, 2000

Reaching a specific goal

During a 4-hour layover in Vegas, what's my best strategy to double a $2000 bankroll? what game, large or small bets, etc.? - Tom from Culver City

First I would take a taxi to the Hard Rock, the closest major casino to the airport. I'm not sure how much odds the Hard Rock allows in craps but I would guess 3-4-5. If that is the case then bet 1/7 of your bankroll on the don't pass bet, or $275 to round down. If a point is established then lay the maximum on the odds, or $1650. If you win you'll be a lot closer to your goal, the amount will depend on the point. Win or lose bet the lesser of 1/7 of your bankroll and 1/7 of how far you are from $4000. If you get close to either extreme just get it over with and bet everything if you're low, or whatever you need to close the gap on $4000 if you're high, and forget about the odds. Four hours should be enough time. However don't dilly dally. The lines at security can get pretty bad. If your outbound flight is in terminal C be sure to ask an agent about the secret entrance. Oct. 26, 2005

Dear Mr. Wizard, If you had $5000 to bet and wanted to win only $200 what game would you play? Please assume European rules and choose only among roulette, black jack, or baccarat.

I would bet $200 on the player bet in baccarat. If it wins, walk, if it loses then bet $400. Then just go into a Martingale until you win your $200 or lose your entire $5000. Jan. 2, 2005

If I flat bet until I either win $100 or lose $1000 what is my probability of reaching both goals?

Ignoring the house edge the probability of achieving the winning goal is the loss marker divided by the sum of the loss marker and the winning marker. In this case 1000/(1000+100) = 1000/1100 = 90.91%. However the house edge will reduce the probability according to the house edge of the game played and the amount bet per hand, the smaller the bet the lower the probability of achieving the winning goal. Nov. 28, 2004

I have been mulling over some money management techniques, and wanted to thank you for some sound advice (more to the point, probability theory) and it seems that win probabilities are based roughly on your starting bankroll. For example, estimating the probability of winning $100 on a $200 buy-in without going broke first. This is very helpful, but without doing the math (shame on me, I know) I get the feeling win limits should be based more on your betting unit, i.e. $1, $5, $10, etc. Basically the idea you will have smaller fluctuations over time using a smaller bet than a larger one. I guess my actual question is this: if I have a given bankroll (say $100) and a stated win limit (say $50), what (if any) betting unit would give the greatest possibility of success? I am thinking too small of a bet would give lower my chances of getting much above the mean and too large would run the risk of bankruptcy. Any advice or comments? - Scott Miller from Saline, Michigan

In your example, assuming a negative expectation game, the best bet size to maximize the probability of reaching your win goal is $50. In a positive expectation game the best bet size is as small as possible. The reason is that the more you play the more the house edge will grind you down in a regular game, or the more you will grind the casino down if you have the edge. March 11, 2001

The Wizard's Challenge

Is there a way you can test my betting system outside of your $2000 challenge?

My fee to do a straight up test would still be $2000. That is the value of my time to do the test. It costs me almost nothing to offer $20,000 if you pass the challenge because it is mathematically nearly impossible that you will win. Nov. 28, 2004

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. Nov. 8, 2004

Bet Sizing

Q: I've been a dealer n Vancouver, Canada for over six years. Having dealt all casino games (except craps) I've decided that a person's best betting strategy would be to bet all their bankroll on a one-shot bet, preferably in bacarrat, and on Banker. My decision is based on the observation that the longer a player gambles, the more likely that the odds will get the best of them, and the more likely it is that they will lose everything. The one-shot method may not be entertaining, but surely it'd be more profitable (or I should say, "less damaging") - Mel C. from Coquitlam, BC, Canada

A: You're absolutely right. The fewer bets actually made the better the odds are of actually winning. The expected loss is also a function of the total amount bet. If the player keeps circulating money back and forth between himself and the dealer the house edge will gradually grind the player down. However maximizing the odds of winning should not be the only objective to gambling. Having fun is also important. Plopping an entire bankroll on the table in one bet may not be as fun as playing it out gradually. It may also may have a greater chance at a large loss depending on how much play is involved in the alternative. If one really wants to cut down the house edge the best thing to do is put your money in the machine marked "change." Aug. 13, 2000

Quitting while you're ahead

I was wondering about house edge: The house always has the advantage of change but one thing is forgotten and is not to be taken lightly when doing the calculations: I choose when to stop. If I see I am losing I may cut my losses. If I see that I won enough (there is no such thing) I can decide to stop betting. The house has no such option How does it affect the calculation. - Yaniv

In the long run this kind of money management will neither help you nor hurt you. By cutting your loses at a certain point and walking away you risk missing a comeback. By walking away with a modest win you risk not turning it into an even bigger win. Of course things could also get worse too. In general you can assume the past does not matter and every hand is a new beginning. The best way to improve your odds is to cut down the house edge as much as possible. I'm not against money management but it won't effect the house edge. July 18, 2001

First of all, I really enjoyed your site. I thought I had the ultimate system until I did the math that your site provided me. My question is this: the Casino is stationary, the gambler is mobile. If everyone went to the Casino and promptly left after winning just one unit would there be any casinos? What I mean by one unit, lets say playing Black Jack is a five dollar Chip, I show up with $50 and leave after winning $5. Is there a ratio that I should use if this is the strategy I want to play ($50 to $5 = 10%)? And most importantly can this work? - Peter Simpson from Vancouver, Canada

The casinos would still win. Most players would indeed walk away with their one unit but some would lose their entire stake, whatever it may be. If you consider only even money games with a probability of winning of p then the probability of winning 1 unit before losing x units is (((1-p)/p)^x- 1) / (((1-p)/p)^x+1 - 1). In the case of double zero roulette with a bankroll of ten units the probability of winning one unit is 85.4268%. For every one of these players the casino will profit 10*(1-0.854268)- 0.854268 = 0.603056 units. Dec. 31, 2000

Q: I was curious - I'm sure I can't get better than house odds - but wanted to test a modest gambling approach -the quit while you're ahead scenario. Say I start with an even $1000. What percentage of time will I leave with $1200 rather than leave with $0 assuming I have to leave once I hit one or the other. Up 20% rather than be down 100% betting on the player in baccarat. - Brian from Denver, Colorado

A: One key piece of information you left out is how much you are betting. I'll assume you are flat betting $1 at a time. The probability that the player will win a single Player bet in baccarat, given that there isn't a tie, is 49.3212%. Let a_i denote the probability if the player has $i he will reach $1200 before losing everything. Let p the probability of winning any given Player bet = 49.3212%.

a₀ = 0

a₁ = p*a₂
a₂ = p*a₃ + (1-p)*a₁
a₃ = p*a₄ + (1-p)*a₂
.
.
.
a₁₁₉₇ = p*a₁₁₉₈ + (1-p)*a₁₁₉₆
a₁₁₉₈ = p*a₁₁₉₉ + (1-p)*a₁₁₉₇
a₁₁₉₉ = p*a₁₂₀₀ + (1-p)*a₁₁₉₈
a₁₂₀₀ = 1

---

Divide the left side into two parts:

p*a₁ + (1-p)*a₁ = p*a₂
p*a₂ + (1-p)*a₂ = p*a₃ + (1-p)*a₁
p*a₃ + (1-p)*a₃ = p*a₄ + (1-p)*a₂
.
.
.
p*a₁₁₉₇ + (1-p)*a₁₁₉₇ = p*a₁₁₉₈ + (1-p)*a₁₁₉₆
p*a₁₁₉₈ + (1-p)*a₁₁₉₈ = p*a₁₁₉₉ + (1-p)*a₁₁₉₇
p*a₁₁₉₉ + (1-p)*a₁₁₉₉ = p*a₁₂₀₀ + (1-p)*a₁₁₉₈

---

Rearange with (1-p) terms on the left side and p terms on the right:

(1-p)*(a₁) = p*(a₂ - a₁)
(1-p)*(a₂ - a₁) = p*(a₃ - a₂)
(1-p)*(a₃ - a₂) = p*(a₄ - a₃)
.
.
.
(1-p)*(a₁₁₉₇ - a₁₁₉₆) = p*(a₁₁₉₈ - a₁₁₉₇)
(1-p)*(a₁₁₉₈ - a₁₁₉₇) = p*(a₁₁₉₉ - a₁₁₉₈)

---

Next multiply both sides by 1/p:

(1-p)/p*(a₁) = (a₂ - a₁)
(1-p)/p*(a₂ - a₁) = (a₃ - a₂)
(1-p)/p*(a₃ - a₂) = (a₄ - a₃)
.
.
.
(1-p)/p*(a₁₁₉₇ - a₁₁₉₆) = (a₁₁₉₈ - a₁₁₉₇)
(1-p)/p*(a₁₁₉₈ - a₁₁₉₇) = (a₁₁₉₉ - a₁₁₉₈)

---

Next telescope sums:

(a₂ - a₁) = (1-p)/p*(a₁)
(a₃ - a₂) = ((1-p)/p)²*(a₁)
(a₄ - a₃) = ((1-p)/p)³*(a₁)
.
.
.
(a₁₁₉₉ - a₁₁₉₈) = ((1-p)/p)¹¹⁹⁸*(a₁)
(a₁₂₀₀ - a₁₁₉₉) = ((1-p)/p)¹¹⁹⁹*(a₁)

---

Next add the above equations:

(a₁₂₀₀ - a₁) = a₁ * (((1-p)/p) + ((1-p)/p)² + ((1-p)/p)³ + ... + ((1-p)/p)¹¹⁹⁹)

1 = a₁ * (1 + ((1-p)/p) + ((1-p)/p)² + ((1-p)/p)³ + ... + ((1-p)/p)¹¹⁹⁹)

a₁ = 1 / (1 + ((1-p)/p) + ((1-p)/p)² + ((1-p)/p)³ + ... + ((1-p)/p)¹¹⁹⁹)

a₁ = ((1-p)/p - 1) / (((1-p)/p)¹²⁰⁰ - 1)

---

Now that we know a₁ we can find a₁₀₀₀:

(a₂ - a₁) = (1-p)/p*(a₁)
(a₃ - a₂) = ((1-p)/p)²*(a₁)
(a₄ - a₃) = ((1-p)/p)³*(a₁)
.
.
.
(a₉₉₉ - a₁₈) = ((1-p)/p)⁹⁹⁹⁸*(a₁)
(a₁₀₀₀ - a₁₉) = ((1-p)/p)⁹⁹⁹⁹*(a₁)

---

Add the above equations together:

(a₁₀₀₀ - a₁) = a₁ * (((1-p)/p) + ((1-p)/p)² + ((1-p)/p)³ + ... + ((1-p)/p)⁹⁹⁹)
a₁₀₀₀ = a₁ * (((1-p)/p)¹⁰⁰⁰ - 1)) / ((1-p)/p - 1))
a₁₀₀₀ = [ ((1-p)/p - 1) / (((1-p)/p)¹²⁰⁰ - 1) ] * [ (((1-p)/p)¹⁰⁰⁰ - 1) / ((1-p)/p - 1) ]
a₁₀₀₀ = (((1-p)/p)¹⁰⁰⁰ - 1) / (((1-p)/p)¹²⁰⁰ - 1) =~ 0.004378132.

Categories for Questions about Betting Systems

FAQ General Questions Martingale Specific Games & Systems Positive-Expectation Situations

It may seem incredible that the chance of reaching $1200 before losing $1000 is less than 1%. But the fact is that if given enough time the odds are likely to catch up to the player in any game of luck and the bankroll will keep going down gradually. However if you were to bet larger amounts your odds would be much better. The following are the odds of winning 20% before losing 100% at various units of bet size.

$5: 0.336507
$10: 0.564184
$25: 0.731927
$50: 0.785049
$100: 0.809914

June 18, 2000

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