Gambling Guide
Betting Systems - FAQ
Brian from Denver, Colorado
Two key pieces of information you left out is how much you are betting and in which game. I'll assume that you are flat betting $1 at a time on the Player bet in baccarat. The probability that the player will win, given that there isn't a tie is 49.3212%.
Let ai denote the probability if the player has $i he will reach $1,200 before losing everything. Let p the probability of winning any given 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.
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
For more on the math of this kind of problem, please see my MathProblems.info site, problem 116.
JJ from Boston, USA
No.
Scott 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, or the more you will grind the casino down if you have the edge.
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.
Emi from Manila, Philippines
There are lots of them, and they are all worthless.
Mike from Mesa, USA
It doesn’t make any difference.
KYK from Hong Kong
As I have said many times, in the long run all betting systems are equally worthless.
Shuck from Las Vegas, USA
No, the number of hands does not affect the house edge. The amount you can expect to lose is the product of the house edge, average bet size, and number of bets.
"Anonymous" .
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.
"Anonymous" .
You’re welcome. It must have taken all day to read my entire site. You are confusing betting systems, which are worthless, to legitimate strategies that give the player an advantage. Two games that can be proven beatable with good rules and proper strategy are blackjack and video poker. So I call a system a worthless method of following trends in games with a house advantage, and a strategy something like card counting in blackjack that is mathematically proven to work. Video poker can be beaten by hunting down the best pay tables and then following a reliable strategy on which cards to keep and which to discard.
"Anonymous" .
Thanks! Yes, I said before that if I had a betting system that had just a 1% advantage I could turn $1000 into $1,000,000 by simply grinding out that edge. This would also be possible in video poker but it would take much longer because the 0.77% advantage game (full pay deuces wild) can only be found in the quarter level. Assuming you can play 1000 hands per hour (a speed few can attain) and played perfectly that would result in an average income of $9.63 per hour. To reach $1,000,000 would require working 11.86 years non-stop. $1000 would also be very undercapitalized to play quarter video poker, so the risk of ruin would be quite high. It would be faster to reach the $1,000,000 with the same edge in a table game because the player can bet more.
Mark, a casino manager
I couldn’t have said it better myself.
Chris
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.
"Anonymous" .
As I’ve said hundreds of times there is no magic number of when you enter the "long run." However the more impressive your results the fewer the hands your need to make a case that they are not just random. In your case the probability of getting 54.5% or better out of 2391 games is just about 1 in 200,000. So I would say that record is worth to be taken very seriously. Here is how I arrived at that number:
Expected wins = 2391/2 = 1195.5
Actual wins above expectations = 107.5
Standard deviation = sqrt(2391*(1/2)*(1/2)) = 24.45
Standard deviations away from expectations = (107.5 + 0.5)/24.45 = 4.4174
Probability of 4.4174 standard deviations or more = normsdist(-4.4174) = 0.000005 = 1 in 200,000
"Anonymous" .
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.
Richard
Personally I don’t see what is so interesting about Parrondo’s paradox but you are not the first to ask me about it so I’ll give you my thoughts on it. The thrust of it is that if you alternate between two particular losing games the player can gain an advantage.
As an example, consider Game 1 in which the probability of winning $1 is 49% and losing $1 is 51%. In Game 2 if the player’s bankroll is evenly divisible by 3 he has a 9% chance of winning $1 and 91% of losing $1. In Game 2 if the player’s bankroll is not divisible by 3 he has a 74% chance of winning $1 and 26% of losing $1.
Game 1 clearly has an expected value of 49%*1 + 51%*-1 = -2%.
In Game 2 you can not simply take a weighed average of the two possibilities. This is because the game quickly gets off of a bankroll remainder of 1 with a win, and often alternates between remainders of 0 and 2. In other words the bankroll will disproportionately play the game with a 9% chance of winning. Overall playing Game 2 only the expected value is -1.74%.
However by alternating two games of Game 1 and two games of Game 2 we break the alternating pattern of Game 2. This results in playing the 75% chance game more and the 9% less. There are an endless number of ways to mix the two games. A 2 and 2 strategy of playing two rounds of Game 1 and two of Game 2, then repeating, results in an expected value of 0.48%.
I should emphasize this has zero practical value in the casino. No casino game changes the rules based on the modulo of the player’s bankroll. However I predict it is only a matter of time before some quack comes out with Parrondo betting system, alternating between roulette and craps, which of course will be just as worthless as every other betting system.
Michael from Los Angeles
I get asked variations of this question a lot. The fact of the matter is I have written hundreds of different simulations. I write the simulations myself in C++ to do exactly what I want. Those writing usually seem to be looking for something to test betting systems. I’m afraid I have nothing and know of nothing that lets the user input how the betting system works and then tests it. If there were something that worked perfectly what you would learn from it is that all betting systems are equally worthless, exactly what I have been saying for years.
"Anonymous" .
I disagree, at least for the reason you state. Under your scenario most people would indeed leave Vegas winners. However, some players would lose the first bet and keep falling deeper and deeper in the hole after that, until they exhaust their entire bankroll. Assuming the same game and player strategy, the overall house edge would remain the same regardless of player money management strategy. In other words, betting systems not only can’t overcome the house edge, they can’t even put a dent in it. Getting back to your question, if everyone quit as soon as they were ahead, there would be a lot less gambling going on. So while the house edge would be the same, it would be applied to less total money bet, which would indeed hurt the casinos financially.
Dennise from Lakewood, CO
I will. Over a long enough period of play 99.9% of system losers will lose and 0.1% will sit there full of self-righteousness, thinking it was skill when it was really just luck.
Harold from Los Angeles
You are right. In the long-run, betting systems and the reasons for a bet don’t matter. With a limited bankroll, ruin is frequently the outcome. However, there is also a cap on the downside, with an almost unlimited upside. In the long run it all averages out and the casino makes close to expectations based on handle and the house edge.
Rob from Vienna
I’ve addressed this before and I disagree with your hypothesis. As I have said many times, all betting systems are equally worthless. Thus, if the casino had no house edge it would neither win nor lose money over the long run. Let’s say every player had the goal of winning $1,000,000 or go bust trying. Most would go broke but the few players that won the $1,000,000 would even things out.
Jonathan from Preston, England
I tried to sign up for an account there to check this out but they block U.S. players. I’m told the minimum bet is £2 and the maximum is £50. Even in a zero house edge game like no-zero roulette there is still no betting system that will get above, or below, that 0% figure. No matter what you do the more you do of it the closer the actual house win will get to 0%.
My question is, is there any way to calculate an optimum playing range of wins/losses? That if a player reaches X number of losses it is highly unlikely that they will recover and should just quit? Likewise, if a player wins X amount, then the player has achieved respectable winnings considering the probabilities of the game and should quit while ahead.
Chris from Tampa
I get asked variations of this question all the time. If you are playing a game with a negative expectation, which is almost always the case, the best strategy to preserve your money is to never play. However, if you are going to play anyway, for the sake of entertainment, there is no best quitting point. The more you play, the more you can expect to go down from wherever your bankroll is at the moment. As I have said many times before, a good time to quit is when you aren’t having fun any longer.
My question isn’t about winning long term with systems, as we know that’s impossible. But might systems have a usefulness in ’tailoring’ the losing experience? For example, player A prefers that each trip to the casino he will either win or lose a moderate amount of money (of course he’ll lose slightly more often than win). Player B prefers a chance to make a little money 4 out of 5 trips, and lose lots of money 1 in 5 trips.
Both will lose money in the long run, but is there a betting system that might help each accomplish his goal?
"Anonymous" .
Yes. While betting systems can not change the house edge, they can be used to improve the probability of achieving trip objectives. Player A wants as little risk as possible. To minimize risk he should flat bet. Player B wants a high probability of a trip win. He should press his bets after a loss. Such a strategy carries the risk of a substantial loss. Although you didn’t ask, a player who wants to either lose a little or win big should press his bets after a win. This kind of strategy will usually lose, but sometimes will have a big win.
Mark from Gatineau, Quebec
Thanks for the kind words. I think I’ve answered this one before, but no, even with zero house edge there is still no betting system that can win, over the long run.
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%.
Since I cannot control the statistics, my question involves something that I can control, session length (and bankroll). Since a million or a billion number of hands is composed of so many “sessions” of, for example, 300 − 1,000 hands, does it not make sense to play until you either a) reach a pre-established target win amount, or b) play until you recover from a losing streak and end the session at break even?
One last question, can you recommend a source for a software simulation system that can handle all rule variations, stop/loss provisions, extracting “sessions” of variable lengths, and variable hit/stand strategy based upon the size of the bet. I would love to give my approach a shot at the computer.
Tom from Bowling Green, KY
Thanks. I get questions like this a lot. Usually I delete them, but since you buttered me up so nicely, I’ll answer this time. As I state many times, all over the site, all betting systems are equally worthless. There is no magic quitting point. I am not opposed to any winning or losing marker for quitting, but the expected value is no better or worse than flying by the seat of your pants. I’m told that Casino Vérité is capable of simulating what you ar asking about. Finally, in blackjack, the hit/stand decision should not depend on the size of the bet. The right play for a $1 bet is right for a million dollars.
Larry
Putting aside the issue that such a system would be impossible, I would charge about 50 million dollars. If I had no buyers, fine, I would just go out and make much more than that on my own.
Thus says Bluejay, "...if you know that the longer you play, the more likely you are to lose, then that means that the shorter you play, the better your chances of winning. And the shortest term you can have is just one play. And so statistically, that’s your best bet: making just one even-money bet, putting all your money on the line at once..."
Does the Wizard of Odds agree with this reasoning?Peter from Sydney
Yes, absolutely! If your goal is to win or lose $x, if limited to even money games, then you maximize your odds by making just one even money bet. This was once the situation, although not limited to even money bets, on a never-aired episode of “The Casino.” There I was consulted on how to maximize the odds of achieving a win of $4,000, given ordinary games and a starting bankroll of $1,000. I had them bet $100 on the pass line and then $900 on the odds in craps. Unfortunately, we lost. Had we won that bet, I would have had them bet enough to get to the $4,000 goal.
However, if fun enters into the equation, you will get more by making smaller bets for a longer period of time. If you simply want to minimize your expected loss, then don’t play at all.
Is Andy correct in that the best way to turn $30 into $1,000 is to put the whole $30 on a single number in roulette?
"Anonymous" .
No, he is not correct. The probability of Andy's single bet strategy is 1/38 = 2.6316%.
After much trial and error, I devised my "Hail Mary" roulette strategy, which will increase the odds of turning $30 into $1,000 to 2.8074%.
Wizard's "Hail Mary"strategy for roulette:
This strategy assumes that bets must be in increments of $1. In all bet calculations, round down.
Let:
b = Your bankroll
g = Your goal
- If 2*b >=g, then bet (g-b) on any even money bet.
- Otherwise, if 3*b >=g, then bet (g-b)/2 on any column.
- Otherwise, if 6*b >=g, then bet (g-b)/5 on any six line (six numbers).
- Otherwise, if 9*b >=g, then bet (g-b)/8 on any corner (four numbers).
- Otherwise, if 12*b >=g, then bet (g-b)/11 on any street (three numbers).
- Otherwise, if 18*b >=g, then bet (g-b)/17 on any split (two numbers).
- Otherwise, bet (g-b)/35 on any single number.
In other words, always try to reach the goal, with just one bet, if you can, without exceeding the goal. If there are multiple ways to accomplish this, then go with the one with the greatest probability of winning.
What about other games, you might ask? According to the Discovery Channel voice-over guy, "Everyone agrees that roulette is the best get rich quick scheme in the casino." Well, I don't. Even limiting ourselves to common games and rules, I find craps to be better. In particular, betting the don't pass and laying the odds.
Following my Hail Mary strategy for craps (explained below), the probability of turning $30 into $1,000 is 2.9244%. This assumes the player may lay 6x odds, regardless of the point (which is the case when 3x-4x-5x odds are allowed taking the odds). This probability of success is 0.117% higher than my Hail Mary strategy for roulette, and 0.2928% higher than the Andy Bloch strategy.
Andy might argue that my argument above relies on an assumption of a minimum bet of $1, which is hard to find in Vegas on a live dealer game. Expecting somebody might say that, I ran through both games under an assumption of a $5 minimum and betting in increments of $5. In that case, the probability of success using my Hail Mary strategy is 2.753% in roulette and is 2.891% in craps. In both cases, greater than the 2.632% under the Andy Bloch strategy.
In all fairness, the Discovery Channel would have never put the insane rant above on the air and was surely looking for something simple that the masses would understand. Andy was surely giving them something they wanted to hear. The basic premise of his advice is that if you want to reach a certain goal, then a hit-and-run strategy is much better than letting the house edge grind you down with multiple bets. That is definitely true and something I've been preaching for 17 years.
Wizard's "Hail Mary" strategy for craps.
This strategy assumes that bets must be in increments of $1 and wins will be rounded down to the nearest dollar. In calculating bets, never bet so much that you overshoot the goal. Also, never make a bet amount that will cause you to get rounded down.
Let:
b = Your bankroll
g = Your goal
- Bet max($1,min(b/7,(g-b)/6)) on the don't pass.
- If a point is rolled, and you have enough for a full odds bet, then lay the full odds. Otherwise, lay whatever you can.
So, I hope Andy and the Discovery Channel are happy. I've spent days running simulations to prove them wrong.
This question was raised and discussed on my forum at Wizard of Vegas.
