# Betting Systems - Martingale

Chris from Andover

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.

Charger

This is a close variation of the Martingale betting system, in which the player doubles after every loss. Usually, the Martingale player will win but occasionally he will have more consecutive losses than he can handle and suffer a major loss. Assuming your friend is betting on the player, the probability that any given bet will begin a streak of nine losses in a row is (2153464/(2153464+2212744))^{9} =~ .001727, or 1 in 579, assuming ties are ignored. There is more information available about the folly of the Martingale in my section on betting systems. However, the more ridiculous a belief is the more tenaciously it tends to be held. It usually takes a big loss to possibly convince a believer in any particular betting systems to stop.

Jack from Desoto, Texas

__Every__ betting system based on a negative-expectation game like craps is doomed to eventual failure. By tripling your bets, you will have bigger single wins, but you will reach your bankroll limit faster and have more losses. It all averages out to the house edge in the long run.

Michael from Fort Worth, Texas

My Java games are based on the random number generator that comes with Visual J++. For personal play, it should be quite fair. I speculate that any bias would only show up over millions of hands. Your results are not the result of a biased random number generator but of both luck and a progressive betting system.

Jim from St. Peters, USA

The probability of getting any number three times out of 4 is 38*4*(1/38)^{3}*(37/38) = 1/5932. However, if you play long enough you almost can't help but notice unusual events like this. This does not nearly rise to the level of being suspicious. Cheating does occur in real casinos. It is usually a rogue dealer who is caught by casino security. There have been some strong cases of cheating made against online casinos but no governmental authority has ever convicted anyone to the best of my knowledge.

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.

John from Crestwood, Illinois

In the long run it doesn’t matter what you do. As I have said numerous times measured by long term results all betting systems are equally worthless. Systems in which you chase loses with bigger bets increase the probability of a modest short term win but at the cost of even bigger losses when your luck is at it’s worst.

Jane from Dayton, USA

The expected loss is 5.26% of total money bet. This is true of ANY betting system based on American roulette rules.

Jay from New Haven, Connecticut

The name for this system is the Martingale. Ignoring ties the probability of a new loss for a hand of blackjack is 52.51%. So the probability of losing 8 in a row is .5251^{8} = 1 in 173.

Andrew from Maitland, Canada

The Martingale is dangerous on every game and in the long run will never win. However it is better to use in baccarat than roulette, just because of the lower house edge. The probability of the player winning 8 times in a row is 0.493163^8 = 1 in 286. Also keep in mind you could win a hand late in the series and still come out behind because of the commission. For example if you started with a bet of $1 and you won on the 7th hand you would win $60.80 ($64*95%), which would not cover the $63 in previous loses.

Eric from Ephrata, Pennsylvania

I maintain that even with an infinite bankroll, betting limits, and time the Martingale still would not beat a negative expectation game like roulette. My reason is that as these elements approach infinity the expected value of the Martingale on roulette is still -5.26%.

Still, mathematicians I respect have disagreed with me. The debate tends to get very abstract and absurd, hinging on the nature of infinity, which is a man-made construct to begin with. There is nothing known in our universe that is infinite. If forced, I think it is a ridiculous question.

"Anonymous" .

No. Some might argue that it would take an infinite number of losses to lose in this situation, which would be impossible. The truth is that 0.5^{infinity} approaches 0 but does not equal zero. If this did happen you would lose $2^{infinity}. The expected return of this strategy is thus 1- $2^{infinity} * 0.5^{infinity} = $1 - 1 = 0. Another more graceful way to look at is that as your bankroll increases the expected value still remains unchanged at zero. So the limit of the expected value as the bankroll approaches infinity is zero. In other words an increasing bankroll doesn’t help your odds, even if it goes to infinity.

**May 22, 2004 column**

**). If I had an infinite amount of money and time, and the casino would take any bet, then could I ensure a profit by playing the Martingale (doubling after every loss until I win) on a fair bet on the toss of a coin? The question writer proposes a random walk on a fair bet. The expected value is indeed zero, as you say. But the probability of ever being ahead is 1, as long as you are willing to quit after being ahead some finite amount. The probability of eventually achieving that finite amount with an infinite bankroll and infinite time is 1.... for ANY finite level of winnings. Even if the game is unfair, infinite bankrolls can ensure that eventually you can receive a positive result... and then quit. Pick a level of winning you want... $1 million. Bet a million. If you lose, bet $2 million. If you lose again bet $4 million. In an infinite number of flips, even with the game as unfair as you like, you will eventually win. Take your $1 million and go home. Come back tomorrow and repeat.**

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Jonathan Falk, actuary

I had a feeling one of my fellow actuaries might disagree with me on this one but I stand by my answer. I see this as a question of expected value rather than probability. The writer used the word "ensure", which is related to the word insurance. An insurance policy would have a fair cost of 1, which is simply the product of the probability (1/2^{infinity}) and amount covered (2^{infinity}). As I said in my original reply, 2^{infinity}/2^{infinity} = 1. So the player would give up his one unit win to pay for the insurance policy. You might argue that the insurance company would never have to pay because they could claim an infinite number of flips have not occurred yet, but I’m assuming a timeless quality in the question. If we did consider time I would be even more right because the player would never live long enough to play an infinite number of flips, and any finite number of losses is definitely possible.

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Dear Mr. Wizard, If you had $5,000 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.
If someone follows such a Martingale system in blackjack, what is the probability of being able to win $200 per day or lose the entire $5,000? Also, does increasing the amount available for total wagering increase the likelihood of winning the $200.
First off let me say unequivocally I understand and agree with your stance on betting systems. It’s quite simple: If you are at a disadvantage for an individual hand, the same holds for multiple hands, regardless of bet amount. End of story. I know the longer I play games in a casino, the higher my chances of leaving without money.
**

"Anonymous" .

I would bet $200 on the player bet in baccarat. If it wins, walk, if it loses then bet $400 (or whatever you lost). Then just go into a Martingale until you win your $200 or lose your entire $5,000.

"Anonymous" .

If you had a game with no house edge the probability of winning $200 with $5000 to risk, using any system, would be 5000/(5000+200) = 96.15%. The general formula for winning w with a bankroll of b is b/(b+w). So the larger the bankroll the better your chances. The house edge will lower the probability of success by an amount that is hard to quantify. For a low house edge game like blackjack, the reduction in the probability of success will be small. It would take a random simulation to know for sure. Forgive me if I don't bother with that. VegasClick did a small simulation about the probability of success with the Martingale.

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.