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The strategy had the gambler double the bet after every loss, so that the first win would recover all previous losses plus win a profit equal to the original stake. We teach you the way to earn 100 EUR each day.

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Gamblers’ goals are as varied as gamblers themselves. Some gamblers just want to have a good time regardless of how much money they lose. Others want to grind out small wins over a long period of time. Some just want to get free buffets and cocktails.

There’s no right or wrong way to gamble, as long as you’re not hurting yourself or other people. But gambling addiction is beyond the scope of this post.

One popular goal that a lot of gamblers set for themselves is to double their money. This post explains some of the best ways to double your money in the casino.

You might not like these strategies, though, because many of them aren’t as fun as other ways of gambling.

The amount of money you put on a single bet compared to the size of your bankroll can be considered your boldness.

For example, if you have a $1000 bankroll, and you’ve found a casino that allows you to play blackjack for a dollar per hand, you’re not being very bold at all. You’re risking 0.1% of your bankroll on each hand.

Take that same $1000 bankroll and bet it all on red on a roulette wheel, and you’re making the boldest bet you possibly could. (That boldness isn’t related to betting on roulette, though—it’s maximum boldness because of the amount, not the game.)

If you want to know what your best chance of doubling your money when gambling is, it depends on whether you’re playing a game where you have the edge or where the casino has the edge.

Most of the time in most casino games, the casino has the mathematical edge. In that case, a maximum boldness strategy offers you the best chance of doubling your money.

Rare gamblers with advantage strategies, like card counters, have an edge over the casino. In this case, a minimum boldness strategy is the approach that will be most likely to provide you with a shot at doubling your money.

Here’s why:

With a negative expectation game, the more bets you make, the closer you get to the long run. And since the negative expectation is a long-term result, the closer you get to the long-term, the more likely you are to lose money. The short run is your friend when it comes to gambling on negative expectation games.

With a positive expectation game, though, the worse the short run is for you. In the short run, the results are unpredictable. This means you could go broke in the short run even if you’re playing with a mathematical edge over the casino.

An Example from the Game of Roulette

Unless you’re a biased wheel specialist—which is beyond the scope of this post—roulette is a game where the house has a clear edge against the player. You have 38 numbers on a standard roulette wheel. The even money bets all have 18 opportunities to win versus 20 opportunities to lose.

Here’s how that works:

The even-money bets on a roulette wheel are as follows:

  • Even or odd
  • High or low
  • Red or black

The numbers on the roulette wheel are 0, 00, and 1 through 36. If you bet on even, there are 18 numbers which can win for you: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, and 36.

Any odd number or any 0 or 00 is a loser for you with that bet.

The high numbers on the roulette wheel are 19 through 36, while the low numbers are 1 through 18. If you bet on high, you have 18 ways to win and 20 ways to lose.

2 of the numbers on the roulette wheel, the 0 and the 00, are green. Half of the rest of the numbers are red, and the other half are black. That’s 18 ways to win if you bet on red.

The probability of something happening is the number of ways it can happen divided by the number of total possible outcomes. So the probability of winning any of these even money bets is 18 divided by 38. (18 ways to win, and 38 total possibilities.)

18/38 = 47.37%

You can express that probability in multiple ways, but most people understand percentages in a reasonably intuitive way. Most people know that 47.37% is a close to 50% shot, which is roughly half the time.

A 47.37% chance of doubling your money isn’t bad, though. Now let’s see what happens to the probability of doubling your money if you divided your bankroll in half and made 2 bet. You’d have to win both bets, right?

The probability of winning 2 bets in a row is the probability of winning each of them multiplied by each other. That’s 47.37% X 47.37%, or 22.44%. That’s a little more than 1 in 5 but not as good as 1 in 4. It’s a far cry from 1 in 2.

Here’s why:

When you place 2 bets on an even-money bet at the roulette table, you have to win both of them. But you have roughly equal probabilities of seeing the following results:

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  1. You’ll win both bets. (22.44%)
  2. You’ll lose both bets. (27.70%)
  3. You’ll win the 1st bet but lose the 2nd bet. (24.93%)
  4. You’ll win the 2nd bet but lose the 1st bet. (24.93%)

Your probability of losing is always slightly greater in these situations, but there are 4 possible outcomes instead of just 2.

Most gamblers do want to do more than double their money on a single bet, though. They want to gamble a lot. And some of them even want to make a living doing it.


These people are called “advantage gamblers.” They’re card counters, pro poker players, and sharp sports bettors. The thing about all these advantage gamblers is that they all operate with a small edge over the house (or the other players).

Remember how the casino has a small mathematical edge over the player?

This is what gives the player a reasonable chance to double his money if he makes a single bet. You’re taking advantage of how unpredictable the results are in the short run.

If you have a small advantage over the casino, though, your goal becomes getting into the long run. By getting into the long run, you’re able to compound that small edge repeatedly. You just have to avoid going broke in the short run.

Let’s assume you’re hosting a roulette game in your home, but your start-up costs are limited. (Think of the Will Ferrell movie The House.) You have $10,000.

A high roller finds out about your game and comes to your house wanting to bet $10,000 on a spin of the roulette wheel. You’re more likely to win that bet than he is, sure. Your probability of winning, since you’re the house, is 52.63%.

But do you really want to take a 47.37% probability of going out of business because of a single bet?

The risk level is too high. This leads to an important principle—when you have an edge, a minimum boldness strategy is correct. When the house has an edge, a maximum boldness strategy is best.

This is why casinos have betting minimums at their casino games. At many casinos, the most you can wager on a single spin of the roulette wheel is $500.

Let’s look a real example of an advantage gambler in action, since most people don’t host roulette games in their homes. Let’s assume you’re a card counter.

Most card counters have an edge of maybe 1% over the house, give or take. But they still have a less than 50% probability of winning a specific hand, even when the count is positive. They get their edge from the greater probability of getting a natural and the corresponding 3 to 2 payout.

But if a card counter bet all or most of his bankroll on a single hand of blackjack, his risk of going broke would be enormous. He’d be out of business until he put together a new bankroll.

Most card counters will limit the size of their bets to 1% of their bankroll or less. As their bankroll grows, the size of their bets grows.

This is true for professional poker players and sharp sports bettors, too. Few poker players sit down with their entire bankroll in front of them. The pros never do that. With sports bettors, you’re likelier to see a bettors limit his wagers to 1% or 2% of his total bankroll.

The risk of ruin drops as the size of your bets compared to your bankroll decreases. It’s not impossible to go broke in those situations; it’s just less likely.

Game Would You Choose? And What Would You Do After Winning that Bet?

To maximize your probability of doubling your money, you need to choose the bet in the casino which offers the highest probability of winning even money. Any bet that pays off at more than even money is off limits to begin with, because it’s going to have a much lower probability of winning.

You won’t find a casino bet that has a greater than 50% probability of paying off at even money, either. That’s not how casino games work.

We’ve already established that the probability of winning an even-money roulette bet is 47.37%.

But you can find some casinos which have single-zero wheels. The probability of winning an even-money bet on one of these roulette tables is better, because you now have 18 ways to win out of 37 possible outcomes. That improves your chances to 48.65%.

Believe it or not, that’s probably your best bet, too. You’d be forgiven for thinking that blackjack would be better because it has a lower house edge, but the probability of winning a single bet on a blackjack hand is actually lower—it’s more like 42.20%.

How is that possible?

It’s because a certain percentage of those wins will be naturals, 2-card hands which total 21, and those pay off at higher odds: 3 to 2. This compensates for your lower win rate. Also, a percentage of your blackjack hands will also result in a “push,” which is a tie result where you neither win nor lose any money.

But believe it or not, there’s a game with a house edge between that of roulette and blackjack where your probability of winning is actually a little higher than 50%. It’s casino war.

If you’ve never played casino war, it’s basically a gambling version of the game you played as a kid. You each flip over a single card, and the person with the higher-ranked card wins.

In casino war, though, you must put up a 2nd bet if you go to war. If you win after going to war, you only get a one-unit payout. If you lose, you’ve lost 2 bets.

So even though you have a slightly better than even money chance of doubling your money, the game has a built-in mathematical advantage. You just lose that much more when you lose. That’s where the house gets its edge.

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You won’t be able to bet your entire bankroll on a hand of casino war, though, or your probability will differ greatly. That’s because if you go to war, you must put up another bet the same size as your initial bet.

The best way to double your money when gambling in a casino—if you’re playing a game where the house has an edge—is to make the maximum boldness move. Bet your entire bankroll on an even-money bet.

The best bets for this purpose are roulette and casino war, but roulette is actually superior. With casino war, you can’t place a bet of your entire bankroll, because if you do, you won’t have the money to bet when you have to go to war.

On the other hand, if you’re an advantage gambler, you should manage your bankroll carefully. You don’t want to go broke before your long-term edge starts to become realized.

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A martingale is any of a class of betting strategies that originated from and were popular in 18th-century France. The simplest of these strategies was designed for a game in which the gambler wins the stake if a coin comes up heads and loses if it comes up tails. The strategy had the gambler double the bet after every loss, so that the first win would recover all previous losses plus win a profit equal to the original stake.

Since a gambler will almost surely eventually flip heads, the martingale betting strategy is certain to make money for the gambler provided they have infinite wealth and there is no limit on money earned in a single bet. However, no gambler possess infinite wealth, and the exponential growth of the bets can bankrupt unlucky gamblers who chose to use the martingale, causing a catastrophic loss. Despite the fact that the gambler usually wins a small net reward, thus appearing to have a sound strategy, the gambler's expected value remains zero because the small probability that the gambler will suffer a catastrophic loss exactly balances with the expected gain. In a casino, the expected value is negative, due to the house's edge. Additionally, as the likelihood of a string of consecutive losses occurs more often than common intuition suggests, martingale strategies can bankrupt a gambler quickly.

The martingale strategy has also been applied to roulette, as the probability of hitting either red or black is close to 50%.

Intuitive analysis[edit]

The fundamental reason why all martingale-type betting systems fail is that no amount of information about the results of past bets can be used to predict the results of a future bet with accuracy better than chance. In mathematical terminology, this corresponds to the assumption that the win-loss outcomes of each bet are independent and identically distributed random variables, an assumption which is valid in many realistic situations. It follows from this assumption that the expected value of a series of bets is equal to the sum, over all bets that could potentially occur in the series, of the expected value of a potential bet times the probability that the player will make that bet. In most casino games, the expected value of any individual bet is negative, so the sum of many negative numbers will also always be negative.

The martingale strategy fails even with unbounded stopping time, as long as there is a limit on earnings or on the bets (which is also true in practice).[1] It is only with unbounded wealth, bets and time that it could be argued that the martingale becomes a winning strategy.

Mathematical analysis[edit]

The impossibility of winning over the long run, given a limit of the size of bets or a limit in the size of one's bankroll or line of credit, is proven by the optional stopping theorem.[1]

However, without these limits, the martingale betting strategy is certain to make money for the gambler because the chance of at least one coin flip coming up heads approaches one as the number of coin flips approaches infinity.

Mathematical analysis of a single round[edit]

Let one round be defined as a sequence of consecutive losses followed by either a win, or bankruptcy of the gambler. After a win, the gambler 'resets' and is considered to have started a new round. A continuous sequence of martingale bets can thus be partitioned into a sequence of independent rounds. Following is an analysis of the expected value of one round.

Let q be the probability of losing (e.g. for American double-zero roulette, it is 20/38 for a bet on black or red). Let B be the amount of the initial bet. Let n be the finite number of bets the gambler can afford to lose.

The probability that the gambler will lose all n bets is qn. When all bets lose, the total loss is

i=1nB2i1=B(2n1){displaystyle sum _{i=1}^{n}Bcdot 2^{i-1}=B(2^{n}-1)}

The probability the gambler does not lose all n bets is 1 − qn. In all other cases, the gambler wins the initial bet (B.) Thus, the expected profit per round is

(1qn)BqnB(2n1)=B(1(2q)n){displaystyle (1-q^{n})cdot B-q^{n}cdot B(2^{n}-1)=B(1-(2q)^{n})}

Whenever q > 1/2, the expression 1 − (2q)n < 0 for all n > 0. Thus, for all games where a gambler is more likely to lose than to win any given bet, that gambler is expected to lose money, on average, each round. Increasing the size of wager for each round per the martingale system only serves to increase the average loss.

Suppose a gambler has a 63 unit gambling bankroll. The gambler might bet 1 unit on the first spin. On each loss, the bet is doubled. Thus, taking k as the number of preceding consecutive losses, the player will always bet 2k units.

With a win on any given spin, the gambler will net 1 unit over the total amount wagered to that point. Once this win is achieved, the gambler restarts the system with a 1 unit bet.

With losses on all of the first six spins, the gambler loses a total of 63 units. This exhausts the bankroll and the martingale cannot be continued.

In this example, the probability of losing the entire bankroll and being unable to continue the martingale is equal to the probability of 6 consecutive losses: (10/19)6 = 2.1256%. The probability of winning is equal to 1 minus the probability of losing 6 times: 1 − (10/19)6 = 97.8744%.

The expected amount won is (1 × 0.978744) = 0.978744.
The expected amount lost is (63 × 0.021256)= 1.339118.
Thus, the total expected value for each application of the betting system is (0.978744 − 1.339118) = −0.360374 .

In a unique circumstance, this strategy can make sense. Suppose the gambler possesses exactly 63 units but desperately needs a total of 64. Assuming q > 1/2 (it is a real casino) and he may only place bets at even odds, his best strategy is bold play: at each spin, he should bet the smallest amount such that if he wins he reaches his target immediately, and if he doesn't have enough for this, he should simply bet everything. Eventually he either goes bust or reaches his target. This strategy gives him a probability of 97.8744% of achieving the goal of winning one unit vs. a 2.1256% chance of losing all 63 units, and that is the best probability possible in this circumstance.[2] However, bold play is not always the optimal strategy for having the biggest possible chance to increase an initial capital to some desired higher amount. If the gambler can bet arbitrarily small amounts at arbitrarily long odds (but still with the same expected loss of 10/19 of the stake at each bet), and can only place one bet at each spin, then there are strategies with above 98% chance of attaining his goal, and these use very timid play unless the gambler is close to losing all his capital, in which case he does switch to extremely bold play.[3]

Alternative mathematical analysis[edit]

The previous analysis calculates expected value, but we can ask another question: what is the chance that one can play a casino game using the martingale strategy, and avoid the losing streak long enough to double one's bankroll.

As before, this depends on the likelihood of losing 6 roulette spins in a row assuming we are betting red/black or even/odd. Many gamblers believe that the chances of losing 6 in a row are remote, and that with a patient adherence to the strategy they will slowly increase their bankroll.

In reality, the odds of a streak of 6 losses in a row are much higher than many people intuitively believe. Psychological studies have shown that since people know that the odds of losing 6 times in a row out of 6 plays are low, they incorrectly assume that in a longer string of plays the odds are also very low. In fact, while the chance of losing 6 times in a row in 6 plays is a relatively low 1.5%, the probably of losing 6 times in a row (i.e. encountering a streak of 6 losses) at some point during a string of 200 plays is approximately 95%. Even if the gambler can tolerate betting ~1,000 times their original bet, a streak of 10 losses in a row has a ~17% chance of occurring in a string of 200 plays. Such a loss streak would likely wipe out the bettor, as 10 consecutive losses using the martingale strategy means a loss of 1,023x the original bet.

These unintuitively risky probabilities raise the bankroll requirement for 'safe' long-term martingale betting to infeasibly high numbers. To have an under 10% chance of failing to survive a long loss streak during 5,000 plays, the bettor must have enough to double their bets for 16 losses (a ~7% chance of occurring during 5,000 plays). This means the bettor must have over 131,000 (2^16-1 for their 16 losses and 2^16 for their 17th streak-ending winning bet) times their original bet size. Thus, a player making $10 bets would want to have over $1.3 million dollars in their bankroll (and still have a ~7% chance of losing it all during 5,000 plays).

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When people are asked to invent data representing 200 coin tosses, they often do not add streaks of more than 5 because they believe that these streaks are very unlikely.[4] This intuitive belief is sometimes referred to as the representativeness heuristic.


In a classic martingale betting style, gamblers increase bets after each loss in hopes that an eventual win will recover all previous losses. The anti-martingale approach, also known as the reverse martingale, instead increases bets after wins, while reducing them after a loss. The perception is that the gambler will benefit from a winning streak or a 'hot hand', while reducing losses while 'cold' or otherwise having a losing streak. As the single bets are independent from each other (and from the gambler's expectations), the concept of winning 'streaks' is merely an example of gambler's fallacy, and the anti-martingale strategy fails to make any money. If on the other hand, real-life stock returns are serially correlated (for instance due to economic cycles and delayed reaction to news of larger market participants), 'streaks' of wins or losses do happen more often and are longer than those under a purely random process, the anti-martingale strategy could theoretically apply and can be used in trading systems (as trend-following or 'doubling up'). (But see also dollar cost averaging.)

See also[edit]


  1. ^ abMichael Mitzenmacher; Eli Upfal (2005), Probability and computing: randomized algorithms and probabilistic analysis, Cambridge University Press, p. 298, ISBN978-0-521-83540-4, archived from the original on October 13, 2015
  2. ^Lester E. Dubins; Leonard J. Savage (1965), How to gamble if you must: inequalities for stochastic processes, McGraw Hill
  3. ^Larry Shepp (2006), Bold play and the optimal policy for Vardi's casino, pp 150–156 in: Random Walk, Sequential Analysis and Related Topics, World Scientific
  4. ^Martin, Frank A. (February 2009). 'What were the Odds of Having Such a Terrible Streak at the Casino?'(PDF). Retrieved 31 March 2012.
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