## The Mirage of the Martingale

Posted by silentarchimedes on May 17, 2008

Like all those people looking to find the edge in Roulette at the casino, I fell into what I call “*the Mirage of the Martingale*“. The Martingale’s premise is in putting money only on 50/50 bets, like the flip of a coin. In Roulette, the closest to this is the RED/BLACK bet, ODD/EVEN bet or 1-18/19-36 bet. It is not a full 50/50 bet because the two green slots (0 and 00) reduces each bet to a 47.3684% winning probability instead of the 50% winning probability in 50/50 bets. (Remember, American Roulette has 38 slots, 18 for RED, 18 for BLACK, and 2 for GREEN. This creates a 2.6316% for landing on any specific slot. The house edge is the two GREEN slots, or 2*2.6316 = 5.2632%)

**The basic strategy of the Martingale is when you lose a 50/50 bet, you double your bet on the next spin. That way, if you win, you cover your last round’s loss and also gain your initial bet’s amount.**

For example, let’s say you bet $5 on BLACK in round 1. The spin turns up RED. You are down $5. In round 2, you now bet $10 on BLACK. If the spin turns up black, you cover the $5 you lost in the first round, and you also make $5. If you lose, you are now down $15. In round 3, you now bet twice the last round’s bet, or $20. And so on and so forth… Assuming you lose 5 BLACK bets in a row until winning on the sixth, your betting sequence is 5, 10, 20, 40, 80, 160, … and your profit sequence is 0, -5, -15, -35, -75, -155, 5.

In short analysis, this sounds like a fool proof idea because you assume that if you keep betting on black, at some point it will land on black, and you will cover your losses.

Upon further analysis, there are some obvious problems with the Martingale. The problems are both in the mathematics, and in the practicality of using such a strategy in a casino.

** 1. First, the mathematics. **At first glance, the probability of a losing streak of 5 seems very small. For every spin, there is a 1/2 probability of winning or losing. Five in a row has a 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 = 3.125% likelihood. If someone asked you if you would bet $5, and the chance of you losing is 3% and the chance of you winning is 97%, you would take that bet in an instant. However, this 3% is nice when you are talking short term (although you might be unlucky). For example, if you stay at the table for 5 rolls, your chance of hitting that losing streak is a small 3%. However, the longer you stay at the table, the more you are exposing yourself to the Law of Large Numbers and the Central Limit Theorem (see images below, courtesy of Wikipedia).

As the number of trials increases, the output plot starts to resemble a bell curve. Even the unlikely outcomes eventually occur, based on their respective probabilities. In gambling, unwanted outcomes will eventually show up if enough bets are placed. |
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.

Eventually, that 3% will show its ugly face. This is also true for losing streaks greater than 5, such as 6=1.56%, 7=0.78%, 8=0.39%, etc. On a positive note, as the length of a losing streak increases, the probability decreases and you can stay longer at the table without fear of the two above theorems/laws. For example, the chance of a losing streak of 8 is only 0.39% for any 8 straight rolls. **However, a losing streak of 8 means you would have to wager $640 on the 8th round, and this is all an attempt to get your original $5 back!! **A person with a limitless purse would not care, because they know eventually they will break this losing streak and cover however big their loss is. **Unfortunately, the casinos know this!** Which leads to our second problem…

**2. Every casino has a minimum bet and maximum bet on the roulette table.** A typical “cheap” table usually has a minimum of $5, and a maximum of $100 or $200, or for short (5,100) and (5,200) table. When I started going to casinos at 18, I always wondered why the maximum bets are so low. Now I know. The maximum limit of a (5,200) table would limit a person using the Martingale strategy to a maximum losing streak of 6. That means if you bet $160 on that sixth spin and lose, you lose $315. You cannot bet $320 to potentially cover your losses because that is over the max limit of the table. **You must carry that $315 loss with you!! All for an original $5 bet!** That risk proposition doesn’t seem so good now.

These two problems might not fully convince some people to not use the Martingale strategy. The first thought is *I will make enough money before I hit a losing streak of 6*. Now we talk about how to bet on winning streaks and why they are not practical to overcome the odds of hitting a losing streak. The most common method is the Anti-Martingale Strategy:

**The Anti-Martingale Strategy **– The betting strategy is the same as the Martingale except you double your bet when you win. However, you must have winning streak level where you bank your profit, and then begin betting at $5 again. Let’s say you make your level at 4. So when you win 4 in a row, you bank 5, 10, 20, 40 = $75. Not bad huh? The problem is a winning streak of 4 only occurs 1/2 * 1/2 * 1/2 * 1/2 = 1/16 = 6.25% Remember, a (5,200) table only allows a max losing streak of 6. If it happens, you lose $315. As mentioned, these losing streaks occur 1.56% of the time. That means win streaks of 4 occur exactly 4 times more than losing streaks of 6. Both of these are the only two instances in which non-even money is exchanged. All other streaks are break even for you. So, for every time you lose $315, you only make 4 * $75 = $300! You are still short $15. And all these computations above don’t include the house edges on the greens. Which would mean a slightly larger loss on your end! No matter how you change the length of the winning streak when you bank your money, you are on the short end. For example, say you bank every time you win two in a row = $15. This occurs 25% of the time. This occurs 16 times more than your losing streak of 6. That means for every time you lose $315, you make only (16 * 15) = $240!

Our **LS6-WS4 Martingale/Anti-Martingale Strategy on a (5,200) American Roulette Table** looks like this:

I tried modifying the Martingale and Anti-Martingale and even ran simulations in Matlab. The numbers never ever add up. **In the end, it still comes down to luck.**

Anybody disagree or have ideas about at least slowing down your losing in a mathematically dependable way? I apologize if the math is somewhat off, as I did this on the spot, but I think no matter what, it’s impossible to use the Martingale effectively in a casino.

****Update 05/21/08 **** – Yesterday’s NBA lottery draft is a perfect example that anything can happen, even with the probability against you. Probability is simply that, the likelihood of something happening and not happening. If something has a nonzero probability (even if it’s 1.7%), it can still happen at anytime. That’s what happened last night. The Chicago Bulls, had a 1.7% chance of landing the #1 pick in the draft. Eight teams were ahead of them (in terms of probability). Miami had the highest probability to land the top pick, with 25%. So guess what happens? That 1.7% happens, and the Bulls get the top pick!! (As a Knicks fan, I was rooting for the Knicks 7.6% chance, but instead of picking even 5th, they ended up 6th)

**Related link**

## rom3o said

Excellent Article.

it has at least given me a idea as appose to blind betting.

thanks.

## silentarchimedes said

Thanks for the comment Rom3o. It’s funny. As much as I know it’s not advised to use the Martingale, I’m still tempted to try it out the next time I’m in Atlantic City. I guess the gambler inside me still thinks I can catch that lucky run.

🙂

## Anonymous said

I personally will not used this martingale method due to reward/risk ratio and also the fact that this is a maximum bet size. What you are calculating here is the chances of a streak breaking. It does not mean I will get an increase in the odds of the independent event. Getting a long streak of black is highly unlikely, just like getting a long streak of heads on a coin flip. But it can happened and can happen longer and more often than you think because the odds of every single spin is still the same since they are all independent events (ie, in outcome does not depends on previous events). It’s like a coin flip. The chances of every coin flip is 50/50. Even on the 100th roll of 100 heads, the chance of a head on the 101th roll is not 0.5^100 = 7.8886e-31 ( 7.8e-29% or close to zero percent), but it will still be 50%, since a flip is an independent event. It purely will not increase my chance of getting a tail because it will not remember what it flipped last time. The odds of the 101 flip is the same as the odds of the 1st flip. So if on the 10th roll, I still have a head, I will need $5120 bet to make my $5; a reward to risk ratio of about 1 to 1000..not a good RR ratio. Just my 2 cents.

## silentarchimedes said

Hi Anonymous,

That is actually a very good point. Something I should have touched more on and what I mean by the Law of Large Numbers and Central Limit Theorem. You are absolutely right, it really makes no difference which color you bet on on each turn because each turn has no memory of the past turns, it’s independent. However, it is because of this independence that makes streaks interesting and thus the bell-curve shape. The thing is, when it comes to the Martingale, your past is a factor in how you bet on the current turn. This is where the flaw of the strategy occurs. You are trying to make the current turn contingent on past turns, and that is very risky because of the independence of each turn. (If that makes sense)

Thanks for bringing it up! Can you think of other strategies to use on the Roulette?

## Anonymous said

If I’m serious about the money, I will not play roulette, but only Blackjack since playing the correct strategy with the standard favorable rules give house edge of only 0.5%. But for entertainment and testing my ‘luck’, I will only play European roulette since the house edge is only about 2.7% compare to like you said 5.26% in American roulette due to the extra 2 zero slot. I will then have 2 strategies –

1) bet 1 black and 1 zero = -2*48.65%, 0*48.65%, 34*2.7% (35 times payout on zero, but -1 from the black) = house edge is 5.5% ( i do this for the heck of it if I ‘feel lucky’ )

2) bet 1 red and 1 even or 1 black and 1 odd, then use the martingale method for each independent bet. If red loses and even wins, I double up on the red and the even bet remains 1; vice versa. This way you have 23.67% of both winning the 2 bets, and 50.04% of break even and 26.37% of losing both = house edge is 2.7% which is even better ( I use this a lot, in fact I should’ve tried it out in las vegas this past weekend >_- )

## silentarchimedes said

I used to solely play Blackjack when I went to AC, but it just becomes to robotic. There was no suspense or luck. I would play for several hours only to break even or make a couple of bucks. Which is fine, but there is something about wanting to go to AC and make some big money. 🙂

I like your second betting strategy on roulette though, although it’s quite expensive. Now you got me tempted to take a weekend trip to AC and try it out. Wait, do they have European roulette in Vegas? no, right?

## Anonymous said

Yes, they have European Roulette in Vegas, but it is ones on the computer by the slot machines. The second betting strategy is ok because you will lose only 26% of the time and will stay at the table longer especially if you do not use martingale method.

## Mike said

Im new to roulette and am not a mathmatician by any stretch. But with standard payout ratios, a finite 37 or 38 possibilities on the board for winning coverage, and multiple betting choices– has no one used a computer to calculate the prime wager at a variety selected ratios and board positions for profit or at least minimizing loss opportunities. I think there is something like this used in horse racing.

## silentarchimedes said

Hi Mike. You know, this is one of the most common questions that people ask regarding multi-place betting, such as Roulette. The difference between a game like Roulette and horse racing is that the odds in Roulette are always the same, where in horse racing, the odds continue to change up until the gun blast. But you are absolutely correct, there have been mathematical studies done on such games. The problem is the casinos know about this, and they hire mathematicians that do exactly that, to make sure there are no loopholes or soft predictable sequences that guarantee steady profit. The casino’s main goal is to guarantee profit over a large number of bets. They are willing to allow short winning streaks to lucky individuals, but they sure won’t tolerate dependable winning strategies. Let’s just say, if there exist such strategies, the casinos either already know about them or word spreads fast and they see stop it right in its tracks.

## No Name said

That was a very comprehensive article on this strategy.

However, there are certain factors you are not considering when looking at this betting strategy. You have only factored in when the strategy is performed by a single person. The maximum bet limit can be removed or more realistically delayed when a group of independent gamblers work together. Not strictly 100% legal I would suggest but done effectively using the ‘outside’ bets and the straight up red numbers on the board, with in the group. This can be achieved by using the roulette computers which feature many players/computers around the same real wheel located with in casinos.

Wouldn’t also using the other outside bets, or 50/50 bets as you call them, after a certain period of time lessen the effect, probability of, Law of Large Numbers and the Central Limit Theorem. Such as switching to odds or evens, or high or low numbers. Or possibly reset the counter as it were.

I am not a mathimatician and only have a basic understanding of this so any feedback/reply would be appreciated.

## Anonymous said

hi this is all bullshit i have seen 14 time black in one row

## Paco said

There are several problems with this paper. First of all the house edge only makes the Martingale strategy worse. If there were no house edge at all (like a coin toss) the laws of statistics say that the longer you play the more and more likely ever longer streaks will occur. After 45 coin tosses you have a 50/50 chance of getting at least 6 in a row. After 90 coin tosses you have a 50/50 chance of getting at least 7 in a row. After 11,500 coin tosses there is a 50/50 chance of getting 14 in a row. If the casino kept records there would be some incredibly long streaks.

= The house limits are not specifically to protect against Martingale, but simply to protect the house from super rich gamblers. A small casino like “Golden Gate” (the oldest hotel in Las Vegas) makes about $10-$12 thousand a day on average from their pit. A lucky gambler playing $10K per hand might walk out with the quarterly profits in an hour. Remember, they are gambling as well, they just don’t want to go in over their heads.

– The Golden Nugget lifted their house limits about 5 years ago when two young men bought the casino. A rich gambler walked away with $8.5 million in a two week period. They decided to restore the limits. In the short run house edge doesn’t matter, not for your, not for the house.

## \bob\finch said

Yep, Martingale is ultimatley nonsense, I’m afraid. A thousand coin tosses is as likely to come out 1000x heads as any other combination. It’s important to remember that if you’ve found a “loophole” with a bookmaker etc. they will have known about it since the dawn of gambling.

## bwc said

On another note, if you toss a coin and it comes out 1000 times heads out of 1000 tries (which is possible esp from Nassim Taleb’s point of view), you will start wondering if the coin is bias. This is where statistical significance analysis come into play. Not completely sure how to calculated stat significance, but if this happens, a non mathematic oriented kid will even suspect that the coin is bias.

There is an article regarding whether the presence of Mike Jordan in the Chicago Bulls games are of any statistical significance or is it just that whole team is already good by itself so there is no need for him during those famous Bull eras. The result was that Jordan was statistical significant. If I recall, he skew the stats in the number of wins. But of course, even a kid can see that without analyzing mathematically like this.. so 1000 times head in 1000 tries will start prompting suspicion since it’s an extreme case.

## et said

Maybe the way I think seems stupid, please give any advice. I am just a new player. The way I use is I will pay attention of 4 tables at the same time. I will bet the opposite, if the outcomes have been so many times like 7 times at least of one side like black, odd or big numbers. If I lose, then I will double 3 more times maximum. More you bet, more chane you lose. So, I walk away, after I win. Wait for another chance like that in one of the 4 tables.

## et said

… I mean more I play at the same table, more chance to lose back later if I am winning now.

## silentarchimedes said

Hi Et,

Thanks for the post. There is one flaw in your strategy that many people won’t admit or believe. In a binary game (red/black), it is true that having one color happen 7 times in a row is of low probability, 1 / [(1/2)^6] (I use 6 here assuming you don’t care about the first spin and base the 6 other spins on the first spin’s color). However, the flaw or important thing to note is that each spin is still independent of the previous spins. So for example, even if a table currently has 7 reds in a row, on the next spin, red and black both have the same probability of happening! The current spin has no dependence on history. The only thing we can guarantee is that yes, eventually a black will show up, but we don’t know when. (That’s why casinos will limit the max you can bet so you can’t cover your losses).

Your selection of which of the four tables to play has no inherent advantage to your earnings outcome. Also, the fact that you will double 3 more times maximum does not guarantee a profit. If you win and you walk away, it is not because of the 7 reds that’s why you won. It is because you got lucky a black showed up. In the end, your strategy is a combination of luck and martingale and does not guarantee long-term profit.

Nice try though. But your kind of thinking is good, just make sure to work out the math. So far all strategies on the Roulette (or binary games) have been proven to not guarantee a profit.

## Skeptic said

After watched the roulette table for the outcome for awhile, I decided to jump in. I saw no 7 or 8 black or red in a row for a long time. On Carnival Triumph, the minimum is $5 and the maximum is $500, which allows you to lose 6 in a row without losing your pants. So, I threw in $640 and got chips.

When the last hand was red, I would bet $5 on black, and when it loses I double my bet. I kept rolling, and rolling, and I was winning, and winning. Before too long I won more than $300.

Like in a baseball game, if you keep hitting the balls, they change the pitcher, and so did Carnival. They changed one after another, and I was still rolling. Then came a stone face croupier who gave me 7 blacks in a row, and I was ready to go bankrupt. I decided to sit out the next round, and he gave one more black. I would have lost everything if I stayed in that. Licking my wound, I decided to look around and changed table. There I saw the result was never so drastic and I jumped in again and after recovered my losses I went to bed.

It turns out there is a different approach, which is not to bet against the trend, but to bet with the trend. That is, if the last one was red, you bet red, and if the last one was black, you bet black. I am going to give that a try the next time I get a chance.

The best place to play roulette is in Europe where some have $1 minimum and $5000 maximum. That allows you to recover even after losing 12 hands in a row.

## nicky said

skeptic ..

i totally agree what you said good luck to you

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## Andrew Gill said

I typically use a similar strategy as the martingale, but it’s modified to play 2 of 3 dozens or columns. I also implement selective betting, when possible, and I’ll have multiple bet runs at a time. These bets are treated as a bet-paid, and a win on either is considered a win on both. Even though winning one requires losing the other, it pays 2:1, so, if I have 1 unit on dozen x and 1 unit on dozen y, you should view it as 2 units on x or y. If the spin is in x, then I lose y – minus 1 unit – I win on x – plus 2 units – netting a win of 1 unit. There is no need to chase the loss on y as it’s payoff is built into the x win.

My ideal bets are placing 1.5 * table min on 2nd and 3rd dozen/column after 2 consecutive spins in the 1st dozen/column. Or, betting table min on one of the 50/50 bets mentioned after 3 consecutive spins in the same segment.

If the conditions are there I may have 5, 6, maybe 7 different bet runs in progress at any given time, and, of course, I chase all losses. One thing worth noting, on dozens/columns, you have to double losing bets plus a chip to fully recoup after a long loss streak.

## Andrew Gill said

I meant bet-pairs above, sorry…

## Andrew Gill said

Skeptic, do you realize that betting the trend long term is no good. Given sufficient sample size, the odds equal the statistics, or in other words, blacks and reds are a wash and you slowly lose at the 0/00 odds rate.

## Gale said

Great Post I am a huge roulette player from Austria

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