Anyone knows that odds for the same event to occur several consecutive times in a series of independent plays of a game are very low. This applies for any game of chance, including roulette.

For example, hitting the same color for *n* times in a row on a single-zero roulette has the probability _{} For *n* = 3, this is 11.51%, for *n* = 5 is 2.72%, while for *n* = 10 it goes down to 0.07%.

Basing on these low probabilities of repetition and some mathematical certainties, players built systems and strategies, among which the martingale is the most practiced.

Assuming *n* consecutive bets are lost and the *n* + 1 – th is won in a martingale, starting with the amount *S* as the stake of the first bet, algebra can easily show that the amount received is bigger than the previous cumulated loss, and the profit is *S*, no matter the value of *n*.

For a player who runs such system once, the following facts are unquestionable:

– For a sufficient number of consecutive plays, the player always has a positive profit (*S*);

– For hitting the favorable color and have that profit, the player must sustain the previous cumulated loss (which is _{}

This cash sustaining is the first issue of the martingale and is a problem of personal money management, because the amounts are not low at all.

As example, for *n* = 9 (sustaining 9 consecutive failures) and *S* = $1 as the first stake, the total amount to lose before the winning bet is $511. For a $2 initial stake, the amount to lose becomes $1022. These are investments for a profit of only $1 or $2. Pretty low profit rate, isn’t it?

Still, this low profit rate is compensated by the low risk of failure. Only 0.07% is the probability of 10 consecutive color failures, which will ruin a player having only $511 as capital.

Relying on this low risk, players usually try to extend the use of the martingale over the long run, with the goal of cumulating small positive profits to make an acceptable overall gain.

The error they make stands in their false intuition about having the same risk over the long run as they had in the isolated use of the martingale. Although the color outcomes are independent, when we talk about sequences of consecutive outcomes in a pre-established number of spins, these sequences are not independent any more, so we cannot extend the probability result from an isolated sequence.

Actually, the probability of failure increases significantly over the long run.

Staying with the same example of 9 consecutive failures to sustain, let us evaluate some probabilities of having 10 consecutive failures (the same colour for 10 times in a row) over a series of 1000, respectively 10000 spins.

The exact calculation for the probability of having the same color 10 times in a row at least once in 1000 spins is very laborious. We provide here an easier estimation, based on some particular sequences of outcomes. If we split the 1000 spins in 100 consecutive sequences of 10 outcomes (spins 1 – 10, 11 – 20, …, 91 – 100), we have that these sequences are independent and we have now a Bernoullian probability distribution, which easily allows us to calculate:

– The probability of having exactly one 10-outcome sequence of the same color (choose red, for example) among all 100 sequences is _{ }

– The probability of having exactly two 10-outcome sequences of reds among all 100 sequences is _{ };

The next probabilities will become lower, so we can state that the overall probability of having at least one sequence of 10 consecutive reds is higher than 7% (100 times bigger than the initial probability of the isolated case).

In fact, the exact probability of having at least one sequence of 10 reds over 1000 spins (among all 990 sequences, which are not independent) is much higher than 7%.

Making the same Bernoullian calculation for a run of 10000 spins, we get:

– The probability of having exactly one 10-outcome sequence of reds among all 1000 is _{ } ;

– The probability of having exactly two 10-outcome sequences of reds among all 1000 sequences is _{};

– The probability of having exactly three 10-outcome sequences of reds among all 1000 sequences is _{ } and the next probabilities become lower.

Adding them together, we find that the probability of having at least one sequence of 10 consecutive reds is higher than 51% (over 700 times bigger than the initial probability of 0.07%). In fact, the exact probability of having at least one sequence of 10 reds over 10000 spins (among all 9990) is much higher than that.

The conclusion is that the real risk of failure must count in any long-run martingale strategy, since it increases significantly from the isolated case. In addition, the low profit rate of the martingale makes any failure to be a potential ruin. Of course, the player might cumulate enough previous successes to sustain a martingale loss and go ahead, finishing the round with profit. In fact, these conclusions submit to the general consequences of the application of probability theory in gambling: any circumstantial winning is possible in any amount, but toward infinity the player will lose cumulatively in favor of the house. In other words, if we would be immortal regular gamblers, we shall be ruined for sure. Because we are not, there are still chances to walk away not broke. This could be one of the advantages of not being immortal.

Catalin Barboianu

Mathematics Department of Infarom Publishing

Author of Roulette Odds and Profits: The Mathematics of Complex Bets

I was playing live roulette @ Coral casino,i was playing on 2 diffrent tables and 10 reds or more came out in a row on both wheels at same time. What are the probalities for that?

The probability if that table being rigged is way much bigger especial if you add the human in the mathematical equation.

You will have to put into account that in real world the human shroud roll the same way for more then 1000 times at a time without pause that is not the case…

I don’t know but its the same as any other sequence.

If you’re playing live, then no way can it be rigged as all you’d have to do is double up on red and you’d be quids in:)

Rigged tables are equipped with magnetic coils under each slot.

The ball is also magnetic with the reverse polarity.

So,if they don’t want the ball to land in a certain slot the turn on the coil and the ball physically can’t land on that number. They can, for instance, do this for all of a color as well.

I played pokerstars.eu roulette and started with balance $50. Used martingale method by adding $5 everytime it hit wrong. I reaced balance $175 before 8blacks came and lost all. I deposited another $50 and 8 blacks came up before any red. a total of 16 blacks came up in a row. now i didnt continue after that as im sure pokerstars have rigged their roulette for defending against maringale.

My believe is that the software recognizes martingale method with an algorithm and changes the outcome.

This is total illegal to do actual.If all roulette schemes are that way.. then is pointless to play the games as is only a way to loose..

Online casino:

Your bets —>control units (Casino side) —>RNG units —> Results

Real world Casinos

Your bets —>RNG units (roulette wheel, shuffled cards etc) —> Result —> control units (cashing, dealer change, etc)

All online casinos are far more dangerous than the real world casinos based on my 22 years professional gambling experience (12 times black in a row only once, same number 5 times in a row only once, 14 times bank in a row on Punto Banco once), I can make gambling for a living in the real world casinos, but I couldn’t make for a living in the online rigged casinos

Never play any online casinos, if you want to gamble, just go to the real world casino. you will stand a much better chance

agreed

I was playing in Vegas with 30k of retirement money, I lost it all using the martingale system. 23 reds in a row showed up. I was devastated for months, it does happen and it can happen.

Hows it going I will get straight to the system am playing 2 dozens the chances of 4 spins/losses is 1,5 percent thats the same dozen coming up 4 times now if I play for 5 sequences does that make my chances of 1 loss 7,5 percent please anyone can help tell me if this is correct and dont tell me if this system can’t work in the long run

No it does not, your chance for each sequence is the same as the first , to ensure this you must still play two dozen but change them up so you get the same odds , don’t play the same two over and over ok

Ok I left a previous comment I am playing a similar method but with one dozen not even chance here I am playing one dozen waiting for 2 dozens to come up and playing that dozen that didn’t come up for example 1st and 2nd dozen come up I play 3rd dozen I play a progression of 6 bets my chances of losing all 6 and that 2 previous numbers makeing it 8 times out of that dozen is 4,3 percent chance now if I play 2 sequences does that make my chances of losing 8,6 percent thats 2 sets of 8 spins?

Can someone inlight me please, with 47.5% of winning what is the possibility to lose 22 and 23 times in a row? Note that the number of spins will be lets say 500,000.

Thank you in advance

just ‘ played ‘ Ladbrokes on line roulette,for the first time ! A change of dealer and he spins twelve black numbers in a row ! Previous to this it was 2red,3black,green,one black,3red etc .Staying with football betting ….man city to beat Napoli ten to nil tomorrow (probable ?)

I watched Ladbrokes roulette for a long time and my system worked. Then as soon as I bet, bam 8 blacks in a row. It’s fucking rigged

It might not be the gambling site but the universe that’s playing us

I am a craps player (20+Years) am in Thailand going to Cambodia where they have casino’s. No crap tables. Only roulette,baccarat,pai gow,etc. On other sites the word Martingale brings out anger and “you idiots” etc. When I have waited for 4 winning pass line bets and start on don’t .Ex: 25,75,175,375,775,1600. I went 21 games once!Before 11 pass line wins would of course break it.And often stood there and watched 17 pass line wins. 24 wins at Binion’s. But when I put in a 20 minute max time at table I end up on positive. So I just say when I buy in for the $3,000 needed. I will pay $3,000 to see 11 wins and then enjoy the fun of craps. I am going to try this on Roulette flat bets. 4 dozens (2-1 odds) before I begin my $25. and then pay $3,000 to see 11 no hits on that dozen. (I have read of all the 15 no hits etc).With a walk away goal of $300.00 (about 15 minutes craps). I’ve already read all the anger at this play. But, if the $3,000 loss is what I’m ready to gamble to see this I’ll enjoy Cambodia as the casino’s are on the beach..And after 7 days back to Thailand and no casino’s and hope “the dragon” doesn’t appear. Cheers,David

Just played maringal for 6+ hours $3 starting bet. Started with $200 and grinded up to $855. Then, 13 in a row hit red. This system can fail and will eventually. I have heard of 25 same color in a row. The only way this works is if you have an unlimited BR and there is no max bet.

Make a pattern of reds and blacks, or odds and evens. Then martengale the pattern. Bet on the last color that came up, when it changes, change with it and double it up. If it fails start your pattern martengale right there. When it wins follow the color it won on for a while.

What about playing a varied version of this. Playing the same bet 3 times in a row before increasing your bet? And then not worrying about doubling to make up your losses in 12 spin, but just an increase of 50% which you stick with for 3 spins. If still nothing after 6 spins you double up that original spin. By this method it would be much easier to handle double digit loses in a row. And obviously you’re gradually increasing your wager so the idea is you will come out slightly on the positive over the long haul.

I also thought you could try doing this doubly during a game. Say a bet on a colour and a wager on high or low or odd or even. And you stick to the lose 3 in a row and increase your wager by 50% of your original first bet.

Any mathematicians out there know how this would play out over the course of 500 or 1000 spins? I would find it hard to see someone being in the hole. Mind you perhaps the gains are so low it wouldn’t be worth it. But I do think a variation of this formula is the best method and as close as your can for winning in roulette.

If the procentages of winning at roulette is 48,64864865% while betting red and black, what is the chance of winning at one of them if you martingale 7 times? Will it be 1-(51,35135135)^7?

Thank you very much for answers.