# What are the odds of the pattern “2 black, followed by 2 red” occurring in a roulette game?

I am working on a system, but I struggle to calculate the mathematical odds of a series of 2 black, followed by 2 red numbers occurring at a roulette table.

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A quick comment - if your 'system' is one designed to win at roulette, you should be aware that rolls of a roulette wheel are random (and biased in favour of the house) and that there is no 'system' that consistently wins at roulette. If, on the other hand, you mean something else by 'system' then feel free to ignore this comment! – Chris Taylor Feb 7 '13 at 14:16
i am aware that there is no consistent wining system at roulette but am interested how often will i lose if i keep a sertain pattern of bets. I am in the casino bussiness for 10 years now so i must be sure that there is no consistently winning system. – alexander Feb 7 '13 at 14:32
@alexander, roulette and dice are radically different from poker and blackjack. Roulette and dice are memoryless systems: each spin or throw is independent of the previous ones, whereas cards are dealt from decks. – alancalvitti Feb 7 '13 at 15:39
To clarify alan's remarks, there is no winning system in roulette, at all, full stop. Anything you do will, on average, lose you money, even in the short term. – Ben Millwood Feb 7 '13 at 15:53
you don't seem to understand me. I know that the ball has no memory but there is also a long term statistic right. So as i was saying i am interested in the odds of how often will i get beaten not if the ball will land here and there. It's like you are one of the guys that are making comments like the " Casino will stop you if woy follow martingale system at 500 max bet on even chance; no it wont you can also bet on the numbers but why dont i see such comments in any post it almost feels like all the gamblers are stupid lol. If you win 95 times 1000 units and lose 5 times 11000 is this + EV – alexander Feb 7 '13 at 15:57

The probability thet the next four events are, in this order, black, black, red, red, is $\left(\frac {18}{37}\right)^4$. The probability that the next event is red, given that the previous three have been black, black, red, is $\frac{18}{37}$.