# Measuring Roulette sequence probability

Statistics has always been my weakest point in math, so please bear with me. I'm not asking how to win, just analyzing the game.

Assuming an American Roulette with two zero slots, the greens, or labeled G here:

P(R) = P(B) = 18/38 = 0.4737
P(G) = 2/38 = 0.0526


I'm trying to calculate the probability of a certain sequence to happen, for example: RRRBRG.

Say, n is the length of the sequence (the number of spins), how do I calculate the probability of any given sequence. RRRRRR, or probably GGGGGG? Red and Black have higher probability than green, obviously the P(RRRRRR) is > P(GGGGGG), but how do I calculate this?

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It's a potty you didn't ask how to win because that is easy - own the wheel :-) – Dale M May 10 '13 at 22:28

Since successive spins of a roulette wheel are independent, you can just multiply probabilities.

For example, the probability of getting a red followed by a black is

$$P(RB) = P(R)P(B) = \frac{18}{38} \frac{18}{38} = 0.224$$

and the probability of getting two zeros in a row is

$$P(GG) = P(G)P(G) = \frac{2}{38} \frac{2}{38} = 0.0028$$

In general, to find the probability of a sequence of independent events, you can use

$$P(A_1\cdots A_n) = P(A_1) \cdots P(A_n)$$

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Ah ok. Thanks! I did come up with that but just wondering why it's so small. – atedja Feb 9 '12 at 0:33