# Roulette outcome probability

I have a question about the probability of a certain Roulette outcome.

In the American Roulette wheel with double zero pockets, what are the chances of rolling black and red alternately for a total of nine times consecutively (i.e. B,R,B,R,B,R,B,R,B)? And also how many rolls will it take so that the probability of (B,R,B,R,B,R,B,R,B) in that order will happen?

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On any trial, the probability of black is $\frac{18}{38}$, as is the probability of red. This is because there are $38$ slots, of which $18$ are black and $18$ are red.

The results of individual trials are independent, so you multiply the individual probabilities.

There is some ambiguity in the problem. Is RBRBRBRBR rolling black and red alternately?

If no, then our answer is $\left(\frac{18}{38}\right)^{9}$.

If yes, then our answer is $2\left(\frac{18}{38}\right)^{9}$.

Added: In a comment, it is mentioned that red first is fine too. Then the answer under yes is the right one. Numerically, it turns out to be approximately $0.0024$, so approximately $0.24\%$.

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RBRBRBRBR or BRBRBRBRB –  Theo Jun 29 '13 at 21:41
You multiplied $\frac{18}{38}$ by $9$. That's very wrong, it gives you a number bigger than $1$! We have to take the $9$th power of $\frac{18}{38}$. And maybe double, depending on how one interprets the problem. –  André Nicolas Jun 29 '13 at 21:42
so the first formula you give is non-alternate i.e. BBBBBBBBB or RRRRRRRRR? –  Theo Jun 29 '13 at 21:43
No, the first formula is alternate, but only B first allowed, so only RBRBRBRBR. It turns out that any string of length $9$ has the same probability. Thus the probability of BBBBBBBBB is also $\left(\frac{18}{38}\right)^{9}$. So is the probability of RRRRBBBBB. –  André Nicolas Jun 29 '13 at 21:47
why is the probability higher for formula #2 if the results of individual trials are independent? –  Theo Jun 29 '13 at 21:48

We have a total of $38$ slots in which any spin of the wheel may land. $18$ of these are red, and $18$ of the slots are black. (Two have no color). For each roll, the probability of landing on red is equal to $\dfrac{18}{38} = \dfrac 9{19}$. For the same reason, on any spin, the probability of landing on black is $\dfrac{18}{38} = \dfrac 9{19}$.

The result of each spin of the wheel is independent of all others, so we multiply the probabilities of each of the 9 results, so the overall probability of getting any one particular prescribed outcome, say RBRBRBR, is $$\underbrace{\dfrac{9}{19}\cdot \dfrac{9}{19} \cdot \cdots \cdot \dfrac{9}{19}}_{\large 9 \;\text{factors}} = \left(\dfrac{9}{19}\right)^9$$

If we want to find the probability of obtaining a string of spins in which it lands alternately on red-then-black-then..., or black-then-red-then..., we multiply by $2$, because we can have two possible overall outcomes where the result of each trial alternates: $RBRBRBRBR$ or $BRBRBRBRB$: $$2\times \left(\dfrac{9}{19}\right)^9$$

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