2

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?

Theo
  • 77

2 Answers2

4

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 $BRBRBRBRB$: $$ 2\times \left(\dfrac{9}{19}\right)^9$$

amWhy
  • 209,954
2

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 $0.24\%$.

André Nicolas
  • 507,029
  • 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
  • Let's say you will be happy if you get alternation, black first or red first, it doesn't matter. Then there are two different patterns that make you happy. Each has probability $\left(\frac{18}{38}\right)^{9}$, so you add these together, or equivalently double. – André Nicolas Jun 29 '13 at 21:51
  • For the result - 0.24% - is it possible to calculate how many rolls are needed in order to get this alternate of 9 outcome (RBRBRBRBR)? – Theo Jun 29 '13 at 21:56
  • I so not understand the question. Maybe you mean how many rolls are needed so that the probabilitity of RBRBRBRBR in that order is $0.0024$. Let it be $n$. Then $\left(\frac{18}{38}\right)^{n}=0.0024$. But one cannot expect this to have an exact integer solution. One can find approximations to similar questions using logarithms. – André Nicolas Jun 29 '13 at 21:59
  • Sorry I meant how many rolls are needed so that the probability of RBRBRBRBR in that order will happen. – Theo Jun 29 '13 at 22:03
  • It is possible that you roll $10000$ times and it doesn't happen. So we need to change your question to something like "will happen with probability at least $99%$. This is a decidedly much more complicated problem than the one you asked! Solvable though, but should be a separate question. – André Nicolas Jun 29 '13 at 22:08
  • Hey, can you help me out with another problem I have about the Fibonacci betting system for Roulette? The person provided an answer, but I'm unsure whether it's correct or not as he did not provide an explanation for my questions. – Theo Jul 02 '13 at 16:44
  • I had seen it. There is some ambiguity about stepping back $2$ if we win. The calculation could be unpleasant. With these sorts of semi-complicated strategies, the best approach is often not analytical. Instead one does extensive simulation. One can write a program that will quickly run through a few million cycles. – André Nicolas Jul 02 '13 at 16:51
  • I actually know a site that provides a simulator for these betting systems. I have tested the Fibonacci system with it several dozen times and it seems like both the Martingale and Fibonacci systems will result in negative gain in the long run, but the Fibonacci system takes longer for that to happen i.e. if the M-system takes 1000 runs to lose everything then F-system will take maybe 2000 runs or so? But I was thinking, since casinos have max-betting limit in place then the F-system will certainly have a higher odds of winning since it allows you to have more bets in a series. – Theo Jul 02 '13 at 17:01
  • this is the simulator I know. – Theo Jul 02 '13 at 17:04
  • what I meant was since casinos have max-betting limit in place then the Fibonacci system will have a higher odds of winning once(initial bet) in a series since it allows you to have more bets in a series. – Theo Jul 02 '13 at 18:12
  • That may be good from the point of view of the House. On average, the more you bet the more you lose. – André Nicolas Jul 02 '13 at 18:21