Roulette betting system probability The Fibonacci is a popular Roulette betting system that is based on a naturally occurring mathematical sequence. The sequence itself is cumulative. In other words, the next number is equal to the sum of the two previous ones. So the first 12 numbers in the sequence are:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

How this system works is:

You progress through the sequence on losing bets and return towards the start with winning bets. Each time you lose, you move on to the next number in the sequence. Each time you win, you step back two numbers.

I was wondering, if I were to use £10 as my initial bet betting on red and black only on the American Roulette wheel (with the double zeros), what are the chances of winning once in a series of 12 bets? That is, what are my chances of winning at least £10 in the total sum of money that I need to put in for the whole series of 12 bets:

£10, £10, £20, £30, £50, £80, £130, £210, £340, £550, £890, £1440

Note that this is different than the Martingale system in that once you win a bet, you don't start over at the beginning of the sequence (£10 initial bet). Instead, you step back two numbers.
 A: It seems that the "American roulette wheel" is one that has $18$ red numbers, $18$ black numbers, and $2$ green numbers (marked "$0$"): a total of $38$ numbers. So for an individual bet on either red or black, the probability of winning that bet is $p = \dfrac{18}{38}$, and the probability of losing that bet is $q = 1 - p = \dfrac{20}{38}$. 
As the different bets are independent, the betting strategy used (Fibonacci or otherwise) does not matter for the probability we want to calculate here:
$$\Pr(\text{winning at least once in a series of $12$ bets})\\
= 1 - \Pr(\text{losing all $12$ bets})\\
= 1 - q^{12}\\
= 1 - \left(\frac{20}{38}\right)^{12} \\
= \frac{2212314919066161}{2213314919066161} \\
\approx 0.999548\dots
$$
So there's a $99.95\%$ chance that you win at least one of the $12$ bets. Note again that this does not depend on the betting strategy you use. Only the actual amount you'll win depends on the betting strategy of how much money to place on each bet.
A: The idea of going back two steps comes from the observation that winning $F_n$ exactly cancels the previous losses of $F_{n-1}+F_{n-2}$, so you are indeed in the same situation as you were two steps before, including your balance.
This implies that your balance is exactly zero if you win the second "1" bet or the "2" bet, is positive (by precisely your betting unit) if you win the first "1" bet, is negative after winning (not to mention losing) any other bet).
So the question is: What is the probability to ever win a first "1" bet?
Or what is the probability that you never do? Since you have an (almost) 50% chance of moving two steps back and a (slightly more than) 50% chance of moving only one step to the right, you may expect to be very often at the left end of the sequence, hence very often have the chance to win such a round.
The precise answer in your concrete question is best answered by using a Markov process and calculating the probabilities of being in one of the following states after $0, 1, 2, \ldots, 12$ rounds For $i=1,2,\ldots$ we have a state "about to bet the $i$th Fibonacci number" and one state $0$: "have made profit".
From state $i>2$ we have a $\frac{20}{38}$ chance of moving to state $i+1$ and a $\frac{18}{38}$ chance of moving to $i-2$.
From state $i=2$ we have a $\frac{20}{38}$ chance of moving to state $i+1=3$ and a $\frac{18}{38}$ chance of moving to $1$.
From state $i=1$ we have a $\frac{20}{38}$ chance of moving to state $i+1=2$ and a $\frac{18}{38}$ chance of moving to $0$.
From state $0$, we always stay in $0$.
What is the probability of being in state $0$ after $12$ rounds, given that we are in state $1$ initially?
Calculating this for 12 rounds gives us
$$ \frac{1896272908066161}{2213314919066161}\approx 0.857$$
as the probability that at one moment within 12 rounds we have a positive balance. The high probability of having made (some) profit implies that the loss in the somewhat rare cases of loss is much more than the 10GBP win you hunt for.
