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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.

Theo
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  • That betting plan says nothing about your probability of winning or your payoff on winning. Are you betting on a color, or a number? – Thomas Andrews Jul 02 '13 at 15:28
  • @ThomasAndrews my mistake I forgot to include that in. I'm going to edit the question. – Theo Jul 02 '13 at 15:30
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    For those without knowledge of roulette: Is it correct that the probability of winning a particular bet is $\frac{18}{38}$? Or is the winning probability something else? Also the payoff: if you bet an amount $x$ and you win, you gain another $x$, while if you lose the bet, you lose your $x$. Is that correct? – ShreevatsaR Jul 02 '13 at 15:48
  • @ShreevatsaR yes the payout for black/red is 1 to 1, but I'm not sure about the probability of winning a particular bet. Actually I think you're correct, the probability of winning a black/red bet is 18/38 since there are two zero slots on the wheel. – Theo Jul 02 '13 at 15:54
  • I know the formula for the Martingale system: 1-(18/38)^12 from the other question I posted, but I'm not sure if it's the same for this system. – Theo Jul 02 '13 at 16:00
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    What happens when you win your first bet? How will you step back two numbers? – Rick Decker Jul 03 '13 at 00:23
  • @RickDecker When you win your first bet you stay at the first bet for your next bet. – Theo Jul 04 '13 at 14:51

2 Answers2

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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.

ShreevatsaR
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  • In a casino where there is a maximum betting limit, I guess this system actually gives you a higher odds of winning versus the Martingale system since this system allows you to have a higher sequence? – Theo Jul 02 '13 at 16:10
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    also, does this calculation taken into account that winning at least one of the 12 bets doesn't necessarily win you money unless you win on your initial bet? Since unlike the Martingale system, it does not start over at the beginning of the sequence once you win a bet, you step back two numbers. – Theo Jul 02 '13 at 16:16
  • @Theo: Your question was, "what is the probability of winning at least one bet of the 12". The above is the answer, for any (feasible) betting system. Now you seem to be asking newer (and very different) questions, such as what if the betting system is itself infeasible, and what the probability of winning some money is. Please be clear about what your question(s) is/are. – ShreevatsaR Jul 02 '13 at 17:42
  • Sorry I must have been confused. What I actually wanted to find out was what is the chance of winning £10(not 1 bet) on a series of 12 bets(since winning one bet does not necessarily win you money unless you win on the initial bet because this is different than the Martingale system it does not start over once you win a bet, instead you step back twice. – Theo Jul 04 '13 at 15:26
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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.

Martin Sleziak
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Hagen von Eitzen
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