Jim goes the Casino to play roulette. He only makes “dozen” bets at each spin ; his probability of winning is therefore $\frac{1}{3}$ every time (to simplify, we neglect the effect of the zeros in the roulette wheel), so that if his stake is $x$ units of currency, when he wins he is returned thrice that amount, the expectation of his gain is $0$. He has also decided to stop playing and leave the casino as soon as he has achieved a positive gain (however small), and has brought with him a total sum of $M$ units of currency. Finally, he decides what amounts he bets according to a precise rule : when he has lost $k$ units of currency ($1 \leq k < M$) he bets $f(k)$ units, where $f$ is a function. We call $f$ a martingale. Note that $f$ must satisfy $k+f(k)\leq M$ for every $k$ (Jim cannot bet more than he already has).

This is a typical Markov Chain situation, with $M+2$ vertices $v_{+},v_0,v_1,\ldots,v_{M}$ (where $v_j$ corresponds to the point where Jim has lost $k$ units, and $v_{+}$ means he has some positive gain). If we denote by $\pi(k)$ the probability for Jim to eventually win from a position where he has lost $k$ units of currency, we have

$$ \pi(k)=\frac{1}{3}\pi(k-f(k))+\frac{2}{3}\pi(k+f(k)) $$

Then, the question is, what is the best strategy (i.e. which $f$ maximizes $\pi(0)$) ?

After looking at a few numerical examples up to $M=20$, it would seem that there is a unique optimal $f$, satisfying $f(k)=1$ when $k$ is even and $f(k)=\lceil \frac{k}{2} \rceil$ when $k$ is odd and $k\leq \lfloor \frac{M+1}{3}\rfloor$ (what happens for the other values of $k$ is unclear).

Does anyone know more about how to solve this exercise ?


2 Answers 2


To maximize the probability of a positive sum, you want each individual bet to give you a total net positive sum if you win. Then your probability of winning is $1 - ({2\over 3})^N$, where $N$ is the number of bets you make. To maximize $N$, bet the smallest amount each time that will produce winning results.

To that end, you would bet: $1$ unit on the first bet. If you win, stop. If you lose, then bet $1$ again, and continue with $2, 3, 4, 6, 9, 14, \ldots$ units. The pattern is: $$ a_i = a_{i-1} \times 1.5 $$ Round fractions down, then up, alternately. Each $a_i$ is thus the smallest integer greater than half what you have already lost.

If you follow that betting pattern, then if you win any spin, you will end up with either $1$ or $2$ units more than you started with.

If you started with $M = 40$ units, then you can make up to 8 bets, the last being $14$. You have $(2/3)^8 \approx 4\%$ chance of losing all $40$ units, and a $96\%$ chance of winning $1$ or $2$ units.

  • $\begingroup$ @Did Your comment is wrong unless I missed something. At every spin, the algebraic gain $G$ of Jim is distributed according to the zero-mean law $P(G=2s)=\frac{1}{3}, P(G=-s)=\frac{2}{3}$ where $s$ is the stake. Winning this particular spin allows Jim to leave the game iff $2s > l$ where $l$ is the amount already lost. So the minimal $s$ is, indeed, roughly equal to $\frac{l}{2}$. $\endgroup$ May 20, 2014 at 7:55
  • $\begingroup$ @EwanDelanoy I am wrong, I somehow forgot that when one wins one receives twice one's bet. Thanks for your explanation. $\endgroup$
    – Did
    May 20, 2014 at 8:16

I don't know mathematics. I play only single dozen. I wait for a dozen not occurred 5 times. Then start betting on that dozen. Progression is as follows:













I have seen maximum 15 games not occurred and lost. But 16th game must win. You need to increase bet as above. Lot of patience required.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .