[Math] Roulette Strategy

probability

I came up with a strategy, which immitates the martingale(doubling) strategy , but it seems that there is a good probability to gain profit, which I want to compute.

So this is how we start:

We do not bet on black/red, but we chose columns, or dozens. Let's say we start with 100 chips.

  • Bet $1$ chip in e.g. a column. The probability to win is $12/37$. If you win, start over again, while if you lose:
  • Bet again $1$ chip to a column. Why again $1$? This is because we will still have profit if we win ( I will have won 3 chips, and lost 2 ). Again, if win go to step $1$, if lose:
  • Bet $2$ chips in a column. Each time we bet the lowest amount of funds, such that we have profit in case we win. In this case, if we win we will have played $4 $ chips , and won $6$, which gives the minimum profit($2 $ chips).

Continuing with this strategy, one has to bet in columns/dozens, the following sequence of chips (each time he loses), in order to have the minimum profit :
$$1,1,2,3,4,6,9,14,21,31,47\dots$$
So, here is my question:

Using this strategy, and starting with $100$ chips, what is the probability of gaining $+50\%$, and what is the probability of doubling?
You continue until you have doubled, and stop if you have less chips than they are required to put the next bet.

*We have to consider that when we gain some profit, e.g. we have reached $140$ chips, we are able to bet more chips of the above sequence. With $100$ chips we can lose up to 9 times, and then bet $31$ chips. But with $140$ chips we can lose one more time , and then bet $46$.

Due to the complexity of the problem, I would also appreciate a numerical solution, but I do not have the knowledge to run a simulation.

Anyone can figure out a solution using probability theory?

Best Answer

We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips; in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.

A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence

  • $p_{k,d}=0$ if $2k<d$
  • $p_{k,1}=1$ if $k\ge 150$
  • $p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$

This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction $$ p_{100,1}\approx0.56097114279613511032732301110367086534$$ should be good enough for us.

Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):

 Start=100;Target=150
 A=vector(Target-Start+1,n,-1);
 getP(k,d)={local(pkd);
    if(2*k<d, 0, if(k>=Target, 1, 
       pkd=if(d==1&&A[k-Start+1]>=0, 
             A[k-Start+1], 
             12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
          );
       if(d==1,A[k-Start+1]=pkd);
       pkd))
    }
 getP(Start,1) +.0

The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.

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