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?