Probability Game Optimal Strategy You're playing a game where you have a .45 probability of winning and .55 probability of losing. You start out with 2000 chips and the winning condition is that if you bet a cumulative total of 10000 chips, then you win. Otherwise if you run out of chips, you lose.
For example, if you bet 20 chips, no matter the outcome, that will contribute 20 to the total. And you'll be left with either 1980 or 2020 current chips.
The min bet size is 1$ and max is as much as you currently have. What is the best strategy to maximize your probability of winning?
What I said was, you'd want to use a strategy akin to betting as much as possible at each round. Since otherwise, from LLN, you are losing in general and just prolonging with small bets will make you lose more in the long run. I said you'd be able to maybe calculate the exact strategy using dynamic programming, with f(s,t) representing what you should bet at each combination of (current money, total cumulative so far). However I do not know if this is correct nor did I have time to fully solve it.
Thank you!
 A: First, see the comment/assertion of lonza leggiera.  I strongly suspect that he is right.  I have added an Addendum to (superficially) explore his assertion.
Assuming so, the answer shown below is wrong.  The answer was based on my lack of intuition re Statistics, and is an example of my going off the rails.

I don't know enough about statistics to rigorously prove my answer.
However, I am fairly certain this is the optimal approach.

*

*Try to bet 2000 three consecutive times.


*If you lose the first bet, you lose.


*If you win the first bet but lose the next two, you lose.


*If however, you win the first bet, and then win either the 2nd or 3rd bet, then you automatically win.


*This is because you will have at least 4000 and you will have already bet 6000.


*Therefore, make the 4th bet for 4000.
Even if you lose, and it busts you, you still win overall because you succeeded in wagering 10000.
Your chance of losing is
$$ (0.55) + \left[(0.45) \times (0.55)^2\right].$$

Addendum
Responding to the comment/assertion of lonza leggiera.
First of all, his assertion hit me in my blind spot.  With my general ignorance of statistics, my intuition was that small bets will almost certainly deplete one's bankroll, given the [55%-45%] underdog constraint.  That intuition is not conclusive however, because the goal here is not to show a profit in the bets, but rather to succeed in wagering \$10,000.
In my initial answer, the probability of successfully wagering a cumulative total of \$10,000 is clearly less than 0.45.  Consider the following strategy that is very similar to the one proposed by lonza leggiera.

Strategy A: 
Bet \$1 until you either go broke or have made the bet 10,000 times.

You start with \$2000.  As a [55%-45%] underdog, your expected return using Strategy A is to convert your \$2000 bankroll into only \$1000.  In order for this strategy to be (somehow) worse than the strategy proposed in my initial answer, your chances of going broke (instead of hovering around \$1000) would have to be greater than 55%.  It is hard to believe that your chances of losing \$2000 (instead of only losing \$1000) would be that high.
Now consider the strategy proposed by lonza leggiera.

Strategy B: 
Bet \$1 until you either go broke or 
(the cumulative amount that you have wagered + your remaining bankroll) 
equals or exceeds \$10,000.  
If you reach the point where the sum equals or exceeds \$10,000 
then bet your remaining bankroll.

Actually, Strategy B is (in effect) equivalent to Strategy A.  
Once you reach the point where 
(the cumulative amount that you have wagered + your remaining bankroll) 
equals or exceeds \$10,000,  
then wagering your remaining bankroll \$1 at a time will (also) guarantee victory.
