3 ms·
Time is money, so my definition of the best strategy is the one that gets you your $250 asap. Potentially giving up if your bankroll gets so small that it would
by im3w1l 5y ago
Time is money, so my definition of the best strategy is the one that gets you your $250 asap. Potentially giving up if your bankroll gets so small that it would take unreasonably long to climb back.
With this idea we can assign a dollar value to every possible bankroll. Those values should satisfy.
V(250) = 250
V(x) = max(x, max over 0<y<=x pV(x+y) + (1-p)V(x-y) - a)
I can't immediately see how to solve it though.
- kqr 5y agoThe Kelly criterion, which maximises long-term growth, also has the nice property that it gets to any particular desired level faster than any other strategy, on average.
- im3w1l 5y agoI don't think this can be true. Consider if you start with x money and want to reach x+epsilon. Then the first bet in an optimal strategy should be to bet epsilon (if you win you reach your goal, and if you dont the setback is smaller than kelly).
- kqr 5y agoBut what do you do if you don't win? Then the target level is two epsilon away -- do you bet that amount? Maybe that works if the probability of success is high enough and the initial epsilon is small enough. If that is not the case, it starts to sound like a martingale type strategy.
- im3w1l 5y agoIf you start super close to the goal, then yes you would double for every loss... for a while. When it starts actually cutting into your bankroll the bets would grow slower and slower and then even start shrinking. It should not surprise me if, far from the goal, you would bet a fraction of your bankroll every time, like Kelly does.