Hi,
here's a short C++ program that computes the optimal fraction of account that one should bet
for a set of user defined probabilities and their payouts:
And the same also in a zip file:
here's a short C++ program that computes the optimal fraction of account that one should bet
for a set of user defined probabilities and their payouts:
Code:
/*
optimal_bet_fraction.cpp
Author: U.M. in Germany (user botpro @ www.elitetrader.com)
2016-02-02-Tu: init
Solution to this online posting:
http://www.elitetrader.com/et/index.php?threads/how-much-should-you-risk.297565/
There is a 35% chance that you win 20% of your bet;
There is a 25% chance that you lose 15% of your bet;
There is a 20% chance that you win 12% of your bet;
There is a 15% chance that you lose 6% of your bet;
There is a 4% chance that you win 50% of your bet;
There is a 1% chance that you lose 100% of your bet.
What percentage of your betting account should you risk to maximize your potential gain?
Compile using a C++11 conformant C++ compiler, here using GNU g++ under Linux:
g++ -Wall -O2 -std=c++11 optimal_bet_fraction.cpp -o optimal_bet_fraction.exe
Run:
./optimal_bet_fraction.exe >data.csv
And the answer is:
MaxGrowthRate=2.87735% @ fraction=77.79000%
Misc:
You can change the pre-set conditions, and also add new conditions or del some conditions,
then recompile and execute the program.
*/
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
using namespace std;
int optimal_betfraction()
{ /* creates a csv file of all iterations (this can be analysed in Excel/LibreOffice-Calc),
and prints the maximum to stderr
*/
// set these input data (vp and vr must both be same size):
const vector<double> vp = { 0.35, 0.25, 0.2, 0.15, 0.04, 0.01 }; // probabilities
const vector<double> vr = { 0.2, -0.15, 0.12, -0.06, 0.5, -1.0 }; // payouts
const double PctStep = 0.01; // 0.001; // 1.0;
double MaxGrowthRate = -1e9, xPctForMaxGR = 0; // will be filled below
printf("r,f\n");
for (double xPct = 0.0; xPct <= 100.0; xPct += PctStep)
{
const double x = xPct / 100.0;
double GrowthRate = 1.0;
for (size_t i = 0; i < vp.size(); ++i)
{
if (vr[i] >= 0.0)
GrowthRate *= pow(1.0 + x * fabs(vr[i]), vp[i]);
else
GrowthRate *= pow(1.0 - x * fabs(vr[i]), vp[i]);
}
printf("%.5f,%.5f\n", (GrowthRate - 1.0) * 100.0, xPct);
if (GrowthRate > MaxGrowthRate)
MaxGrowthRate = GrowthRate, xPctForMaxGR = xPct;
}
fprintf(stderr, "MaxGrowthRate=%.5f%% @ fraction=%.5f%%\n", (MaxGrowthRate - 1.0) * 100.0, xPctForMaxGR);
return 0;
}
int main()
{
optimal_betfraction();
return 0;
}And the same also in a zip file:
