Monty Hall Problem Something wrong with counter

Question

int main()
{
    srand(time(0));
    int counter = 100000;
    int WinSwitchCounter = 0;
    int WinStayCounter = 0;
    int decision;

    cout << "Number of Times Staying Was The Correct Strategy: "
        << WinStayCounter << endl;
    cout << "Number of Times Switching Was The Correct Strategy:"
        << WinStayCounter << endl;

    for (int decision = 0; decision <= counter; ++decision)
    {
        int prize_door = (rand() % 3 + 1);//random choice from 1-3
        int stay_choice = (rand() % 3 + 1);
        int switch_choice = (rand() % 2 + 1);

        if (stay_choice == prize_door)
        {
            ++WinStayCounter;
        }
        else if (switch_choice == prize_door)
        {
             ++WinSwitchCounter;
        }
//Compare Stay to Switch to determine best strategy
         if (WinStayCounter > WinSwitchCounter)
         {
             cout << "Therefore, the best thing to do is to Stay " << endl;
         }
         else
         {
             cout << "Therefore, the best thing to do is to Switch " << endl;
         }
    return 0;
    }
}

v

Vulpes · Answer

Are you sure you've copied it correctly as I've just recompiled and run it again with the following results: Number of Times Staying Was The Correct Strategy: 33368 Number of Times Switching Was The Correct Strategy:66632 Therefore the best thing to do is to Switch When I ran it before I always obtained similar results which is what you'd expect as switching should of course be the correct strategy twice as often as staying. As you'll have gathered I'm numbering the doors 0 1 and 2 rather than 1 2 and 3 as this makes life easier when dealing with zero-based arrays. The array doors simply stores what's actually behind the doors. It's initialized to contain all zeros (goats) and then the prize (car) is placed at random behind one of the doors. The player also chooses a door at random. Finally what's behind one of the doors is shown. This shouldn't be the same as the player's choice or be the car - if it is the do/while loop continues until an acceptable door is found. You then figure out whether you'd have won by staying or switching and add it to the appropriate counter.

Q F · Answer

Number of Times Staying Was The Correct Strategy: 33409 Number of Times Switching Was The Correct Strategy:33409 Therefore the best thing to do is to Switch This is not right I am slightly confused about the arrays and what you put in them

Vulpes · Answer

Try this: #include #include #include using namespace std; int main() { srand(time(0)); int counter = 100000; int WinSwitchCounter = 0; int WinStayCounter = 0; int decision; for(int decision = 0; decision { int doors[] = {000}; // 0 is a goat 1 is a car int prize_door = rand() % 3; // random choice from 0 - 2 doors[prize_door] = 1; // put car in prize door int stay_choice = rand() % 3; // choose any door int shown; //the shown door do { shown = rand() % 3; } // don't show the car or the choice while (doors[shown] == 1 || shown == stay_choice); WinStayCounter += doors[stay_choice]; // if you won by staying count it // the switched (last remaining) door is (3 - choice - shown) because 0+1+2=3 WinSwitchCounter += doors[3 - stay_choice - shown]; } cout cout //Compare Stay to Switch to determine best strategy if (WinStayCounter > WinSwitchCounter) { cout } else { cout } return 0; }

Monty Hall Problem, Something wrong with counter