Flip a coin problem

Question

I have been playing around and wrote this little piece of code. I am trying to flip a coin defined number of times and then count how many tails and heads I am getting. So here it is:

private void Start_Click(object sender, EventArgs e)
{
    int headss = 0;
    int tailss = 0;
    int random2, g;
    string i = textBox1.Text;
    int input2, input;
    bool NumberCheck = int.TryParse(i, out input2);

    if (textBox1.Text == String.Empty) // check for empty string, when true
        MessageBox.Show("Enter a valid number between 0 and 100000.");
    else // check for empty string, when false
        if (!NumberCheck) // number check, when false
        {
            textBox1.Text = String.Empty;
            MessageBox.Show("Enter a valid number between 0 and 100000.");
        }
        else
        {
            input = Convert.ToInt32(textBox1.Text);

            for (g = 0; g < input; g++)
            {
                Random random = new Random();
                random2 = random.Next(2);

                if (random2 == 0)
                {
                    headss++;
                }
                else if (random2 == 1)
                {
                    tailss++;
                }
            }
        }

    heads.Text = Convert.ToString(headss);
    tails.Text = Convert.ToString(tailss);
}

The problem is that I keep getting problems while displaying the content. It's not even close to display they right result. Any ideas?

EDIT. Solution: move following line 3 lines up :D

Random random = new Random();

The display seems fine, but the flipping is not fine. `Random`'s default seed is based on the system clock, whose resolution is not that great. By creating a new `Random` object each time in the loop, you'll get the same seed over and over for many iterations, so your flips will not be very random. — mqp, Apr 25 '11 at 05:47

bendewey · Accepted Answer · 2011-04-25T06:02:03.670

6

Instead of

for (g = 0; g < input; g++)
{
   Random random = new Random();
   random2 = random.Next(2);
}

Declare a single Random for use throughout:

private Random randomGenerator = new Random();
private void Start_Click(object sender, EventArgs e)
{
    // ...
    for (g = 0; g < input; g++)
    {
        random2 = randomGenerator.Next(2);
    }
    // ...
}

edited Apr 25 '11 at 06:02

answered Apr 25 '11 at 05:49

bendewey

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122

Thank you. This work for me. By any chance is there performance difference between them? – HelpNeeder Apr 25 '11 at 05:57
This new way is better, but probably not measurable. – bendewey Apr 25 '11 at 06:01
1

As a rule of thumb, would you say that recreating an object over and over is slower then reusing one object? – jb. Apr 25 '11 at 06:43

score 4 · Answer 2 · answered Apr 25 '11 at 05:46

4

You should use only one Random object to generate good (as good as default Random does) random sequence.

answered Apr 25 '11 at 05:46

oxilumin

4,565
2
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score 3 · Answer 3 · answered Apr 25 '11 at 05:48

The default constructor for random take the systmem time as seed. Therefore, if you generate lots of them in a short amount of time they will all generate the same sequence of random numbers. Pull the random object out of the loop and this effect will not occur.

score 0 · Answer 4 · edited Jan 20 '18 at 20:03

With RandomGenerator. This code will count how many times coin has been flipped. It will end with 3 consecutive HEADS.

private RandomGenerator rgen = new RandomGenerator ();

public void run () {

    int value = 0;
    int total = 0;
    while (value != 3) {
        String coinFlip =  rgen.nextBoolean() ? "HEADS" : "TAILS";
        println (coinFlip);
        if (coinFlip == "HEADS") {
           value+=1;
        } else {
           value=0;
        }
        total +=1;
    }
    println ("It took "+total+" flips to get 3 consecutive heads");     
}

Flip a coin problem

4 Answers4