Please elaborate more on the specific problem.

It's generally considered polite to post full questions/answers on the forum so everyone can benefit from everyone else.

navaji,

I can guarantee you, that no one on this board will laugh at your code. You must remember, that even the best coders out there, once where complete beginners too and had go write "simple" code and make many mistakes to reach the expertise they have now.

Even if you do have flaws or misconceptions in your code, the users here will gladly help you point them out and tell you how you can improve your program.

Yeah, don't worry about people laughing. 'Tis all good

Ok, I'm not sure what exactly is what you need to do to get this bellcurve of flip coins. I mean, I know what a bellcurve is, but I'm not sure exactly what do you want to acomplish.

What I can suggest is, try to think of the logical steps to solve your problem *before* you program anything. That way you can see if you need for loops, while loops, etc.

What I mean is, for example: "I need to first write down the state of each of ten coins (write down if they are tails or heads), then, count each tails and each heads and add them toghether. Then divide them by PI and multiply by 2..."

Or whatever you need to do. Obviously mine was an absurd example.

If you put the problem this way, I can help you out on what exactly do you need. Specifically, here's an example of a nested for loop:

for(int i=1;i<=10;i++){
  for(int j=1;j<=10;j++){
      println("i * j = " + i*j);
  }
}

This will print the multiplication table, i.e. 1x1=1...3x4=12...up to 10x10=100

Ok, this is not as hard as you might think.

You need to do two things. Generate the array of "flips", and display the array in a BellCurve.

If you need 10 flips each trial, with 1000 trials, then you need an array to store the coins like

Code:

boolean[][] coins=new boolean[1000][10];
 

Then you need to fill it, like:
Code:

float r;
for(int trial=0;trial<1000;trial++){
     for(flip=0;flip<10;flip++){
		 r=random(0,1);
		 if(r<0.5){
			   coins[trial][flip]=true;
		 }
		 else{
			   coins[trial][flip]=false;
		 }
	    
     }
}
 

Each array location represents a "heads" if its true, or false otherwise (or whatever you like). So then you just have to work out the math for building the bellcurve, and you are done.

**Sorry for the code formatting, but its hard to indent in the forums for some reason : / **

acapulco, I still need to converse...contact me by using the sites message system.

Please =/