Shuffling an Array is putting the same integer in more than one place

**katecook** · September 11th, 2014, 01:06 PM

I have to shuffle a deck (array) of 52 integers but I started with 3 for testing if it was an even shuffle and it will place the same integer in more than one spot in the random array. I'm not sure what I'm doing wrong.

import java.util.Random;
 
public class shuffleDeck 
{
	public static void main(String[] args) 
	{
		int[] Deck = new int[3];
 
		for (int i=0; i<3; i++)
		{
			Deck[i] = i;
		}
 
		Random r = new Random();		
		int[] RandomDeck = new int[3];
 
		for (int i=0; i<3; i++) 
		{
		    int randomPosition = r.nextInt(3);
		    RandomDeck[randomPosition] = Deck[i];
		}
 
		System.out.println("Original Deck \t Random Deck");
 
		for (int i=0; i<3; i++)
		{
			System.out.println(Deck[i] + "\t\t " + RandomDeck[i]);
		}
	}
}

**KevinWorkman** · September 11th, 2014, 01:12 PM

Think about what this for loop is doing:

for (int i=0; i<3; i++) 
{
   int randomPosition = r.nextInt(3);
   RandomDeck[randomPosition] = Deck[i];
}

You're going through the original deck, and copying each element of the original deck to a random index in the new deck. What happens if your random positions are all zeros? Or two 1s and a 2?

**katecook** · September 11th, 2014, 01:16 PM

Should I be doing it the other way around?

**KevinWorkman** · September 11th, 2014, 01:34 PM

What happened when you tried?

What you need to do is figure out *what you're trying to do* first. How would you do this in real life, with a real deck of cards?

Pretend you have a friend who has no idea how to shuffle cards. Your job is to tell that friend to take a deck of cards and then *copy* that deck of cards (your friend loves drawing) into a new shuffled deck of cards. Write down instructions that your friend can follow- remember, your friend has no idea how to shuffle cards, so the steps you write have to be very small and easy to follow. If a step is too complicated, then break it down into smaller steps.

When you have a list of instructions that your friend can follow, then you'll have an algorithm that you can think about implementing.

**GregBrannon** · September 11th, 2014, 02:58 PM

Another way to think of it is that dealing from any place in the deck randomly chosen is the same as shuffling before dealing the cards in order.

Also posted here.

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