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    <ul> <li>We have to count occurance of each number and then visualize it as histogram. </li> <li>We actually should use dictionary for each number and count its stats. </li> <li>But in our case we can use array, too.Cause we know that our numbers will be within (1-5) <code>{1,2,3,4,5}</code>.So we will make 5 length array.Then we will just store <code>number</code> stats at <code>(number-1)</code> index.</li> <li>Then when using we will know that stats at <code>array[index]</code> will be our <code>number //number=(index+1)</code> stats</li> </ul> <p>Now the same things with code for 100 tries:</p> <pre><code> //make randomized 100 tries with {1,2,3,4,5} numbers Random rnd = new Random(); int[] RandomArray = new int[100]; for (int i = 0; i &lt; 100; ++i) RandomArray[i] = rnd.Next(1, 6); // //Now lets count of occurance of each number int[] Count = new int[5]; for (int i = 0; i &lt; 100; ++i){ // Count[RandomArray[i]-1]++; int number=RandomArray[i];//get our number from randomized box int index=number-1; // number stats will store at (number-1) index //in our case we will increase our number count ++Count[index] ;//Count[index]+=1 } </code></pre> <p>And when visualizing those as histogram we will get stats to our number using (index+1). </p> <pre><code> for (int i = 0; i &lt; 5; ++i){ //number=i+1 and its stat will be at Count[i] Console.Write("{0,2} Adet {1,2} &gt;&gt;&gt;&gt;", Count[i], i+1); for (int j = 0; j &lt; Count[i]; j++) { Console.Write("*"); } Console.WriteLine(); } </code></pre>
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    1. POAssistance with arrays
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    1. COActually i didnt understand why this 1 is deducted... and then at the end, this "i++"
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    2. CO@user3080201 I edited my answer.Check to see if it answers your question
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    3. COExtremely helpful to understand... many thanksss
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