VB.NET - Removing a number from a random number generator - vb.net

Question

I am trying to create a lottery simulator. The lottery has 6 numbers, the number generated must be between 1 - 49 and cannot be in the next number generated. I have tried using the OR function but I'm not entirely sure if I am using it properly. Any help would be great. Thanks.
Public Class Form1
Private Sub cmdRun_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles cmdRun.Click
''#Creates a new Random class in VB.NET
Dim RandomClass As New Random()
''####################################
Dim RandomNumber1 As Integer
RandomNumber1 = RandomClass.Next(1, 49)
''#Displays first number generated
txtFirst.Text = (RandomNumber1)
''####################################
Dim RandomNumber2 As Integer
RandomNumber2 = RandomClass.Next(1, 49)
If RandomNumber2 = RandomNumber1 Then
RandomNumber2 = RandomClass.Next(1, 49)
End If
''#Displays second number generated
txtSecond.Text = (RandomNumber2)
''####################################
Dim RandomNumber3 As Integer
RandomNumber3 = RandomClass.Next(1, 49)
If RandomNumber3 = RandomNumber2 Or RandomNumber2 Then
RandomNumber3 = RandomClass.Next(1, 49)
End If
''#Displays third number generated
txtThird.Text = (RandomNumber3)
''####################################
Dim RandomNumber4 As Integer
RandomNumber4 = RandomClass.Next(1, 49)
If RandomNumber4 = RandomNumber3 Or RandomNumber2 Or RandomNumber1 Then
RandomNumber4 = RandomClass.Next(1, 49)
End If
''#Displays fourth number generated
txtFourth.Text = (RandomNumber4)
''####################################
Dim RandomNumber5 As Integer
RandomNumber5 = RandomClass.Next(1, 49)
If RandomNumber5 = RandomNumber4 Or RandomNumber3 Or RandomNumber2 Or RandomNumber1 Then
RandomNumber5 = RandomClass.Next(1, 49)
End If
''#Displays fifth number generated
txtFifth.Text = (RandomNumber5)
''####################################
Dim RandomNumber6 As Integer
RandomNumber6 = RandomClass.Next(1, 49)
If RandomNumber6 = RandomNumber5, RandomNumber4, RandomNumber3, RandomNumber2, RandomNumber1 Then
RandomNumber6 = RandomClass.Next(1, 49)
End If
''#Displays sixth number generated
txtSixth.Text = (RandomNumber6)
End Sub

Answer 1

Instead of "If", use "While" - in other words, keep generating random numbers until you find a new one. Currently if you get a duplicate and then get a duplicate on the second attempt, you'll keep going.
Also, while I'm no VB expert, I believe you'll need to specify each comparison in full, so instead of this:
If RandomNumber3 = RandomNumber2 Or RandomNumber2 Then
RandomNumber3 = RandomClass.Next(1, 49)
End If
you need:
While RandomNumber3 = RandomNumber1 Or RandomNumber3 = RandomNumber2 Then
RandomNumber3 = RandomClass.Next(1, 49)
End While
There are alternatives here - such as generating the numbers 1-49, shuffling them, and then fetching the first 6 results... or keeping to the "pick until there's a new one" but keep the results in a set. Either way you could avoid having quite so much code duplication.

Answer 2

You don't just need a random number generator here, you need one in conjunction with a shuffling algorithm.
Create an array of N items (we'll use seven for our example), each containing the integer relating to its position:
+---+---+---+---+---+---+---+
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
+---+---+---+---+---+---+---+
<pool(7)
and set the pool size to 7.
Then generate your random number, based on the pool size (i.e., get a number from 1 to 7). Let's say your generator returns 3.
Pull out the value at position 3 then replace that with the top value, then reduce the pool size:
+---+---+---+---+---+---+---+
| 1 | 2 | 7 | 4 | 5 | 6 | 7 | -> 3
+---+---+---+---+---+---+---+
<pool(6)
Then you just keep doing this until you've gotten the quantity of values required. If our lotto was 5 from 7:
+---+---+---+---+---+---+---+
| 1 | 2 | 7 | 4 | 5 | 6 | 7 |
+---+---+---+---+---+---+---+
<pool(7)
rnd(7) returns 3
+---+---+---+---+---+---+---+
| 1 | 2 | 7 | 4 | 5 | 6 | 7 | -> 3
+---+---+---+---+---+---+---+
<pool(6)
rnd(6) returns 1
+---+---+---+---+---+---+---+
| 6 | 2 | 7 | 4 | 5 | 6 | 7 | -> 1
+---+---+---+---+---+---+---+
<pool(5)
rnd(5) returns 5
+---+---+---+---+---+---+---+
| 6 | 2 | 7 | 4 | 5 | 6 | 7 | -> 5
+---+---+---+---+---+---+---+
<pool(4)
rnd(4) returns 2
+---+---+---+---+---+---+---+
| 6 | 4 | 7 | 4 | 5 | 6 | 7 | -> 2
+---+---+---+---+---+---+---+
<pool(3)
rnd(3) returns 1
+---+---+---+---+---+---+---+
| 7 | 4 | 7 | 4 | 5 | 6 | 7 | -> 6
+---+---+---+---+---+---+---+
<pool(2)
and there you have it, 5-from-7 numbers (3,1,5,2,6) extracted with no possibilities of duplicates and an efficient O(n) method for getting them. Any solution that relies on just getting random numbers and checking if they've already been used will be less efficient.

Answer 3

Here's another option using LINQ if you have VB2008:
Dim rnd As New Random()
Dim randomNumbers = From n in Enumerable.Range(1, 49) _
Order By rnd.Next() _
Select n _
Take 6
'Do something with the numbers here
This is a simple way to do it. If using the Random class is not random enough, then you may have to choose an alternative method.

Answer 4

You have to change the name of the textbox I'm using to the one you're using.
Dim rand As New Random
Dim winnum As New List(Of Integer)
Dim num, counter As Integer
Dim result As String = ""
Do
num = rand.Next(1, 49)
If winnum.Contains(num) Then
Do
num = rand.Next(1, 49)
Loop Until winnum.Contains(num) = False
End If
winnum.Add(num)
counter += 1
Loop Until counter = 6
'Extracting and displaying the numbers from the array
For n As Integer = 0 To 5
result = winnum(n) & " " & result
Next
'The textbox I'm using to display the result is result.text
result.Text = result

Answer 5

You can also use code as another fellow suggested above. In the code below, numbers are generated randomly and are removed from a pool of numbers until the required quantity is reached. Then the numbers left are then displayed. However this is not such a good way for generating numbers for lottery as the sequence of numbers are somehow predictable, but they are unique. Here is the code:
Dim rand As New Random, winnum As New List(Of Integer)
Dim num As Integer, result As String = ""
For n As Integer = 1 To 49
winnum.Add(n)
Next
Do
num = rand.Next(1, 49)
If winnum.Contains(num) Then
winnum.Remove(num)
End If
Loop Until winnum.Count = 7
For n As Integer = 0 To 5
result = winnum(n) & " " & result
Next
a.Text = result

Answer 6

I'd go for something like (in C#)
public static IEnumerable<int> Lotto(int max)
{
var random = new Random((int)DateTime.Now.Ticks);
var numbers = new List<int>(Enumerable.Range(1, max));
while(numbers.Count > 0)
{
int index = random.Next(1, numbers.Count) - 1;
yield return numbers[index];
numbers.RemoveAt(index);
}
}
static void Main(string[] args)
{
var lotto = Lotto(49).GetEnumerator();
lotto.MoveNext();
int r1 = lotto.Current;
lotto.MoveNext();
int r2 = lotto.Current;
lotto.MoveNext();
int r3 = lotto.Current;
Console.WriteLine("{0} {1} {2}", r1, r2, r3 );
}

Answer 7

Instead of picking random numbers and check for duplicates, you can simply loop through the numbers and check the odds for each number to be picked against a random number:
Dim count As Integer = 6 ' How many numbers to pick
Dim pos As Integer = 1 ' Lowest value to pick from
Dim items As Integer = 49 ' Number of items in the range
Dim rnd As New Random()
Dim result As New List(Of Integer)()
While count > 0
If rnd.Next(items) < count Then
result.Add(pos)
count -= 1
End If
pos += 1
items -= 1
End While
The list result now contains six numbers without duplicates, randomly picked from the range 1-49. As an extra bonus the numbers in the list are already sorted.

Answer 8

I think shuffling is the fastest alternative too. But easier to read is your approach in combination with a collections's contains function:
Dim numbers As New List(Of Int32)
For i As Int32 = 1 To 6
Dim containsNextNumber As Boolean = False
While Not containsNextNumber
Dim rnd As New Random(Date.Now.Millisecond)
Dim nextNumber As Int32 = rnd.Next(1, 50)
If Not numbers.Contains(nextNumber) Then
numbers.Add(nextNumber)
containsNextNumber = True
End If
End While
Next
numbers.Sort() 'sort the numbers from low to high