dont want a random number to show twice

Question

I am trying to save random numbers in a list, the same number can not come twice. It´s a BINGO game where this method is used to display ex B12, then the user hits enter and a new number will show. This will keep on until the user writes q.

This works, BUT the number can show up twice...

static void bingo()
{
    Random rnd =new Random();
    List<int> check = new List<int>();
    string choice = "";

    while (choice != "Q")
    {
        int number = rnd.Next(1, 76);

        while (!check.Contains(number))
        { 
            kontroll.Add(number); 
        }
        if (number <=15)
        {
            choice = Interaction.InputBox("B" + number);
            choice = choice.ToUpper();
        }
        else if(number <= 30)

        etc.

so generate a list of all possible letters/numbers first and then shuffle that list and then iterate through the list. you'll never get a repeat as long as you keep moving "forward" in the list. — Marc B, Mar 18 '14 at 20:15
possible duplicate of [Randomize a List in C#](http://stackoverflow.com/questions/273313/randomize-a-listt-in-c-sharp) — Alexei Levenkov, Mar 18 '14 at 20:27

score 2 · Answer 1 · answered Mar 18 '14 at 20:17

2

Something like this should work (if I'm reading your question correctly)

Enumerable.Range(1,76).OrderBy(n => rnd.NextDouble())

answered Mar 18 '14 at 20:17

Kelly Robins

7,168
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this answer hints at a very good solution, but you hasn't followed it through all the way yet. – Sam I am says Reinstate Monica Mar 18 '14 at 20:18

score 0 · Answer 2 · answered Mar 18 '14 at 20:18

There are a couple of ways to do this:

Keep track of what numbers have been "called" - if a number is in the list, pick a different one
Remove numbers that have been called from the original list, then pick a new one at random each time.
Sort the list of possible values by a random number and just work though the list

score -1 · Answer 3 · answered Mar 18 '14 at 20:23

-1

The easiest way to accomplish this is to use a HashSet.

var usedNumbers = new HashSet<int>();
...
int number;
do {
    number = rnd.Next(1, 76);
} while (usedNumbers.Contains(number));
usedNumbers.Add(number);

answered Mar 18 '14 at 20:23

Olivier Jacot-Descombes

104,806
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-1: This is one of the worst implementation of shuffle since picking last numbers will take a bit less than forever (there will be 70+ items selected and random number would have to hit one of remaining 5- slots). – Alexei Levenkov Mar 18 '14 at 20:26
It depends how many numbers you will pick. If you are picking ten numbers out of 76. There will be no problem. (Letting pick the user all the numbers of the set would let him win all the time.) – Olivier Jacot-Descombes Mar 18 '14 at 20:30
According to the question "It´s a BINGO game" so there is a good chance that all numbers will need to be picked (or at least significant portion). – Alexei Levenkov Mar 18 '14 at 20:32
@AlexeiLevenkov It's still just O(72^2) which just about any computer can handle – Sam I am says Reinstate Monica Mar 18 '14 at 20:40
According to [Wikipedia](http://en.wikipedia.org/wiki/Bingo_(U.S.)) "The most chips one can place on a Bingo board without having a Bingo is 19". – Olivier Jacot-Descombes Mar 18 '14 at 20:42
1

@OlivierJacot-Descombes Have you ever played bingo and have had a number called out that wasn't on your sheet? – Sam I am says Reinstate Monica Mar 18 '14 at 20:46
I do not play bingo, but those who play do not always win. In fact they rarely win, which means that only a small part of the possible numbers is picked at each round. – Olivier Jacot-Descombes Mar 18 '14 at 20:49

dont want a random number to show twice

3 Answers3