Distributed probability random number generator

Question

I want to generate a number based on a distributed probability. For example, just say there are the following occurences of each numbers:

Number| Count               


        

Answer 1

I know this is an old post, but I also searched for such a generator and was not satisfied with the solutions I found. So I wrote my own and want to share it to the world.

Just call "Add(...)" some times before you call "NextItem(...)"

/// 
 A class that will return one of the given items with a specified possibility. 
///  The type to return. 
///  If the generator has only one item, it will always return that item. 
/// If there are two items with possibilities of 0.4 and 0.6 (you could also use 4 and 6 or 2 and 3) 
/// it will return the first item 4 times out of ten, the second item 6 times out of ten. 
public class RandomNumberGenerator
{
    private List> _items = new List>();
    private Random _random = new Random();

    /// 

    /// All items possibilities sum.
    /// 
    private double _totalPossibility = 0;

    /// 

    /// Adds a new item to return.
    /// 
    ///  The possibility to return this item. Is relative to the other possibilites passed in. 
    ///  The item to return. 
    public void Add(double possibility, T item)
    {
        _items.Add(new Tuple(possibility, item));
        _totalPossibility += possibility;
    }

    /// 

    /// Returns a random item from the list with the specified relative possibility.
    /// 
    ///  If there are no items to return from. 
    public T NextItem()
    {
        var rand = _random.NextDouble() * _totalPossibility;
        double value = 0;
        foreach (var item in _items)
        {
            value += item.Item1;
            if (rand <= value)
                return item.Item2;
        }
        return _items.Last().Item2; // Should never happen
    }
}