Another Casual Coder

Here is the problem

The key here is to use a simple count and always decide on the following:
- If when picking a random number between 0 and count (inclusive) gives you 0, then pick the new element
- Otherwise, don't change elements

The probability that you will switch elements is always 1/(count+1), which over time will give you a fair and equal probability across all the elements. Code is below and on Github: https://github.com/marcelodebarros/dailycodingproblem/blob/master/DailyCodingProblem09292018.cs

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace DailyCodingProblem
{
class DailyCodingProblem09292018
{
private string randomElement;
private int count;
private Random rd;

public DailyCodingProblem09292018()
{
randomElement = "";
count = 0;
rd = new Random((int)(DateTime.Now.Ticks % Int32.MaxValue));
}

public string RandomElement
{
get
{
return randomElement;
}
}

public void IncomingFeed(string element)
{
randomElement = rd.Next(0, ++count) == 0 ? element : randomElement;
}
}
}