Random function and calculating percentage - math

Question

Using a random library with these functions:
randomChance(p) Returns true with the probability indicated by p.
randomInteger(low, high) Returns a random integer in the range low to high, inclusive.
what is the easiest way to implement a "random selector" that takes consideration of percentage, 1/4 or 1/3 etc... I got a array with key/value pairing. For example "a" migth have the value 2 and "b" have the value 2. 1/2 chance for both.
The max value will be the size of the array, cause it only contains unique items. The randomChance() function ranges between 0.0 - 1.0 where 1 = 100%. If my array size is, say 4. What is the best way of "letting 4 be 1".

Answer 1

Lets say you have:
a = 2, b = 2, c = 1, d = 3
now make it:
a = 2, b = 4, c = 5, d = 8
Create a random number from 1 to MaxVal (value of the last key, 8 in this example). Select the first Key where Value >= RandomNum
EDIT
I made a small VB.Net to show the algorithm and how it works. The code is not meant to be: Good, elegant, performant or readable.
Module Module1
Private Class Value
Public vOrg, vRecalc, HitCount As Integer
Public Key As String
Public Sub New(s, v1, v2, c)
Key = s : vOrg = v1 : vRecalc = v2 : HitCount = c
End Sub
End Class
Sub Main()
' set initial values
Dim KVP() As Value = {New Value("A", 2, 0, 0),
New Value("B", 2, 0, 0),
New Value("C", 1, 0, 0),
New Value("D", 3, 0, 0)}
' recalc values
For i = 0 To KVP.Length - 1
If i = 0 Then KVP(0).vRecalc = KVP(0).vOrg Else KVP(i).vRecalc = KVP(i).vOrg + KVP(i - 1).vRecalc
Next
' do test
Dim r As New Random
Dim runs As Integer = 1000 * 1000, maxval As Integer = KVP(KVP.Length - 1).vRecalc
For i = 1 To runs
Dim RandVal = r.Next(1, maxval + 1)
Dim chosen As Integer = (From j In Enumerable.Range(0, KVP.Length) Where KVP(j).vRecalc >= RandVal Take 1 Select j)(0)
KVP(chosen).HitCount += 1
Next
' ouput results
For Each kv In KVP
Console.WriteLine("{0} was chosen with {1:F3} propability, expected was {2:F3}", kv.Key, kv.HitCount / CDbl(runs), kv.vOrg / CDbl(maxval))
Next
Console.ReadLine()
End Sub
End Module
An output sample:
A was chosen with 0.250 propability, expected was 0.250
B was chosen with 0.251 propability, expected was 0.250
C was chosen with 0.124 propability, expected was 0.125
D was chosen with 0.375 propability, expected was 0.375

Answer 2

just multiply the randomChance() outcome and the array length together. It'll give you the index in the range [0,array_length-1] which you can use to access the array
array_index = (unsigned int)(randomChance(p) * (array_length - 1));
maybe you mean "letting 3 to be 1" (not 4) in your example. The last index of an array of length 4 is 3.