BenAlbahari comments on Drawing Two Aces

[BenAlbahari]

Here’s code to compute the probability empirically (I got an answer of 0.2051384 for 10000 draws. It’s written in C# but can be readily converted to other functional languages such as Haskell).

var cards = new[] { "AH", "AS", "2C", "2D" };  
var pairs = cards.Combinations (2);  
var pairsWithAce = pairs.Where (p => p.Any (c => c.Contains ("A")));  

var drawsWithAce =  
(  
    from i in Enumerable.Range (1, 10000)  
    let randomPairWithAce = pairsWithAce.ElementAt (rand.Next (pairsWithAce.Count()))  
    let isPairOfAces = randomPairWithAce.All (c => c.Contains ("A"))  
    let randomAce = isPairOfAces  
        ? randomPairWithAce.ElementAt (rand.Next (2))  
        : randomPairWithAce.Single (c => c.Contains ("A"))  
    let isRandomAceASpade = randomAce == "AS"  
    select new  
    {  
        isPairOfAces,  
        isRandomAceASpade  
    }  
).ToArray();  

var conditionalProbability =  
    (float)drawsWithAce.Count (d => d.isPairOfAces && d.isRandomAceASpade) /  
    (float)drawsWithAce.Count (d => d.isRandomAceASpade);  

Notice that isPairOfAces is not a function of isRandomAceASpade. In other words, the suit of the random ace doesn’t affect the probability of there being a pair of aces. Computer programs don’t suffer Information Bias. OTOH I do so let me know if I screwed up… ;)

For those interested in the complete working code:

using System;  
using System.Collections.Generic;  
using System.Linq;  

static class Program  
{  
    static void Main()  
     {  
         ...<put code above here>...  
     }  

     public static Random rand = new Random();  

     // Returns all possible combinations of size k from a sequence of elements  
     public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int k)  
     {  
         return k == 0 ? new[] { new T[0] } :  
             elements.SelectMany((e, i) =>  
                 elements.Skip(i + 1).Combinations(k - 1).Select(c => (new[] { e }).Concat(c)));  
     }  
 }