Probabilities are hard. As an example, there's a known puzzle: a family has two children; if one of them is a girl, what is the probability that the other one is also a girl?

The answer, un-intuitively, is **not** 1/2 but 1/3. There are various explanations but – as with the Monty Python puzzle years ago – I wanted to write code to check it out, so I wrote the following using LinqPad:

void Main() { var rnd = new Random(); // Generate a random set of families with two children; true means girl, false means boy var all = Enumerable.Range(1, 10000).Select(_ => new Pair(rnd.Next(2) == 0, rnd.Next(2) == 0)).ToList(); // Extract only the families with at least one girl var oneGirl = all.Where(it => it.First || it.Second).ToList(); // Out of those families, how many have two girls? The result should be 1/3rd var otherGirl = oneGirl.Where(it => it.First && it.Second).ToList(); Console.WriteLine((decimal) otherGirl.Count / oneGirl.Count); } // Define other methods and classes here public class Pair { public bool First { get; private set; } public bool Second { get; private set; } public Pair(bool first, bool second) { First = first; Second = second; } }

Conditional probabilities (that is, probabilities where we have additional information – in this case, knowing that one child is a girl) are surprisingly tricky. A way to express the condition more clearly is: what is the probability that a family with two kids has two girls, **given that** they have at least one girl? Probabilities are, at base, expressions of uncertainty. If we ask "what is the probability that a family with two children has two girls?" we're in a situation of maximum uncertainty: all outcomes (BB, BG, GB and GG) are equally probable so the best we can do is 1/4 – that is the mathematical equivalent of "no clue". However, if we add some information – namely that one of the children is definitely a girl – then we removed some uncertainty from the problem: the BB case is no longer possible. This lowers the uncertainty of the GG case to 1/3.

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