McKinney's Generalization of the Birthday Problem: Part II

In Part I of the article, we considered the expression for computing the probability that at least r people will have the same birthday, given n people are selected at random.

In this article, we will consider a special case to illustrate the use of formula numerically: Suppose five people are selected at random, what is the probability that at least three people will share the same birthday?"

In this example, the Frobenius equation n1 + n2 = 5 has three sets of solutions and they can be tabulated as follows:


Following McKinney, we define the general form of the probability P (n; n1, n2) as


It follows that P(E) can be computed as follows:


The required probability that at least three people are sharing their birthdays is therefore:


Alternatively, we could also arrive at the same result if the problem is approached from another direction, but we would require a slight modification on McKinney's formula.


The required probability is thus the summation of the following fractions:


This is approximately 0.007 percent. From the probability table above, we noted that if we were to select five people at random, chances are 97.3% that they will all have unique birthday, this is consistent with our intuition.

Comments

Anonymous said…
I read this article, thanks for writing it and put it online

reabiagio.wordpress.com
Anonymous said…
I read this article, useful to me, thank you for writing it and put it online.
I'm trying to write an article on the same subject.

reabiagio.wordpress.com

Popular posts from this blog

The Mathematics of Honeycomb: Part I

雪洁健康养生机(一)

The Mathematics of Honeycomb: Part II