Question

Person #1 has 12 female cats.
Person #2 has 4 female cats and 8 male cats.

Assuming equal probability of a cat being male or female (no selection by the persons) and cats of the same sex are indistinguishable, which person's cat home has greater entropy?

Answer 1

Let X be the number of female cats respectively. Then the possible values of X are {0, 1, ..., 12} and,

Entropy =

where P(X) is the PMF for X

Probability of female cats = 0.5 . Then X ~ Bin(n = 12, p = 0.5)

For Person #1 has 12 female cats.

P(X = 12) = ¹²C₁₂ * 0.5¹² * 0.5^12-12 = 0.5¹²

Entropy = - P(X) log(P(X)) = -0.5¹² * log(0.5¹²) = -12 * 0.5¹² * log(0.5) = 0.002030705

Person #2 has 4 female cats and 8 male cats.

P(X = 4) = ¹²C₄* 0.5⁴ * 0.5^12-4 = 495 * 0.5¹² = 0.1208496

Entropy = - P(X) log(P(X)) = -0.1208496 * log(0.1208496) = 0.2553804

Thus, Person 2 cat home has greater entropy.