Kangaroo problem There is a statement in the book Data Analysis by D.S. Sivia that I presently not understand.
The statement occurs during the consideration of an example concerning kangaroos, first proposed by Gull and Skilling (1984), in an introduction of the principle of maximum entropy.
The example goes as follows:
Information:
A third of all kangaroos have blue eyes and
a quarter of all kangaroos are left-handed

Question:
On the basis of this information alone, what
proportion p of kangaroos are both blue-eyed 
and left-handed?

In the discussion that follows, the book states that our common sense would draw us towards p=1/12, because any other value would indicate a correlation between handedness and eye colour. 
Why is it so, that the choice of p=1/12 implicates no correlation between handedness and eye colour and can we explicitly show that for this example?
 A: Basically, what it says is that we assume that both features are independent, so by definition of independence their joint probability is the same as multiplying their marginal probabilities
$$ \begin{align}
P(A \cap B) & = P(A)\, P(B) \\
 & = 1/4 \times 1/3 = 1/12
\end{align} $$
What Sivia and Skilling argue is that such assumption maximizes entropy, what is a further argument for assuming independence if we have no external arguments for other choice.
A: The question here is whether independence of two binary variables is equivalent to no correlation between those variables.
Let's see they are equivalent.
For binary variables, Pearson's correlation becames Phi coefficient:
$$\phi = \frac{n_{11}n_{00}-n_{10}n_{01}}{\sqrt{n_{1\bullet}n_{0\bullet}n_{\bullet0}n_{\bullet1}}}$$
And $\phi=0$ means that both variables are uncorrelated. For convenience, let's express Phi coefficient as a function of probabilities instead of absolute frequencies:
$$\phi = \frac{p_{11}p_{00}-p_{10}p_{01}}{\sqrt{p_{1\bullet}p_{0\bullet}p_{\bullet0}p_{\bullet1}}}$$
From here:
$$\phi=0 \Leftrightarrow 0=p_{11}p_{00}-p_{10}p_{01}$$
On the other hand, as Tim stated, the definition of independence is:
$$P(A \cap B)  = P(A)\, P(B)$$
Or in the notation I'm using here:
$$p_{11}=p_{\bullet1} \cdot p_{1\bullet}$$
Let's compute $p_{\bullet1} \cdot p_{1\bullet}$ (in your example, the probability of a kangaroo having blue eyes mutiplied by the probability a kangaroo being left handed):
$$p_{\bullet1} \cdot p_{1\bullet} = (p_{10}+p_{11})\cdot(p_{01}+p_{11}) = \\
=p_{10}\cdot p_{01}+(p_{11})^2+p_{10}\cdot p_{11}+p_{10}\cdot p_{11}= \\
=p_{10}\cdot p_{01}+p_{11}\cdot (p_{10}+p_{01}+p_{11})= \\
=p_{10}\cdot p_{01}+p_{11}\cdot (1-p_{00})= \\
=p_{11}+p_{10}\cdot p_{01}-p_{11}\cdot p_{00}= \\
=p_{11}- (p_{11}\cdot p_{00}-p_{10}\cdot p_{01})$$
Therefore:
$$\text{no correlation} \Leftrightarrow  p_{11}\cdot p_{00}-p_{10}\cdot p_{01}=0 \Leftrightarrow p_{11}=p_{\bullet1} \cdot p_{1\bullet} \Leftrightarrow \text{independence}$$
Update with the numbers in the question:
Here $p_{1\bullet}=\frac{1}{3}$ and $p_{\bullet1}=\frac{1}{4}$. Therefore, by above, $p_{11}=\frac{1}{12}$ implies both independence and no correlation.
