# Simple calculation in imprecise probability urn example

In Miranda and de Cooman's chapter 3, "Structural judgements", in Augustin et al.'s Introduction to Imprecise Probability, example 3.4 on p. 65 shows that independence in the selection (type-2 independence) does not imply strong independence (type-3 independence) for lower previsions. One of the numerical values in the example seems incorrect to me. I'm not sure whether there is a typo (there is an obvious trivial typo at the top of the same page, so this seems possible), or whether I don't fully understand the example (quite plausible).

There are two possible configurations, or compositions, for a pair of urns containing red ($$R$$) and green ($$G$$) balls, but we don't know which configuration is actual. Given a configuration, draws from each urn are random, i.e. independent.

Configuration 1:    Urn 1: $$\{R, R, G\}$$     Urn 2: $$\{R,R,G\}$$

Configuration 2:    Urn 1: $$\{R, G, G\}$$     Urn 2: $$\{R,G,G\}$$

Let $$X_k$$ be the r.v. representing the outcome of the draw from the $$k$$th urn. The text claims that given this setup, the lower prevision $$\underline P$$ of $$X_1=R$$ & $$X_2=G$$ is $$\begin{equation} \underline P(X_1 = R, X_2=G) = \frac{4}{9} . \end{equation}$$ It seems to me that this value should be 2/9. Since the draws are independent within each configuration, the probability (linear prevision) corresponding to configuration 1 is $$\begin{equation} P(X_1 = R, X_2=G) = P(X_1=R) \times P(X_2=G) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} . \end{equation}$$ Similarly, the probability corresponding to configuration 2 is $$\begin{equation} \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} . \end{equation}$$ Since the lower prevision is the lower envelope of the probabilities in this example, it seems to follow that $$\underline P(X_1 = R, X_2=G) = 2/9$$. However, I suspect that I've interpreted the example incorrectly in some respect or have some more basic misunderstanding.

(I believe that if 2/9 were the correct value here, the example would still illustrate the point that independence in the selection doesn't imply strong independence. For that point all that's needed is that $$\underline P(X_1 = R, X_2=G) > 1/9$$.)

Yes, indeed, $\tfrac 29$ appears to be the correct value for that setup.