Four boxes with black and white pearls and a posteriori probabilities.

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Boxes $S = \{A, B, C, D\}$ contain black $B$ and white $W$ pearls. Here are the counts:

$A = \{2\cdot W, 3\cdot B \}$

$B = \{1\cdot W, 3\cdot B \}$

$C = \{2\cdot W, 1\cdot B \}$

$D = \{3\cdot W, 1\cdot B \}$

Rules of the game: We get a pearl out of $A$ and put it in $B$, then out of $B$ into $C$ and then out of $C$ into $D$. One decides to get out the pearl out of the box $D$ and we all see the pearl that got out is white $W$.

What is the probability that we have moved a black pearl out of the box $B$ into $C$?

I have calculated all the conditional probabilities but I have a problem of understanding what exactly changes in the chain and how it affects the desired value of probability.

What I'm aware is that the a posteriori probabilities of hypothesis change and I guess the value of solution is one of these but I really can't figure out which.

Solution: $\frac{12}{17}$, have no idea how to get that :D

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Here are the numbers of possibilities for the various sequences ending in $W$:

$$ \begin{array}{r|l} WWWW&2\cdot2\cdot3\cdot4=48\\ BWWW&3\cdot1\cdot3\cdot4=36\\ WBWW&2\cdot3\cdot2\cdot4=48\\ BBWW&3\cdot4\cdot2\cdot4=96\\ WWBW&2\cdot2\cdot1\cdot3=12\\ BWBW&3\cdot1\cdot1\cdot3=9\\ WBBW&2\cdot3\cdot2\cdot3=36\\ BBBW&3\cdot4\cdot2\cdot3=72 \end{array} $$

The total is $357$, and the total for the sequences with $B$ in the second slot is $252$, so the probability is $252/357=12/17$.