Expected number of times of choosing a word out of a given vocabulary when words are grouped Two players (player C and player G) are playing a (modified) word guessing game. Both players share the same vocabulary $V$ and words in $V$ are grouped into $K$ bins, denoted as $b_1$, $b_2$, ..., $b_{K}$. Furthermore, we know that $b_{i} \subset V$ and $\cup_{i=1}^{K} b_i = V$. Note here we do not have $b_i \cap b_j = \emptyset$ for $i \neq j$. 
The game protocol is described as follows: 


*

*Player C uniformly chooses a word $w$ from the vocabulary $V$. Player G does not know which word $w$ is. 

*Player G chooses one bin and asks Player C whether his/her chosen word $w$ is in the bin. If it is, the game ends. Otherwise, Player G will choose another bin. 
Questions: What is the best bin choosing order and what is the expected number of times of choosing the bin, according to the best possible order? 
Example:
Suppose we have a vocabulary consisting of ten words $V = \{w_1, w_2, ..., w_{10} \}$ and three bins $b_1 = \{w_1, w_2, ..., w_5\}$, $b_2 = \{w_6, w_7 \}$, and $b_3 = \{w_8, w_9, w_{10} \}$. 
One possible bin choosing order is $b_1 \rightarrow b_3 \rightarrow b_2$ and the expected number of times of choosing the bin is $\frac{1}{2}*1 + \frac{1}{2}*\frac{3}{5}*2 + \frac{1}{2}*\frac{2}{5}*\frac{2}{2}*3 = 1.7$. I suspect this is the best bin choosing order but how can we prove this result? 
Thanks.
Note
A related question (which has additional non-overlapping constraints on the bins) is asked in MO and in its comment, the user @DavidG.Stork gives a good answer (an intuitive proof of best ordering) for the case when those bins have no overlap. 
 A: Not an answer but too long for a comment.
For a given ordering of bins, lets say $a_{1}, a_{2}, ..., a_{K}$ (which is a permutation of the $b_i$'s), let $f(w)$ be the discovery time for word $w$.  Then $f(w) = j$, i.e. it  will be discovered during the $j$th bin, iff $w \in c_j$ where $c_j = a_{j} - \bigcup^{j-1}_{i=1} a_i$.  I.e. $c_j$ are the new words introduced by bin $a_j$.
To calculate the expected number of turns, sum up all the discovery times $f(w)$ and then divide by the number of words.  So the expected number is minimized when $\sum_{w \in V} f(w)$ is minimized.
In the OP's original example all $b_i$'s are disjoint, so it's pretty clear that $\sum_{w \in V} f(w)$ is minimized when you assign the most number of them as $1$s, then the next-most number of them as $2$s, etc.  In this example the expected number of turns $= {5\times 1 + 3 \times 2 + 2 \times 3 \over 10} = 1.7$.
When the $b_i$'s are not disjoint, then as shown by the excellent example of @Jens in the comments, the greedy strategy can be suboptimal - it assigns the numbers $1,1,1,1,1,1,2,2,3,3$ for a total of $16$ when the optimal strategy assigns the numbers $1,1,1,1,1,2,2,2,2,2$ for a total of $15$.
