, 2017

Solution:

When the party distribution is $6$-$6$, the situation can change (to $3$-$9$) only when a group of three contains three people from the same party, and the remaining three are distributed evenly across the other three groups (to be converted).

To compute the probability, we assume that the groups and the members of the group are ordered (so there are $12!$ ways of grouping). There are $2$ ways to choose the party, $4$ ways to choose the group, $6 \cdot 5 \cdot 4$ ways to choose the three members of the group, $9 \cdot 6 \cdot 3$ ways to place the other three members of the party, and $6!$ ways to fill in the members of the other party. The probability is then
$$\frac{2 \cdot 4 \cdot 6 \cdot 5 \cdot 4 \cdot 9 \cdot 6 \cdot 3 \cdot 6!}{12!} = \frac{2 \cdot 4 \cdot 6 \cdot 5 \cdot 4 \cdot 9 \cdot 6 \cdot 3}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7} = \frac{18}{77}$$
This means that the shift in distribution will happen in $\frac{77}{18}$ hours on average.

When the distribution is $3$-$9$, the situation can change (to $0$-$12$) only when the three members of the minority party are all in different groups. Using the similar method as above, there are $12 \cdot 9 \cdot 6$ ways to place the three members and $9!$ ways to place the rest, so the probability is
$$\frac{12 \cdot 9 \cdot 6 \cdot 9!}{12!} = \frac{12 \cdot 9 \cdot 6}{12 \cdot 11 \cdot 10} = \frac{27}{55}$$
This means that the shift in distribution will happen in $\frac{55}{27}$ hours on average.

By linearity of expectation, we can add up the two results and get that the expected value is $\frac{77}{18} + \frac{55}{27} = \frac{341}{54}$ hours.