Twelve Zombies

Let (m,n) be the state representing $m$ blue zombies and $n$ green zombies.

Let $E_{mn}$ represent the expectated number of minutes to have all the zombies be the same color from the state (m,n).

So, from (6,6) you can go to (3,9) , (9,3) or return to (6,6).

From (9,3) you can go to (9,3) or (12,0)

And from (3,9) you can go to (3,9) or (0,12).

By symmetry we can define,

• $p_1 =$ Probability you will go from (6,6) to (9,3) or (3,9)

• $p_2 =$ Probability you will go from (9,3) to (12,0) or (3,9) to (0,12)

This gives the equations:

$E_{66}= 1 + p_1E_{39} + p_1E_{93} + (1-2p_1)E_{66}$ $E_{39} = 1 + (1-p_2)E_{39}$ $E_{93} = 1 + (1-p_2)E_{93}$

Now for the probabilities...

The total number of ways to split the twelve zombies into four groups of three is:

$N = \frac{12!}{(3!)^4\times4!} = 15400$

The only way they can move on to (9,3) is if you have 3 green ones in one group, with the other 3 groups having one green one each.

The number of ways of picking zombies in this way is given by:

$n_1 = \frac{\binom{6}{3}\times3\binom{6}{2}\times2\binom{4}{2}}{3!} = 1800$

This is because there are $\binom{6}{3}$ ways to pick 3 green ones in one group, then 3 $\binom{6}{2}$ ways to choose the one green and two blues in the next group, and finally 2 $\binom{4}{2}$ ways of picking one green and two blues in the third group. The fourth group is just the left overs so only one choice there once the other three groups are chosen. Finally, we need to divide by $3!$ to account for that many ways the three identical groups can be arranged.

So,

$p_1 = \frac{n_1}{N} = \frac{1800}{15400}$

Finally, to go from (9,3) to (12,0) you would need to have all three of the greens in different groups.

The number of ways to pick this is:

$n_2 = \frac{3\binom{9}{2}\times2\binom{7}{2}\times\binom{5}{2}}{3!} = 7560$

This is because if you look at the groups, you will have $3\binom{9}{2}$ ways to choose the first group with a one green, $2\binom{7}{2}$ ways to choose the second group with one green, and $\binom{5}{2}$ ways to choose the third group with one green. (The fourth group just gets the left over blues) Again we divide by $3!$ to account for that many ways the three identical groups can be arranged.

Therefore,

$p_2 = \frac{n_2}{N} = \frac{7560}{15400}$

Solving,

$E_{66} = \frac{341}{54}$

$341+54 =\boxed{395}$

Here is slightly easier way of calculating the probabilities $p_1$ and $p_2$ . Apart from this, my solution technique is the same as Geoff's.

Suppose the zombies display a slight modicum of intelligence, and form the groups by lining up in a row of $12$ . Zombies numbered 1-3 form the first group, those numbered 4-6 form the second group, those numbered 7-9 form the third group, those numbered 10-12 form the fourth group. We can assume that, other than in matters of colour, zombies are indistinguishable. Geoff's solution has distinguishable zombies.

To calculate $p_1$ , note that the $6$ blue zombies are equally likely to end up anywhere in the line, and so there are $\binom{12}{6} = 924$ ways of arranging the blue zombies. There will only be $3$ blue zombies after a line-up if three blue zombies end up in one group, and one blue zombie ends up in each of the other three groups. There are $4\times3\times3\times3 = 108$ ways of doing this. Thus $p_1 = \tfrac{108}{924} = \tfrac{9}{77}$ .

To calculate $p_2$ , there are $\binom{12}{3} = 220$ ways of arranging $3$ blue zombies in a line. We can only end up with no blue zombies if there is one blue zombie in each of three of the four groups. There are $4\times3\times3\times3 = 108$ ways in which this can happen. Thus $p_2 = \tfrac{108}{220} = \tfrac{27}{55}$ .