Best of seven with different home/away win rates? I'm having some issues with a question from my intro probability course and I'm hoping you can help. Here's the question:
In the World Series in baseball and in the playoffs in the National Basketball Association and the National Hockey Association, the winner is determined by the best of seven games. Most teams have better records at home. Assume that the two teams are evenly matched and each has a 60% chance of winning at home and a 40% chance of winning away. In principle, both teams should have an equal chance of winning a seven game series. Determine which pattern of home games is closer to giving each team a 50% chance of winning. Consider the two common patterns: (a) two home, three away, two home; and (b) two home, two away, one home, one away, one home.
Obviously, If the probabilities stayed the same I could use the binomial distribution and add the probabilities of winning after 4, 5, 6 and 7 games. since they're not, I'm not quite sure how to proceed. Any hints?

The ordering of the games doesn't matter with this stripped down model (with no "momentum" factor, say).
To see, this, note that we can always assume that all seven games are played.  Even though the series is probably decided before game $7$, playing the series out correctly determines the winner by simple majority.  Thus the winner is determined by counting the games $Home$ wins at home and those $Home$ wins away (let's say team $Home$ has the home team advantage here).  
$$p(Home)=\,\sum'\binom 4a
\binom 3b.6^a.6^{4-a}4^b.6^{3-b}=\sum' \binom4a\binom 3b\,.6^{3+a-b}.4^{4+b-a}$$  Where the sum is taken over pairs $a,b$ with $0≤a≤4,\,0≤b≤3,\,a+b≥4$  Here, of course, "a" denotes the number of home games $Home$ wins, and $b$ denotes the number of away games $Home$ wins.
That sum is easily evaluated (with mechanical assistance) and we get $$p(Home)=\fbox {0.532032}$$
Note:  For modeling, it is more interesting to introduce some path dependence.  That is, assume that a team gets a probability boost for having won the prior game.  Of course then the order matters a great deal (though in this case it would be tempting to simply evaluate all $2^7$ strings individually.  there are only $128$ of them, after all).
