Team $A$ facing Team $B$. Team $A$ Hosts the first and Second and if needed fifth and seventh games in the best of seven contest. Team $A$ has a 65% chance of beating Team $B$ at home in the first game. After that they have a 60% chance of a win at home if they won the previous game but a 70% chance if they are bouncing back from a loss. Similarly team $A$ chances of victory at Team $B$ are 40% after a win and 45% after a loss. What is the probability that Team $A$ will win in exactly 6 games?
The sequences that lead to a 4th win by A in game 6 are
These are 10 sequences because the 6th slot is fixed as a win and the arrangements of the other 5 slots is $^5$C$^3$ =5! /(3! 2!)=5 (4)/2 = 10.
The game H = home for team A and R = road for team A sequence for 6 games HHRRHRH Then assuming the Markov property these 10 probabilities are
p(1)= (.65)(.60)(.40)(.60)(.30)(.45)=0.012636 based on the stated conditional probabilties for wins and loss on home and road games after a win and after a loss and the probability of a win in the first game.
and you can do the rest.