Another way (perhaps simpler) to answer this would be the following. 

Assume that there are exhibition games totaling 8 complete games played after the series in which the loser of the series must always win. Thus if the games went ABBABB with B winning the series, the extended version would be ABBABBAA, where A must win the last two games. This would be the case regardless of how quickly the series ended or who won the series (it could be AAAABBBB or ABABABAB).

To calculate the amount of combinations in which either A or B will win four of those eight games, we can use a simple combination, 8C4. 

8!/(4!)(4!) = 8*7*6*5/4*3*2*1= 70 possible combinations.