Let's call the former team of nine player team 1 and the one of six player team 2. The number of ways you could pick three players from team 1 and two from team 2 is combin(9,3)Ã—combin(6,2) = 1,260. The total number of ways to pick five out of 15 players is combin(15,5) = 3,003. So, the probability the first team is split 3/2 in favor of team 1 is 1,260/3,003 = 41.96%.
If that happened, then team 1 will have six players left and team 2 four players. The number of ways you could pick three players from team 1 and two from team 2 is combin(6,3)×combin(4,2) = 120. The total number of ways to pick five out of 10 players left is combin(10,5) = 252. So, the probability the second team is split 3/2 in of favor team 1, given that the first team is already split 3/2 that way, is 120/252 = 47.62%.
If the first two new teams are split 3/2, in favor of the former team 1, then the final team will be split 3/2 among the leftovers.
Thus, the answer to your question is 41.96% × 47.62% × 100% = 19.98%.
Formulas:
combin(x,y)=x!/((y!*(x-y)!)
x! = 1*2*3*...*x
This question is raised and discussed in my forum at Wizard of Vegas.