Maybe it might be simpler if we decreased the number in the group. Say it was a group of 3: CC, SP, and TM and for these purposes, we say the probability we are looking for is the probability of any two players picking the same team of 6.
In a group of 3, there are 3 possible pairings of people (CC and SP, CC and TM, SP and TM).
What is the probability that CC and SP choose the same team? 1 in 4096
What is the probability that CC and TM choose the same exact specific team? 1 in 4096
What is the probability that SP and TM choose the same exact specific team? 1 in 4096
What is the probability that CC and SP or CC and TM or SP and TM choose the same exact specific team? 1 in 4096 + 1 in 4096 + 1 in 4096 = 3 times 1 in 4096
or to put it another way:
(3 choose 2) times 1 in 4096
BTW, 3 choose 2 = (3*2*1)/(2*1)= 3
Now if we up our group size to 8, we would have to count the number of subgroups of 2 out of a group of 8. The short cut way is to use 8 choose 2.
