Although the games have already happened, a friend of mine had a bet

(1/2)*(1/2*1/2) = 3/4

While 3/4 might be a tempting answer here, jonblazon's solution is actually correct, and the answer is 5/8. The problem with the other solutions is that they double-count when *both* VCU wins and when UCONN wins. He doesn't win the bet twice when that happens!

You can break down all of the possible events as follows:

VCU wins AND UConn wins it all: (1/2)*(1/4) = 1/8 --> he wins the bet
VCU wins AND UConn does not win it all: (1/2)*(3/4) = 3/8 --> he wins the bet
VCU loses AND UConn wins it all: (1/2)*(1/4) = 1/8 --> he wins the bet
VCU loses AND UConn does not win it all: (1/2)*(3/4) = 3/8 --> he does not win the bet

Adding you can see the answer should be 5/8, though jonblazon's solution is a more elegant approach.

Of course all of this assumes that each game is essentially like a coinflip, something which isn't true in real life, since it's not just luck that determines which team wins a basketball game.

# of different ways your friend can win the bet -
1) VCU wins while UConn loses semi-final = 1/2*1/2 = 1/4
2) VCU wins while UConn loses final = 1/2*1/2*1/2 = 1/8
3) VCU loses while UConn wins = 1/2*1/2*1/2 = 1/8
4) VCU wins and UConn wins = 1/2*1/2*1/2 = 1/8
Add all
1/4+1/8+1/8 +1/8= 5/8

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