jonblazon wrote:

Although the games have already happened, a friend of mine had a bet where he could win one of two ways from the final four.

1. VCU wins over Kansas

2. UCONN wins the whole tournament. (wins against UK, and then wins the final)

What's the probability of him winning the bet?

A. 1/2

B. 3/4

C. 5/8

D. 1/8

E. 1/4

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.

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