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

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Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 04 Apr 2011, 22:38
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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
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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 04 Apr 2011, 22:39
I think the answer would be 5/8 1/2 (VCU) + 1/4( UCONN) - 1/2*1/4 (VCU and UCONN) = 5/8
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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 04 Apr 2011, 22:42
I'm getting 3/4.

1/2 (VCU wins over Kansas) + 1/2 (UCONN wins against UK)*1/2(UCONN wins the final) = 1/2 + 1/4 = 3/4

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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 05 Apr 2011, 17:29
(1/2)*(1/2*1/2) = 3/4
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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 05 Apr 2011, 19:44
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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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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 09 Apr 2011, 23:38
# 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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Re: Although the games have already happened, a friend of mine had a bet  [#permalink]

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New post 24 Jan 2019, 21:30
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Re: Although the games have already happened, a friend of mine had a bet   [#permalink] 24 Jan 2019, 21:30
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