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# Two equally skilled teams play a four games tournament. What

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Intern
Joined: 08 Nov 2009
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Two equally skilled teams play a four games tournament. What [#permalink]

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10 Dec 2009, 20:50
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Two equally skilled teams play a four games tournament. What is the probability of the tournament ending at the fourth game with a winner?

[bonus question: how about a 40 games tournament ending on the 40th game with a winner?]

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Intern
Joined: 03 Oct 2012
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GMAT 1: 620 Q39 V38
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Re: 4 games tournament - probability [#permalink]

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05 Dec 2012, 13:16
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Marco83 wrote:
Two equally skilled teams play a four games tournament. What is the probability of the tournament ending at the fourth game with a winner?

[bonus question: how about a 40 games tournament ending on the 40th game with a winner?]

First, we have to determine what we're looking for. If we are looking for a probability that after playing 4 games there will be a winner than we must consider a case of a tie. The statement does not tell us that the teams should either win or lose. It implies there might be ties.
Therefore:
1) The total possible options are WLT: 3*3*3*3=81

2) In order to be a winner, a team can do 4W, 3W+L&T, 2W+L+T, 2W+2T, 1W+3T
a)4W=1
b)3W=4!/3!=4 and we have T or L as options=4*2=8
c)2W+L+T=4!/2!=12
d)2W+2T=4!/2!*2!=6
e)1W+3T=4!/3!=4

3)Total for one team to become a winner is 1+8+12+6+4=31

4)Total for any of two to become a winner is 31*2=62

5)The probability is 62/81

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Intern
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Re: 4 games tournament - probability [#permalink]

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11 Dec 2009, 05:17
Wins for first team and for the second team are w1 and w2 respectively.

_ _ _ w1 : exactly 2 w1 in those 3 spaces.

Therefore, # of possibilities are 3! / 2! = 3

Similarly for w2, it’s 3

So, probability = (3+3)/ 16 = 6/16 = 3/8

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Re: 4 games tournament - probability [#permalink]

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13 Dec 2009, 00:28
Abin wrote:
Wins for first team and for the second team are w1 and w2 respectively.

_ _ _ w1 : exactly 2 w1 in those 3 spaces.

Therefore, # of possibilities are 3! / 2! = 3

Similarly for w2, it’s 3

So, probability = (3+3)/ 16 = 6/16 = 3/8

There could be all 3 winning games from the first 3 games too for each of the team.
_________________

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Re: 4 games tournament - probability [#permalink]

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13 Dec 2009, 17:50
Probability of one team winning exactly three games, 4c3.

4!/(3!*1!) = 4 options
Each option has a probability of $$(1/2)^4$$ = 1/16

Probability of T1 winning exactly 3 games is 1/4
Probability of T2 winning exactly 3 games is 1/4

Probability of one team winning all four games:

$$(1/2)^4$$ = 1/16

Combining the probability of all four scenarios:

1/4 + 1/4 + 1/16 + 1/16 = 5/8

Alternatively you can find out the probability of either team winning exactly two games:

4!/2!*2! = 6 * 1/16 = 3/8

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Intern
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Re: 4 games tournament - probability [#permalink]

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23 Dec 2009, 23:04
The question asks for "the probability of the tournament ending at the fourth game with a winner" ...

... if I'm not wrong, it implies that the final match determines the winner of the tournament. That means that at the end of the 3rd match, the score card is 2-1 in favor of either team.

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Re: 4 games tournament - probability   [#permalink] 23 Dec 2009, 23:04
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