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Re: Two equally skilled teams play a four games tournament. What [#permalink]
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: Two equally skilled teams play a four games tournament. What [#permalink]
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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Re: Two equally skilled teams play a four games tournament. What [#permalink]
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: Two equally skilled teams play a four games tournament. What [#permalink]
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