Probability In A Five-Team Tournament

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Shihab
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Probability In A Five-Team Tournament

Unread post by Shihab » Thu Dec 27, 2012 4:14 pm

In a five-team tournament, each team plays one game with every other team. Each team has a 50% chance of winning any game it plays. (There are no ties.) Compute the probability that the tournament will produce neither a undefeated team nor a winless team.
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kfoozminus
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Re: Probability In A Five-Team Tournament

Unread post by kfoozminus » Thu Dec 27, 2012 5:43 pm

$\frac{17}{32}$
there will be total $^5C_2=10$ matches and there can be $2^{10}$ results. let's count in how many cases there will be an undefeated team and/or winless team.

there can't be more than one undefeated team or winless team in a single case. so, number of cases in which one team wins all matches is $^5C_1\cdot2^6$. so is for the cases in which one team loses all.

so one team wins all or one team loses all in $^5C_1\cdot2^6+^5C_1\cdot2^6$ cases.

but we counted the cases twice in which there's an undefeated team AND a winless team. number of such cases are $^5P_2\cdot2^3$.

so probability$=\dfrac{2^{10}-(2\cdot^5C_1\cdot2^6-^5P_2\cdot2^3)}{2^{10}}$
Last edited by kfoozminus on Fri Dec 28, 2012 7:56 pm, edited 1 time in total.

Shihab
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Re: Probability In A Five-Team Tournament

Unread post by Shihab » Fri Dec 28, 2012 10:41 am

You made a little mistake. In the third case, (which produce one undefeated team and one winless team) i think you have to permute the two teams. That means there are 5P2 such choices not 5C2.
God has made the integers, all the rest is the work of man.
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kfoozminus
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Re: Probability In A Five-Team Tournament

Unread post by kfoozminus » Fri Dec 28, 2012 7:57 pm

Shihab wrote:You made a little mistake. In the third case, (which produce one undefeated team and one winless team) i think you have to permute the two teams. That means there are 5P2 such choices not 5C2.
yeah, you're right... edited :)

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