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I'm studying Probability and can't solve this question:

Random tournaments: consider a sport tournament in which n teams participate. Each team plays one match against each other team. Each such match ends with one of the two teams winning (there are no draws). Assume that all teams are equally strong, so each match has a random winner. We say that a team is the clear winner of the tournament if it won all its n − 1 matches (this team beats all other teams). What is the probability that the tournament ends without a clear winner for n = 10?

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closed as off-topic by Did, Claude Leibovici, kingW3, hardmath, zhoraster Feb 18 '17 at 18:36

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    $\begingroup$ Every possible outcome is equally likely, so we just need to count the number of tournaments with a clear winner and the total number of tournaments. $\endgroup$ – usul Feb 6 '17 at 9:26
  • $\begingroup$ This might be difficult for you because of difficulty in understanding the problem statement or for other reasons connected to how you approached the problem. The lack of context (how do you understand the problem? how did you approach it?) leaves Readers without much basis to help with your learning. Some will get an impression this is a "pass through" of an assignment, without your digesting the exercise. If nothing else, please try to work a simplified version of the problem. $\endgroup$ – hardmath Feb 18 '17 at 15:50
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If you tackle this problem in a slightly different way, it becomes quite intuitive.
To find the probability of not having a clear winner, we'll look at what's the probability of having a clear winner and then subtract that from 1.
So if team 1 wins all its 9 matches, it'll be a clear winner. The probability of doing that is $\bigr{(}\frac{1}{2}\bigr{)}^9$.
And any one of 10 teams can be a clear winner, so we multiply it by 10. Thus, the probability of having a clear winner is $10\bigr{(}\frac{1}{2}\bigr{)}^9$.
Substract that from 1 and you have your answer, which is $1-10\bigr{(}\frac{1}{2}\bigr{)}^9$
I hope I've made myself clear.

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  • $\begingroup$ Thanks a lot, now It became clear. $\endgroup$ – Alex Ermolaev Feb 6 '17 at 9:36
  • $\begingroup$ No, because they are not independent events (the teams not winning all games). For example, the first team not being a clear winner means it had to lose at least one game, which alters the probabilities for the other teams in their game against the first team. $\endgroup$ – Tony Mulieri Feb 6 '17 at 20:28
  • $\begingroup$ Can you explain a bit more? Isn't probability of losing atleast one game means 1-P(winning all games) $\endgroup$ – Sum-Meister Feb 6 '17 at 20:30
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Each team plays $9$ matches, with $Pr = \dfrac12$ of winning each match,

thus P(Team A is a clear winner) $= \dfrac1{2^9}$

Only one team can be a clear winner, and each team has equal chances,

thus P(There is a clear winner) $=\dfrac{10}{2^9}$

and P( no clear winner) = $1 - \dfrac{10}{2^9}$

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