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$16$ players $P_1, P_2, ..... P_{16}$ take part in a tennis knockout tournament. The order of the matches is chosen in random. Lower suffix player is better than higher suffix, the better wins. What is the probability that the eighth best reaches the semifinals?

Basically I think that we just have to keep selecting only those players who are below $P_8$. For the first round the probability of choosing a player below $P_8$ would be $\frac{8}{15}$. But then I find it difficult to arrange the players who got selected in the next round.

Sourav Suman
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2 Answers2

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In order for a particular player to reach the semifinals, where there are $4$ players, they must be in a group of $\frac{16}{4}=4$ players wherein they are the best: our chosen player will play in a first-round match, then in a second-round game against the winner of another first-round match.

So the question is, how many ways are there to pick four players such that the best player is $P_8$? We must then compare this to the number of ways to pick four players where $P_8$ is included but not necessarily best.

Obviously, one of the four must be $P_8$; this leaves $3$ players to choose from. There are $8$ players worse than $P_8$, so we have $\binom{8}{3} = 56$ ways to choose the remaining three so that $P_8$ is best.

Similarly, there are $15$ players other than $P_8$, so there are $\binom{15}{3} = 455$ ways for $P_8$ to contend for a semifinal slot at all.

So the final probability is $\frac{56}{455} = \frac{8}{65}$ that $P_8$ will reach the semifinals.

Dan Uznanski
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  • Can we calculate the number of ways in which $P_8$ reaches semi-finals – Maverick Apr 10 '17 at 00:33
  • That's $56 \times 3$ - you need the $3$ to determine who $P_8$ is going to play against in the first round. For instance, one of the $56$ is that in $P_8$'s section of the bracket, there are $P_{10}$, $P_{13}$, and $P_{14}$; $P_8$ could face any one of them in the first round, and the other two will face each other. Note that this doesn't do anything at all with the rest of the bracket. If you want to know those, I think it's $12!/2^{12}$ but that's just me fiddling with it in my head – Dan Uznanski Apr 10 '17 at 00:37
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Thinking about it round-by-round is not the easiest way to think about the problem. Instead, think about it this way: the semifinals are between the four players who were best in their quarter of the bracket. What's the probability that $P_8$'s quarter of the bracket doesn't have any of the players better than $P_8$ in it?

There are $\binom{15}{3}$ ways to choose who else is in the same quarter of the bracket as $P_8$, and $\binom{8}{3}$ ways to choose only players from $\{P_9, P_{10}, \dots, P_{16}\}$. So the probability is $\binom{8}{3}/\binom{15}{3} = \frac{8\cdot 7\cdot 6}{15\cdot 14 \cdot 13} = \frac{8}{65}.$

Misha Lavrov
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