What is the probability of 4th player to make it to the final?

Question

There are 8 players in a tennis tournament. At the beginning of the tournament, the players are matched at random. For each subsequent round, the losers are eliminated and the remaining players are matched at random. What is the probability that the 4th make it to the finals?

The 4th one only reach the final when it is not allowed to play with first 3 players and the two second best players get eliminated in the second round, and it will only happen when in the first round the three best players play with each other so the number of favourable cases are $$\frac{4C1}{7C1} \cdot \frac{3C2}{3C2+3C1 \cdot 3C1} \cdot \frac{1}{4C2}$$

Answer 1

The only way possible is that 4th player plays with P5 or P6or P7 or P8 and with 1 of them in the next round So select any one out of these 4 (4C1 ) Now from these remaining 3 select 2 that will play together and winner out these will play P4 in next round (3C2) The possible number of remaining matches played in this round by P1,P2,P3 and one remaining from P5,P6,P7,P8 is 3 .so total favourable matches in round 1 =4C1•3C2•3 Total matches in this round(8!/((2!)^4•4!)) In the next next round P4 has to play the winner of the match between any 2 of P5,P6,P7 ,P8 which we decided earlier in 3C2 ways .so there is one favourable case as P4 cannot play any from first 3 players . Total matches in this round 3 Probability= (4C1•3C2•3 )•(1)/((total matches in round 1)(total matches in round2) )=4/35

Answer 2

Let's begin by counting the number of possible ways the players could be paired in the first round. The top-ranked player could be paired with any of the seven lower-ranked players. That leaves six players. The top-ranked player among them could be paired with any of the five lower-ranked players left. That leaves four players. The top-ranked player among them could be paired with any of the three lower-ranked players left. The final two players must play each other. Therefore, there are $$7 \cdot 5 \cdot 3 \cdot 1$$ possible ways to match the players in the first round.

To make it to the finals, the fourth-ranked player must be paired with one of the four lower-ranked players in the first round, then be paired with another of those players in the second round. For that to happen, two of the other three players ranked beneath the fourth-ranked player must play each other in the first round.

With that in mind, we count the favorable cases in the first round. The fourth-ranked player must be paired with one of the four lower-ranked players. Two of the other three lower-ranked player must play each other. That leaves four players, including the top-ranked player. The top-ranked player could be paired with any of three remaining lower-ranked players. The remaining two players must play each other. Hence, the number of favorable ways to match the players in the first round is $$4 \cdot \binom{3}{2} \cdot 3 \cdot 1$$

Hence, the probability of a favorable outcome in the first round for the fourth-ranked player is $$\frac{4 \cdot \binom{3}{2} \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1}$$

Assuming that this favorable scenario occurs, there are four players left, including the fourth-ranked player, two higher-ranked players, and one lower-ranked player. To reach the final, the fourth-ranked player must draw the lower-ranked player, which occurs with probability $1/3$.

Hence, the probability that the fourth-ranked player reaches the final is $$\frac{4 \cdot \binom{3}{2} \cdot 3 \cdot 1}{7 \cdot 5 \cdot 3 \cdot 1} \cdot \frac{1}{3} = \frac{4}{35}$$

Why is your answer incorrect?

If I understand what you are doing correctly, the first term represents the probability that the fourth-ranked player plays a lower-ranked player in the first round, the second term represents the probability that two of the other three lower-ranked players play each other in the first round, and the third term represents the probability that the fourth-ranked player plays the winner of that match in the second round.

Your first term is correct.

For the second term, There are $5 \cdot 3 \cdot 1$ ways to pair the players who have not already been assigned. There are $\binom{3}{2}$ ways to select which two of the three remaining lower-ranked players play each other and three ways to select which of the higher-ranked players the other lower-ranked player will draw. That gives $$\frac{\binom{3}{2} \cdot 3}{5 \cdot 3 \cdot 1}$$ However, it looks like you wanted to add the case in which no two of the three remaining lower-ranked players play each other to the case in which that does occur. If no two of the three remaining lower-ranked players play each other, each must play one of the three highest-ranked which can occur in $3!$ ways. That gives $$\binom{3}{2} \cdot 3 + 3! = 5 \cdot 3 \cdot 1$$

In the second round, four players are left, including the top player. The top player can be matched with any of the three remaining lower-ranked players. The two remaining players must play each other. Hence, there are $$3 \cdot 1$$ ways to pair the players in the second round. If those players include the fourth-ranked player, two higher-ranked players, and one lower-ranked player (the favorable case for the fourth-ranked player), the probability that the fourth-ranked player draws the lower-ranked player is $$\frac{1}{3 \cdot 1}$$

Answer 3

It is not clearly stated, but I take it that the assumption is that a higher ranked player will always defeat a lower ranked player

Since player ranked $4$ has to clear two rounds to reach the final, it should be obvious that $3$ lower ranked players must in effect be in the same half despite random matchings later. Place "our" player anywhere, and place $3$ lower ranked players in the same half,

Then we can solve very simply as $Pr=\frac 4 7\cdot\frac 3 6\cdot\frac2 5=\frac 4{35}$