Amongst 16 players playing a knockout tournament find probability only $P_1$ reaches the final given that both $P_1$ and $P_2$ both reach semifinal

Question

$16$ players $P_1,P_2,...,P_{16}$ play in a knockout tournament. Assuming that players are paired at random in each round. If all players are of equal strength then find the probability that $P_1$ reaches the final under the condition that $P_2$ reaches the semi final but not in final.

My Attempt $P_1$ and $P_2$ both reach the semifinal so they are two among four persons who reached semi-final whose probability is $$\frac{\binom{14}{2}}{\binom{16}{4}}$$. Now in finals $P_1$ must be there but $P_2$ should not reach finals so $P_1$ is among the two persons reaching the final and there is another player(other than $P_2$) who reaches the final whose probability is $$\frac{\binom{2}{1}}{\binom{4}{2}}$$.

So the required probability =$$\frac{\binom{14}{2}}{\binom{16}{4}}\times \frac{\binom{2}{1}}{\binom{4}{2}}=\frac{1}{60}$$

Is my answer correct

Answer 1

THIS ANSWERS THE QUESTION IN THE OP'S QUESTION STATEMENT

(NOT THE QUESTION IN THE TITLE - for which the answer is obviously $\frac{1}{3}$)

Assume that $P_2$ is a losing semi-finalist. Then there are 15 people who are equally likely to be finalists.

The conditional probability that $P_1$ is a finalist is therefore $\frac{2}{15}$.

NOTE

It is irrelevant what $P_2$ does providing they are not a finalist.

Answer 2

• We want the probability that $P_1$ reaches the finals and $P_2$ loses in the semi-finals
• The crux of the matter is that since players are of equal ability, each has a $Pr$ of $\frac12$ of winning or losing agalnst any opponent in any match, it is only their position in the table that can produce the outcome we desire
• Without loss of generality, we place $P_1$ at the top of the table (i.e. $1$)
• $P_2$ then has to be in position $5$ downwards ($12$ among $15$ left), with $Pr=\frac{12}{15} = \frac 45$
• Whether a person loses or wins the semi-finals, her probability is simply $\frac12\frac12\frac12 = \frac18$, and so this applies to both $P_1$ and $P_2$
• Putting the pieces together, final ans $= \frac45\frac18\frac18 = \frac1{80}$

Note:

AS there is a confusion about what the question is asking, I have tajen the interpretation that both $P_1$ and $P_2$ reach the semifinal and $P_1$ reaches the final.

Answer 3

You wrote (cit.) "given that both P1 and P2... reach semifinal", which implies we have four semifinalists, with $P_1$ and $P_2$ among them.

There is a $(\frac{2}{3})$ chance that $P_1$ and $P_2$ do not meet in the semi, times $(\frac{1}{2})^2$ that $P_1$ advances, but not $P_2$,
and there is a $(\frac{1}{3})$ chance that $P_1$ and $P_2$ do meet in the semi, times $(\frac{1}{2})$ that $P_1$ advances, but not $P_2$.


This gives $$p = \frac{2}{12}+\frac{1}{6}=\frac{1}{3}$$

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You could however have meant the overall probability of the desired outcome that $P_1$ makes the final and $P_2$ loses his semi, evaluated before the tournament starts.

There is a $(\frac{14}{15})$ chance that $P_1$ and $P_2$ do not meet in R16, times $(\frac{1}{2})^2$ that both advance.
Then, there is a $(\frac{6}{7})$ chance that $P_1$ and $P_2$ do not meet in R8, times $(\frac{1}{2})^2$ that both advance.

So far, we have $p_{SF}=\frac{14\times 6}{16\times 15\times 7}$.

From here, there is a $(\frac{2}{3})$ chance that $P_1$ and $P_2$ do not meet in the semi, times $(\frac{1}{2})^2$ that $P_1$ advances, but not $P_2$,
and there is a $(\frac{1}{3})$ chance that $P_1$ and $P_2$ do meet in the semi, times $(\frac{1}{2})$ that $P_1$ advances, but not $P_2$.

This gives $$p=\frac{14\times 6\times 2}{16 \times 15\times 7 \times 6}=\frac{168}{10080}= 0.01\bar6$$.

Answer 4

I'm going to answer the question in the title.

If we assume that $P_1$ and $P_2$ both reach the semifinals, then since each player is equally likely to win any match and each player is equally likely to face any other surviving player in the semifinals, by symmetry each pair of semifinalists is equally likely to reach the finals.

There are $6$ possible pairs of semifinalists. Of those, $2$ include $P_1$ but don't include $P_2$. Therefore, given that both $P_1$ and $P_2$ are semifinalists, the probability that $P_1$ reaches the finals but $P_2$ does not is $\frac 13$.