Annette, Babette, Colette playing in a $16$ person single elimination tournament

Question

Annette, Babette, Colette, and $13$ other girls are playing in a $16$-player, single-elimination tennis tournament. The $16$ players are placed at random in the first column of the bracket shown in the figure to play $8$ games in Round $1$. The winners of very match are then written into the bracket and play in Round $2$, and so on. The tournament proceeds until only one player remains. Given that Annette is the best player of the $16$ in the tournament, Babette is $2$nd best, and Colette is $3$rd best, and that the best player always wins each match, find the probability that: (a) Annette wins the tournament. (b) Babette is the runner-up of the tournament (she gets to the finals, but loses). (c) Colette is the runner-up of the tournament.

For (a), I ended up getting $1$ since Annette defeats everyone in her way.

For (b), I ended up getting ${8\over{15}}$ since if Annette's position is fixed on one half of the brackets in Round $1$, then there's 15 positions remaining and Babette has to occupy one of the $8$ positions in the other half of the brackets in Round $1$, hence ${8\over{15}}$.

For (c), we need Annette and Babette to be on one half of the brackets in Round $1$, and Colette to be on the other half. Fix Annette's position, then there's a ${7\over{15}}$ chance of Babette being on Annette's half, and then a ${8\over{14}}$ chance of Colette being on the other half, so the probability that Colette is the runner-up of the tournament is ${7\over{15}}{8\over{14}} = {4\over{15}}$.

Is this correct?

Answer 1

Well, It's been a while and you figured out the answer, nevertheless, I want to give a constructive way of finding the probability for newcomers.

In round $1$ there will be $8$ matches between $16$ girls and $8$ winners. In round $2$, $4$ matches between $8$ girls and $4$ winners, in round $3$, $2$ matches between $4$ girls and $2$ winners.Lastly in the final round $1$ match and $1$ winner (of course Annette).In every round, a person can play with anyone but with oneself, e.g. in the $2nd$ round, Annette can play with $7$ girls.

$(b)$ For Babette to be runner-up, she has to win in round $1,2$ and $3$.She will lose in the final round anyway.

In round $1$, she can play with $15$ girls but she cant play with Annette because she will lose.So $\text{P(of winning in round 1)}$ = $\frac{14}{15}$.

Similarly, in round $2$, she can play with $7$ girls but she can't play with Annette because she will lose. So $\text{P(of winning in round 2)}$ = $\frac{6}{7}$.

Similarly, $\text{P(of winning in round 3)}$ = $\frac{2}{3}$.

$\text{P(Babette is the runner-up of the tournament)}$ = $\frac{14}{15}\times\frac{6}{7}\times\frac{2}{3}$ = $\frac{8}{15}$.

$(c)$ Collette will win if the following cases are true.

$\fbox{Case 1}$: Collette wins round $1$ and Babette wins round $1$ AND Collette wins round $2$ and Babette wins round $2$ AND Collette wins round $3$ and Babette loses round $3$. Babette can't win round $3$ because Collette has to be the runner-up.

If Colette needs to win the $1st$ round, she has to play with $13$ girls among $15$ girls, excluding Annette and Babette. So her probability of winning the $1st$ round is always $\frac{13}{15}$. Now when Collette wins the $1st$ round there are $14$ girls left $($Colette won, she won't play anymore and the girl who lost with her, was eliminated$)$, and for Babette to win the $1st$ round, she has to play with $12$ girls among $13$ girls, excluding Annette. So Babette's winning probability is $\frac{12}{13}$.

Similarly, Colette will win in the $2nd$ round if she plays with any of the $5$ girls among $7$ girls $($excluding Annette and Babette$)$ with a probability $\frac{5}{7}$. After Collette wins, $6$ girls left, and Babette will win if she plays with any $4$ of the girls, excluding Annette with a probability $\frac{4}{5}$.

Collette will win for the $3rd$ round if she plays with only $1$ girl among $3$ girls, so her winning probability is $\frac{1}{3}$. After Collette wins, $2$ girls left, Annette and Babette. So Babette's losing probability is obviously $1$.

So probability of Case $1$ is :$\frac{13}{15}\times\frac{12}{13}\times\frac{5}{7}\times\frac{4}{5}\times\frac{1}{3}\times1$ = $\frac{16}{105}$

$\fbox{Case 2}$: Collette wins round $1$ and Babette wins round $1$ AND Collette wins round $2$ and Babette loses round $2$ AND Collette wins round $3$.

This is very similar to Case $1$. In round $2$, after Collette wins, $6$ girls left, and Babette will lose if she plays with Annette with a probability $\frac{1}{5}$.

In the $3rd$ round, Collette's winning probability is of course $\frac{2}{3}$ because Babette is already lost in round $2$ and she can play with any of the $2$ girls $($but not with Annette$)$ and can win.

So probability of Case $2$ is : $\frac{13}{15}\times\frac{12}{13}\times\frac{5}{7}\times\frac{1}{5}\times\frac{2}{3}$ = $\frac{8}{105}$

$\fbox{Case 3}$: Collette wins round $1$ and Babette loses round $1$ AND Collette wins round $2$ AND Collette wins round $3$.

Colette wins round $1$ with a probability $\frac{13}{15}$ . Babette loses round $1$ with a probability $\frac{1}{13}$ because after Collette wins, $14$ girls left and Babette will lose if she plays with Annette. Consequently, Collete will win round $2$ with a probability $\frac{6}{7}$ $($she can play with every girl excluding Annette$)$ and round $3$ with a probability $\frac{2}{3}$. So probability for Case $3$ : $\frac{13}{15}\times\frac{1}{13}\times\frac{6}{7}\times\frac{2}{3}$ = $\frac{4}{105}$

$3$ cases are mutually exclusive, So $\text{P(Colette is the runner-up of the tournament)}$ = $\frac{16}{105} + \frac{8}{105}+\frac{4}{105} = \frac{4}{15}$