Probability in knock-out torunament

[chrisyuen]

Homework Statement

Albert, Bobby and six other players take part in a table-tennis knock-out tournament. The winner of each match can proceed to the next round as shown in the following figure and the loser is knocked out. The players are randomly assigned to the eight positions in the first round. Suppose the eight players are equally skillful.

(a) What is the probability that Albert will play Bobby in the first round?

(b) What is the probability that Albert will ever play Bobby in a match during a tournament?

(Answers:
(a) $$\frac{1}{7}$$
(b) $$\frac{1}{4}$$)

Homework Equations

Permutation and Combination Formulae

The Attempt at a Solution

Part (a), total combinations = $$\frac{8!}{6!}$$ = 56

P = $$\frac{8}{56}$$ = $$\frac{1}{7}$$.

Am I correct?

Part (b), I don't know how can I start this part.

Can anyone tell me how to solve it?

Thank you very much!

 

[mighty2000]

Yes, you are correct for part (a). For part (b), you can approach it by considering the different possible scenarios where Albert and Bobby could play each other.

One scenario is that they could play each other in the first round, which we already calculated the probability for in part (a).

Another scenario is that they could play each other in the second round. In order for this to happen, both of them would have to win their first round matches. Since the players are equally skillful, the probability of each of them winning their first round match is \frac{1}{2}. Therefore, the probability of both of them winning and playing each other in the second round is \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.

Similarly, they could also play each other in the third round if they both win their second round matches. The probability of this happening is again \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.

Therefore, the total probability of them ever playing each other in a match during the tournament is \frac{1}{7} + \frac{1}{4} + \frac{1}{4} = \frac{3}{7}.