3
$\begingroup$

Suppose that we have a tennis tournament with 32 players. Players are matched in a completely random fashion, and we assume that each player always has probability 1/2 to win a match. What is the probability that two given players meet each other during the tournament.

  • 0
    I am trying to teach myself probability. Just to clarify, not exactly a student looking for homework solutions - but still a student2012-01-23

3 Answers 3

2

I get $ \frac{1}{31} +\frac{30}{31}\frac{1}{4}\left(\frac{1}{15} +\frac{14}{15}\frac{1}{4}\left(\frac{1}{7} +\frac{6}{7}\frac{1}{4}\left(\frac{1}{3} +\frac{2}{3}\frac{1}{4}\right)\right)\right) $ which equals $\dfrac{1}{16}=6.25\%$. Not sure if this is the most elegant solution, though.

  • 0
    See my later answer for a more general and less arithmetic solution.2012-01-23
4

Hint 1: Consider the number of other players a particular player meets with what probability: one other with probability $1/2$; two others with probability $1/4$; three others with probability $1/8$; etc.

Hint 2: What is the expected number of other players a particular player meets?

Hint 3: How does Hint 2 relate to the original question?

Answer: $ \dfrac{1 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2^2} + 3 \times \dfrac{1}{2^3} + 4 \times \dfrac{1}{2^4} + 5 \times \dfrac{1}{2^4}}{31} = \dfrac{1}{16}$

  • 0
    OK. Try it your way, but you should be adding 1 each time: $\dfrac{1 \times \dfrac{1}{1} + 1 \times \dfrac{1}{2} + 1 \times \dfrac{1}{2^2} + 1 \times \dfrac{1}{2^3} + 1 \times \dfrac{1}{2^4}}{31} = \dfrac{1}{16}$2012-01-24
4

Easy general answer for $n$ players in a knockout tournament (and here $n=32$):

There are $\dfrac{n(n-1)}{2}$ potential pairs for matches.

To have one winner, $n-1$ players must be knocked out, so there are $n-1$ actual matches.

So the probability that a particular pair actually have a match is $\dfrac{n-1}{{n(n-1)/2}} = \dfrac{2}{n}$.

  • 0
    yes it seems to be most elegant one.2012-01-24