0
$\begingroup$

I have a 8 person tournament.

For the sake of this problem let's say odds of winning are 50% for each player.

What is the formula to figure out the odds of any 2 players meetings at any point in the tournament.

$\endgroup$
1
  • $\begingroup$ I think the answer is 1/7 + 1/14 + 1/28 or 25%, but i don't know the formula $\endgroup$ Oct 16, 2013 at 17:52

4 Answers 4

3
$\begingroup$

There will be $7$ games. There are a total of $\binom{8}{2}$ ways to choose $2$ people. So the probability that $2$ specific people meet is, under our assumption of symmetry, equal to $\frac{7}{\binom{8}{2}}$.

$\endgroup$
2
  • $\begingroup$ So if i had a 64 player tournament chances would be 63/ (32 * 63) or 1/32... Excellent. thanks a lot $\endgroup$ Oct 16, 2013 at 18:05
  • 1
    $\begingroup$ That's right. One can do also a round by round analysis, like you did. That can be handled in general. I am also missing the way to see the answer with no computation at all. Busy for now, but will write it down unless somebody else does. $\endgroup$ Oct 16, 2013 at 18:10
2
$\begingroup$

with random seeding each pair is equally likely to play, but $2^n - 1$ actually do play, one in each game, hence $\frac{2^n - 1}{{2^n} \choose 2}$. For example, with 4 players, $n=2$, B , C,D are equally likely to play A in the first, so B plays him in first woth probability $\frac 13$. With probabilty $\frac 23$ they both have to win their 3rd round to play, so $\frac 13 + \frac 23 \cdot \frac 12 \cdot \frac 12 = \frac 12$.

$\endgroup$
0
0
$\begingroup$

To the author of the first answer: I understand that the probability of A meeting B on the first round is $\frac{1}{7}$, and that A will meet B on the second round is $\frac{6}{7}$ ( that they didnt meet on the first round) $\cdot\frac{1}{2}\cdot\frac{1}{2}$(their respective winning probabiliy) $\cdot\frac{1}{3}$ (that A will meet B this round) $=\frac{1}{14}$, but I dont understand how to get the third term, $\frac{1}{28}$. Can you shed some light?

$\endgroup$
0
$\begingroup$

Work in progress

Chance of the players meeting in the first round:

Since there are 8 players, the chance of meeting is 1/7 (the chance that they are together).

Chance of the players meeting in the second round (given that they haven't met in the first round, which is 6/7):

Since there are 4 players left, the chance of meeting is 6/7 (refer to above) * 1/2 (the chance of person 1 making it through) * 1/2 (the chance of person 2 making it through) * 1/3 (the chance that they are together), which is 1/14.

Chance of the players meeting in the third round (given that they haven't met in the first two rounds, which is 13/14):

Since there are 2 players left, the chance of meeting is 13/14 (refer to above) * 1/2 (the chance of person 1 making it through) * 1/2 (the chance of person 2 making it through) * 1/1 (the chance that they are together), which is 13/56.

1/7 + 1/14 + 13/56 is the answer.

= 2/14 + 1/14 + 13/56

= 3/14 + 13/56

= 12/56 + 13/56

= 25/56

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.