# Odds of 2 players meetings in a 8 person single elimination tournament

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.

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

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 $\frac{7}{\binom{8}{2}}$.

• So if i had a 64 player tournament chances would be 63/ (32 * 63) or 1/32... Excellent. thanks a lot Oct 16, 2013 at 18:05
• 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. Oct 16, 2013 at 18:10

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 $\frac 13 + \frac 23 \cdot \frac 12 \cdot \frac 12 = \frac 12$.

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?

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