Hi.

I'm struggling to get my head round this issue I'm discussing with friends.

It regards the chances of any of the top 4 teams from last season in a league playing each other on the same day.

So, assuming 20 teams (they play twice, home and away) who each play on the same day once a week for 38 weeks, what is the chance that the teams who finished in positions 1-4 last season, will play each other on the same day at any point in the season given a random fixture draw.

I've been told 17% but that seems too low to me.

Can anyone explain how this would be calculated?

Thanks.

2. Originally Posted by McGraw
Hi.

I'm struggling to get my head round this issue I'm discussing with friends.

It regards the chances of any of the top 4 teams from last season in a league playing each other on the same day.

So, assuming 20 teams (they play twice, home and away) who each play on the same day once a week for 38 weeks, what is the chance that the teams who finished in positions 1-4 last season, will play each other on the same day at any point in the season given a random fixture draw.

I've been told 17% but that seems too low to me.

Can anyone explain how this would be calculated?

Thanks.
Hi McGraw,

Let's say the 4 teams are A, B, C, and D, and let's assume the matches are chosen by random draw each week. On any given week, then, the probability that A will play one of B, C, or D is 3/19, because there are 19 other teams and they are all equally likely to be chosen. Let's say, for example, that A plays B. Then the probability that C will play D is 1/17 by similar reasoning. So the probability that the 4 teams will play each other on any given week is (3/19) * (1/17) = 3/323.

Since there are 38 weeks, the probability that the 4 teams will NOT play each other at all is $(1 - 3/323)^{38}$, and the probability that they will play each other at least once in 38 weeks is
$1 - (1 - 3/323)^{38} = 0.2985$.

3. Thanks very much.