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Four French teams make it to the final eight of the Champions League.

The draw for the round is made at random.

What is the probability that exactly one pairing has two french teams?

... At first, I recognised that there could only either be 0 matching teams, 1 matching team, or two matching teams. If we denote french as $f$ and not french as $nf$, we can see that with 0 matching teams, the order of the teams must be:

  • $f$ v $nf$, $f$ v $nf$, $f$ v $nf$, $f$ v $nf$,

with one matching team:

  • $f$ v$f$, $f$ v $nf$, $f$ v $nf$, $nf$ v $nf$,

and two matching teams:

  • $f$ v $f$, $f$ v $f$, $nf$ v $nf$, $nf$ v $nf$,

I then proceeded to calculate the probability of 0 matching teams and two matching teams occuring, and subtracting them from 1 to calculate the probability for 1 matching team. However, I was not sure what approach to take when calculating this probability.

Thanks, Field

Mr Field
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1 Answers1

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There are $$\frac{8!}{2^4\cdot4!}=105$$ possible ways for the teams to be paired. To see this, imagine lining up the teams in a row and then pairing, the first two teams, the second two teams, and so on. In each pair, it doesn't matter which team comes first, so we must divide by $2^4$. Also, it doesn't matter what order the pairs come in, so we must divide by $4!$.

Now, if there is to be only one match with two French teams, there are $\binom{4}{2}=6$ ways to choose those teams. Then there are $4$ ways to choose which non-French team the third French team plays, and $3$ ways to choose which non-French team the last French team plays. That gives $6\cdot4\cdot3=72$ ways in all. Since all pairings are equally likely, the probability is $$\frac{72}{105}\approx.685714.$$

saulspatz
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