# What is the probability that the 3rd best club will appear in the final stage?

16 clubs are divided into 4 groups with the winner of each group entering the final stage of the competition.

What is the probability that the 3rd best club will appear in the final stage?

All the results of matches reflect the true abilities of the clubs involved.

There was another question that said: what is the probability that the second best club will not appear in the final?

The answer to that question is $\frac{3}{15}$ because of the other players in his group, if the best player is one of those 3, he will not make it to the final.

Please can you help me devise a strategy for this question.

If "the third best club" means that it'll win all the other clubs except for the "best club" and "second best" club, then you just have to compute the chance that none of those clubs are in the same group:

$$\frac{13·12·11}{15·14·13} = \frac{22}{35}$$

This comes from the following reasoning:

In the group where the third best club is, we need to fill the remaining three clubs in that group. Out of a pool of $15$ remaining clubs ($16-1$ because the third best is already selected) we need to choose one of the $13$ clubs that are not one of the two best.

Once we've done that, we have $14$ remaining clubs, and we have to select one out of the $12$ that are not the two best ones.

Finally, there are $13$ clubs and we have to select one out of the $11$ that are not the two best clubs.

• It's the right answer but please explain how you worked out the chance that none of the clubs are in the same group? – StephanCasey Aug 12 '14 at 9:52
• @StephanCasey added the explanation. Do you understand it now? – Darth Geek Aug 12 '14 at 9:59
• Oh yeah. I see it now. It's like you placed the three best in separate groups and then place the rest :) thank you – StephanCasey Aug 12 '14 at 10:04

The $3$rd best will appear in the final stage iff the group where it belongs to does not contain the best and does not contain the second best.

There are $16$ spots. Here $1,2,3,4$ form a group, $5,6,7,8$ form a group, et cetera. Start by placing the $3$rd best on spot number $1$. Now place the best on one of the remaining spots. The probability that it will get a spot with number $>4$ is $\frac{12}{15}$. If this occurs, and the second gets a spot now of the $14$ left then there a probability of $\frac{11}{14}$ that it will get a spot with number $>4$.

This together gives us $\frac{12}{15}\frac{11}{14}=\frac{22}{35}$ that the third best will be the best of group $1,2,3,4$.

Alternative: $$\binom{3}{0}\binom{12}{2}\binom{15}{2}^{-1}=\binom{12}{2}\binom{15}{2}^{-1}=\frac{22}{35}$$ If a vase contains $3$ blue balls and $12$ red balls then what is the probability that two balls taken out (without replacement) are both red?

The blue balls stand for the spots $2,3,4$ and the red balls stand for the spots $5,\dots,16$.