This is why I don't major in math
Solution 1:
Suppose that the first team from the first pair is weak. (odd: 0.5) If this happens, the other team MUST be strong, or a derby will occur in another pair (if the other team is weak, then 4 strong teams will have to be allocated in 3 pairs, which means that at least 1 pair will have 2 strong teams - pigeonhole principle). The odds for this is 4/7. Now, if the first team is strong (odd: 0.5) the other team MUST be weak (or we have a derby), in an odd of 4/7.
If the 4/7 odd happens, 3 strong and 3 weak teams remain for 3 pairs.
Doing this all the time, the total odds are:
(4/7) * (3/5) * (2/3) = 0.228
Solution 2:
We have to arrange 8 teams, in 4 pairs of 2, in which no derby occurs.
The total number of ways to arrange the 8 teams is 8 factorial, or 8! (n! = 1*2*...*n).
Since every strong team has to go in a pair among 1 and 4, and they can be either the first team of the pair or the second one, we have 4! (the ways that the 4 strong teams can be between the 4 pairs), multiplied by 4! (the ways the 4 weak teams can be between the 4 pairs), multiplied by 2^4 (the ways that every strong team can be either first or second in a pair).
In total, this gives
(2^4)*(4!)*(4!)/(8!) = 0.228
The correct answer is indeed 576/2520. Boo has got it right.
There will be given two complete solutions for the problem tomorrow, around 1PM - 2PM eastern american time.
deletedabout 7 years
am i misinterpreting the question? its asking the probability that two strong teams do not get paired together right
