perplexus.info :: Probability : A world cup problem

A world cup problem (Posted on 2014-08-26)

Suppose there are 4 superior teams among the 32 in soccer's World Cup. If the 32 are placed randomly into eight groups of 4, what is the probability that the superior teams will be in different groups?

No Solution Yet Submitted by Danish Ahmed Khan

Rating: 1.5000 (2 votes)

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A little more precision (spoiler)

| Comment 2 of 5 |

The probability that the 2nd team is in a different group than the 1st is 28/31 (because there are 31 available spaces, of which 28 are on a different group).

The probability that the 3rd team is in a different group than the 1st two is 24/30.

The probability that the 4th team is in a different group than the 1st three is 20/29.

Overall probability is 20*24*28/(29*30*31) = 448/899, which is approx .498, or even more approximately 1 in 2

Posted by Steve Herman on 2014-08-26 13:21:34

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