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If you have a random draw for the semi-finals of a competition (say football), with teams A, B, C & D (equal probability that each is drawn), whereby each team is drawn from a hat one after another - what are the odds that team A will play team D?

Any help would be much appreciated.

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Do you mean the odds that team A will play team D in the semi-finals, or that team A will play team D in the rest of the competition? For the latter, you'd also need the probabilities for the teams to beat each other. – joriki Feb 25 '13 at 11:48
Just that A will play D in the semi-final. – Sandy Feb 25 '13 at 11:54

One possibility is to list all the possible cases:

  1. A against B and C against D
  2. A against C and B against D
  3. ....

and so on. There are exactly 24 cases. You can then simply count (all cases are equi-probable).

Another option is to use symmetry arguments: the probability that A will play with D is the same as the probability that A will play with B. The same goes for C. Since it will play with one of them, the sum of all these should be 1. See where this is headed?

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My instinct said 1/3, for obvious reasons, but then wasn't sure if there is a really simple formula. Can you use conditional probabilities in this case i.e. pr(A|D) + pr(D|A) or will it not work because of the two possible semi final places. – Sandy Feb 25 '13 at 12:01
The answer is indeed 1/3 as you said. What does $\operatorname{pr}(A|D)$ stand for? – yohBS Feb 25 '13 at 15:36

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