7
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N balls with numbers 1,2,3...N are filled arbitrarily into 3 urns A,B and C - but no urn may remain empty. We write P(A,B) for the probability that a ball chosen randomly from urn A shows a higher number than a ball chosen randomly from urn B. We call the tree urns intransitive if P(A,B)>1/2, P(B,C)>1/2 and P(C,A)>1/2.

  1. Give the smallest possible N and the corresponding contents of the urns such that the three urns are intransitive.
  2. Give the smallest possible N and the corresponding contents of the urns such that additionally P(A,B)=P(B,C)=P(C,A).
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1 Answer 1

3
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N=7 (1,5,6,),(4),(2,3,7)

N=9 (2,6,7),(1,5,9),(3,4,8) P=5/9

or

(1,6,8),(3,5,7),(2,4,9)

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2
  • $\begingroup$ Correct! Is there an easy argument to see that the solution for question 1 is unique? I have none. Question 2 has 2 solutions btw. $\endgroup$ Jun 7 at 11:59
  • $\begingroup$ Nope, I have nothing elegant. $\endgroup$
    – loopy walt
    Jun 7 at 12:03

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