Two judges: how many pairs do they order the same way?

Question

Two judges each independently rank the same 7 contestants in a completely random order. For each pair of contestants, the judges either agree on which of the two is better or disagree.

On average, on how many of the pairs do the two judges agree?

Accepted Answer

Put one indicator on each pair of contestants and add. There are \binom{7}{2} = 21 pairs. Fix one pair. Each judge, independently and at random, ranks that pair one way or the other. The two judges agree exactly when their two independent coin-flip-like choices match, which happens with probability 	frac12: P(	ext{judges agree on a given pair}) = \frac12. Summing over all 21 pairs, E[	ext{agreeing pairs}] = \binom{7}{2} 	imes \frac12 = 21 	imes \frac12 = \frac{21}{2} = 10.5. Takeaway. Comparing two independent random rankings, each pair is ordered the same way with probability 	frac12, …

Two judges: how many pairs do they order the same way?

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