For small N, things are more interesting

]]>Say there are n+1 judges and n is even. Otherwise just add some floor and ceiling functions.

It only helps if the judge convicts if the other ones are exactly tied, with probability p^{n/2} (1-p)^{n/2}times n choose n/2 . It only helps if the judge acquits if the other ones are exactly unanimous, with probability p^{n}. It’s sufficient to show that the ratio of these is >1 for p<x and x. The ratio is just ((1-p)/p )^{n/2} times n choose n /2, so it clearly satisfies this where x is the ( n choose n/2 )^{2/n}/ ( 1+ ( n choose n/2 )^{2/n} ) n choose n/2 is 2^n divided by a function that grows slower than exponential, so its nth root is 2 – o(1), so the asymptotic is 2^2/ (1 + 2^2) =4/5.

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