Contributing Editor Anirban DasGupta writes on the previous problem, which was about phase transitions:

If the common probability that each observer tells the truth on any given instance is $p$, and if there are $m$ such observers, and if there are $n$ options (colors) to choose from, then by using Bayes’ theorem, the probability that the true color is the universally stated one (purple) given that all $m$ observers said so is

\[ \frac{p^m\, \frac{1}{n}}{p^m\, \frac{1}{n}+(n-1)\,(1-p)^m\,\frac{1}{n (n-1)^m}} \]

\[ = \frac{1}{1+\frac{(\frac{1}{p}-1)^m}{(n-1)^{m-1}}}. \]

If $m = n = 20$, this equals $\frac{1}{n} = 0.05$ if $p = \frac{1}{n} = 0.05$, and it equals $0.00049$ if $p = 0.04$ (just slightly smaller than $p = \frac{1}{n}$).

If $\frac{1}{p} = n-\alpha \log n$, and $m = \gamma \, n$, then the expression reduces to

\[ \frac{1}{1+(n-1-\alpha \, \log n)(1-\frac{\alpha \, \log n}{n-1})^{\gamma \,n-1}} \]

\[ = \frac{1}{1+(n-1)\,n^{-\alpha \, \gamma }\,(1+o(1))}, \]

which converges to $0, \frac{1}{2}$ and $1$ according as $\alpha \,\gamma$ is less than 1, equal to 1, or greater than 1.