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.