So if the boffin does not vote last, the reliability of the jury will be $p_{c}$, which is close to $3/4$. On the other hand, if the boffin votes last, the verdict may already be decided before he comes to vote, if the two yokels vote alike, in which case the verdict will be right with probability $1/2$, as the yokels have no useful information. The yokels will vote alike with some probability $q$, which is at least $1/2$, since they vote alike when they get signals of the same sign, which occurs with probability close to $1/2$. Hence the reliability when the boffin votes last is approximates $$q\left(  \frac{1}{2}\right)  +\left(  1-q\right)  p_{c}<\left(  \frac{1}{2}\right)  \left(  \frac{1}{2}\right)  +\left(  \frac{1}{2}\right)
p_{c}<p_{c},$$ as $p_{c} \approx 3/4$. Hence the boffin cannot vote last in an optimal voting order.

If the boffin votes first, the reliability is exactly the probability $p_{c}$ that he gets it right. But suppose the boffin’s signal is very very close to $0$, so he still views $A$ and $B$ as equiprobable. In this case he would prefer to vote second so that he could follow the vote of the first voter, of ability $x\in\left\{  a,b\right\}  $, and make the correct 