I came across the "green-eyed dragons" puzzle (alternatively known as the "blue eyed villagers" puzzle). The typical proof uses a straightforward inductive strategy. I came up with a probabalistic proof, and I wondered if there are holes in it.
The puzzle in its (possibly original) form can be found here:
https://www.physics.harvard.edu/uploads/files/undergrad/probweek/prob2.pdf
and the inductive proof here:
https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol2.pdf.
Theorem: All 100 dragons transform at midnight on the first night.
Proof: It suffices to prove that each dragon independently concludes
$$\Pr[\textrm{exactly 100 dragons have green eyes}] = 1$$
Upon departing, we tell the dragons collectively that at least one of
them has green eyes. Specifically, we impart the knowledge that
$$\Pr[\textrm{at least 1 dragon has green eyes}] = 1$$
Define $L_i$ to be the event that at least $i$ dragons have green eyes, and $E_i$ to be the event that exactly $i$ dragons have green eyes.
Then the above probability can be restated as:
\begin{align*}
\Pr[L_1] = {100 \choose 1} \Pr[E_1] + {100 \choose 2} \Pr[E_2] + \dots + {100 \choose 100} \Pr[E_{100}]
\end{align*}
Because each dragon observes the eye color of every other dragon except himself, each dragon concludes independently that:
\begin{align*}
\Pr[E_0] = \Pr[E_1] = \Pr[E_2] = \dots = \Pr[E_{98}] = 0
\end{align*}
Therefore, each dragon concludes
\begin{align*}
\Pr[L_1] = {100 \choose 99} \Pr[E_{99}] + \Pr[E_{100}] = 1
\end{align*}
Note that
\begin{align*}
\Pr[E_{99}] &= 1-\left(\Pr[E_0] + {100 \choose 1}\Pr[E_1] + \dots + {100 \choose 98}\Pr[E_{98}] + \Pr[E_{100}]\right) \\
&= 1- \Pr[E_{100}]
\end{align*}
Therefore,
\begin{align*}
{100 \choose 99} \Pr[E_{99}] + \Pr[E_{100}] &= 1 \\
100(1-\Pr[E_{100}]) + \Pr[E_{100}] &= 1 \\
100 – 99\Pr[E_{100}] &= 1 \\
\Pr[E_{100}] &= 1
\end{align*}
Thus, each dragon individually concludes that with probability 1, all
100 dragons have green eyes (including himself). This reasoning
happens simultaneously across all dragons, and thus they all transform
together on the first midnight, QED.
Best Answer
This is a problem about knowledge, not about probability. When the traveller leaves, they all know that at least one dragon has green eyes. If there had been only one such dragon, that dragon would know they had green eyes and would transform at midnight.
The next morning all the dragons know that no dragon transformed at midnight. They therefore also know that there are at least two dragons with green eyes. If there were exactly two, they would transform at midnight. It doesn't happen, so the next morning the dragons know that there are at least three dragons with green eyes.
The information that no dragons transform in the night is additional information not present on the first day. A dragon does not know whether or not it has green eyes until ninety nine nights have passed. If it hasn't got green eyes, the others will all transform on the ninety ninth night. If it does have green eyes, they will all transform on the hundredth night.
Since these two possibilities are live, no dragon can conclude on the first night that all dragons have green eyes.