Tigers and a sheep

We determine the outcome by backward induction on the number of tigers, $n$. The key observation is that a tiger will eat the sheep only if doing so guarantees its own survival after it turns into a sheep.

• $n=1$: The lone tiger eats the sheep, becomes a sheep, and faces no remaining tigers. It survives, so the sheep is eaten.
• $n=2$: If a tiger eats the sheep, it becomes a sheep, leaving one tiger and one sheep. By the $n=1$ case, that remaining tiger will eat the new sheep, so the first tiger would die. Since tigers value survival, neither will eat. The sheep survives.
• $n=3$: If a tiger eats, it becomes a sheep, leaving two tigers and one sheep. From $n=2$, the two tigers will not eat the sheep (they would die), so the eating tiger survives. Hence some tiger will eat, and the sheep is eaten.
• $n=4$: Eating leads to the $n=3$ case, where the sheep is eaten, so the eating tiger would die. No tiger eats; the sheep survives.

Continuing this logic, the sheep is eaten if and only if $n$ is odd. For $n=100$ (even), no tiger will eat the sheep.