[The Saucepan Man]

Originally Posted by Morsul the Dark

I mean we have to kill three(or is it four?)wolves to eliminate their nightly kill

Good point. I assume that the friendly Wolf counts in the Wolves’ numbers for the purposes of determining victory, just as the Cobbler counts for the Innocents. But, he is, of course, playing for us, so it would be somewhat ungracious to kill him, if we can possibly avoid it.

Originally Posted by A Little Green

But actually, I think we shouldn't concentrate solely on the Bear, and yet we shouldn't forget him either.

Quite so. Killing the Bear early would be good from the point of view of limiting Nightly kills, but the Bear will be much more difficult to spot. So, we should not lose sight of either enemy.

I wonder whether it might be worth pondering the dynamic between the Wolves and the Bear further. From the point of view of both, double kills at Night is good because, the more innocents that die, the better their chance of victory. On the other hand, while the Bear is alive, there is always a reasonable chance that he may kill a Wolf at Night. Accordingly, while I presume that the Wolves don’t need to kill the Bear to win (although it is not entirely clear), they will no doubt want to get him out of the way at some point. In a funny kind of a way, therefore, the Innocents and the Wolves have a mutual interest in finding and killing the Bear. The same goes for the Innocents and the Bear, I suppose, since the Bear needs to stay alive to win and is as vulnerable as anyone, I think, to Wolf attacks. What do people think? Is there any way that we might use these dynamics to our advantage?