That was actually what I was wondering about; we have one shiny ourselves, so, how do we account for that?
Actually, hmm....
@NSMS
@TheMaskedReader
@OriginalName
@Tykan
Are any of you the crow with the shiny. You probably will not get robbed tonight, as ravens doing it would just end up outing either both or leave us knowing exactly where the one was anyway, but you will probably get robbed tomorrow in a last ditch effort to win.
Let's call the ravens A and B.
1. Right now, A has a shiny, B does not.
2. This plan ensures crows rob A 100%
3. This plan also ensures that crows steal from B.
4. Ergo, if B takes a shiny in the night, A and B will both be outed and crows will take both.
5. If A takes a shiny in the night, only A will be outed
6. However, if B takes a second shiny, B will be outed and end the night with 1 shiny.
This means a few things. 1. Entering tomorrow, ravens will, at most, have 1 shiny on raven A. IF they do have a shiny, crows will know where it is.
1. Raven B can take raven A's shiny, hiding it. Ravens retain 1 shiny. Raven A can steal up to one shiny, but will get robbed again twice for safety if Ravens have 1 shiny.
2. Both ravens can steal, ravens may hold 3 shinies before crows steal, however, if they do, 2 are on B. Crows steal 2 from B, ensuring ravens end the night with 1 shiny.
3. Jackdaw might start D6 with 2 shinies, however, crows can simply peck the jackdaw in this situation.
Does this scan?