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And just to be clear, I interpreted the strategy to be each player gets an assigned ones digit, while each player picks the tens digit and keeps it to themselves. For example, I get assigned a 5, so I pick 5, 15, 25, etc. I fail to see how this runs an increased risk of equidistant picks.

[That means there is 27% chance two (or more) people pick the same number]

Apologies for the absence last night and this morning.

I've thought a lot about this, and I've yet to think of any meaningful strategy on picking numbers.

As discussed, if no one announces any of their choices, the wolf has no information and cannot target anyone. However, we do run the risk of a double (or more) wolfing as a result of a duplicate pick or equidistant picks.

I did a little research, and here's what a found. The birthday problem can be modeled by the equation: p(different) = e^-((n^2)/(2T)), where n is the number of people, and T is the total choices. Subbing in 8 for n and 100 for T (8 humans, 100 numbers), we get p(different) = e^((8^2)/(2(100))), which gives p(different) = .73. That means there is 27% chance two (or more) people pick the same number. That seems high, and at first thought, I don't want to take the chance.

However, there are some other things to consider. First, two people picking the same number isn't costly unless the wolf's choice is closest to it. I don't know that stats on that, but I'm sure it dramatically lowers the risk from 28%. Second, everyone picking different numbers does not defend against equidistant picks. I have no idea how to model that statistically. But again, that's only relevant if the wolf picks a number between the equidistant picks.

All that being said, I do like the track that Toby is on of avoiding duplicate picks. Having each player assigned a ones digit (as suggested) does accomplish that. Furthermore, I'm not convinced it really gives the wolf that much information, if any at all. So what if the wolf knows that my number will end in 5. I could pick 5, 65, 85, etc.

Let the games begin. Looking forward to this!

This is a very specific question, but with 3 players remaining (HH, PNS, and human), who wins if a wolfing would cause both the PNS (by a swap condition) and HH (by equal humans and HH) to win?

(bold text) In!

Yeah, Im good at forums.

Howdy all, first post. I'm TZPs irl friend, who has pleasantly roped me into the world of secret identity games.

I appreciate the simplicity of BDS's game; just a straight up number draw to determine the wolfing. The addition of the Number Swapper adds an interesting twist. Curious to see what kind of debates/bluffs emerge from such a simple game design.

Gadd, the unknown nature of Hylia makes your game a bit of a mystery, but that's by design could really make the game fun. I do agree that the Yellow Wizzrobe needed a second win condition should the thunderstorm miss, which it looks like you added. With two wolves, I think having at least 7 humans makes it balanced, although the lack of information about Hylia makes that a bit of a question mark. I guess we'll find out eventually (or one of us will)!

TZP's is by far the most complex, which could mean it's extremely fun to play, although it could be hard to balance. Ganon, Link, and Zelda having two lives is a cool twist, along with the presence of items. Ganon and the Analysis Wolf will have to work together to locate/wolf players, while Zelda's seer abilities will yield important information but may paint herself as a target.

Look forward to playing with you all!