Wednesday, November 22, 2006

The Random Pairing Problem

After some more discussion in chat at BGDF, we'v come up with an alternate pairing solution which, while not solving all the problems, might be better (or may not).

The problem pertains to Loose Triggers, Smoking Barrels. The exact problem is this:

Each player is secretly assigned two other players; one is a Weasel, one is a Pal. In addition to helping yourself, you want to also help your Pal, and you want to keep your Weasel down, so to speak. The big problem is, how do yo divvy out the Pals and Weasel secretly while making sure that the player dos not get his own name as a Pal or Weasel. And then, additionally, making sure that the Pal and Weasels are also different players.

The way I currently have this implemented is simple enough; if a player draws his own name, he puts it back into the mix and draws a new one. However, this creates a situation where players will know that Player X has just put his name back in the hat; it also doesn't solve the problem that the last player to draw can get screwed with having his name be the only one left, causing a complete redraw for everyone.

Now, I don't think redraws are truly that much of an issue in this game, it would just be nice to get rid of it altogether. Also,I'm pretty sure that it will become fairly obvious through the way the players play the game to determine who everyone's Pals and Weasels are. I think. A fairly slick player probably could hide it well. So, I'm not too worried about a single player redrawing because of him picking his name oout of the hat.

Still, it would be nice if a more elegant solution could be derived that could solve all these problems.

The thread generated a some good responses, but they all break down on some level. However I think Yogurt solved it pretty well; I can't find any really big holes in it,aside from players trying to memorize various "if player X and player Y put stuff together in one cup, then..." issues. So CONGRATS TO YOGURT!

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