I can imagine that there is a point at which a single player would always be wise to no longer advance. I see it being a function of all of the various combinations. The problem is, I've never quite been able to figure out all of the probabilities of a given combination coming up. Does anyone out there know what the optimal value is where you are more likely to lose then to continue onward? BTW, the version I have is this one, not sure if it makes a difference... Thanks!
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This is a much more complex issue than it at first seems. The game of pass the pigs is derived from an older, simpler game called Pig. In Pig each player takes turns with a six sided die and may roll as many times as they like, or until they roll a 1. If they roll a 1 they score nothing for the round, otherwise they score the total rolled.
Pass the Pigs is roughly equivalent to this. Think of a game with a weighted n-sided die, with sides "Pig out", "Double trotter", "Razorback", etc. A player may roll this die until they roll "Pig out", and their score for the round is the cumulative score, or zero if a "Pig out" was rolled. The difference is the inclusion of "Making Bacon" (reset your score to 0), and "Piggy Back" (player is out of the game).
In Pig each player's optimal play is dependent only on the current scores. This means that when you receive the dice, you can calculate what your target score for the turn should be. If you reach that score then you should bank it, and pass the dice on.
So finally, it should be possible to apply the same methods to Pass the Pigs if you knew the probabilities of the various pig rolls. These probabilities have been calculated here (see Table 4), using a data set of 6000 throws. They have also calculated the value you should aim for to maximize your expected score in a turn in Section 3.2 (the target is about 21 or 22). I can't find a similar analysis corresponding to multiplayer Pass the Pigs though.