# Optimal strategy for slice weighing game

I watched an interesting contest on a Swedish game show the other night. I have tried to find an english name of the contest but haven't found any. Two contestants were each given one large sausage with equal weight, say 1 kg. Without looking at each other, they both cut one slice of their sausage and put the two slices on a scale. The contestant with the heavier slice was awarded one point. This was repeated until one contestant reached five points and was declared winner.

Since I watched the show, I've been thinking about an optimal strategy for winning this contest. It all boils down to winning the rounds by the least amount, so no weight is wasted. One way to make things harder for the opponent is to deliberately loose a point by cutting a really thin slice, leaving you with more sausage left to later rounds.

By observing the slices cut previously by the opponent, one can approximate how much sausage the opponent has left. If your current score is four points and you are certain that you have more sausage left than your opponent, then it is safe to use all your remaining sausage for the last round.

One problem in finding an optimal strategy is that it might take anywhere between five and nine rounds to reach five points, so you can't just divide your sausage into five equal parts. Other schemes, like doubling of halving the slice weight for each round is easily beaten too. Maybe an optimal strategy doesn't even exist for this game.

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Some scattered observations: Optimal strategies are usually probabilistic. You need to deliberately behave randomly in order to make it hard for your opponent to guess and counter what you'll do. In a symmetric game like that, no strategy can guarantee you more than even chances at worst (just imagine it being pitted against itself). – Henning Makholm Oct 26 '11 at 16:44
This question is related, but the aspects of a variable number of rounds and successive decisions are new here. – joriki Oct 26 '11 at 17:05
Wurst-case analysis might be useful here. – TonyK Dec 14 '11 at 10:28