How to play
Threes is played with five dice where the objective is to get the lowest score. Someone starts by rolling five dice. On each roll, you are required to keep at least one of your dice, and you re-roll the remaining ones. You do this until all five dice have been kept. So at most you will roll five times. For example, you can choose to keep anywhere from one to five dice on the first roll. If you kept two of them, then you'd re-roll the remaining three dice, and you can choose to keep one to three of them. And, so on...
The sum of the five selected dice becomes your score, with one caviat: threes are worth zero. Therefore a minimum score is zero by rolling 3-3-3-3-3, and a maximum score is thirty by rolling 6-6-6-6-6.
When beginning, the order is determined randomly. After a round has been played, the highest score goes first and the lowest score goes last. Clearly, going last has the advantage due to the ability to know when to simply just stop trying for a lower score while taking on the risk of getting a higher one. Say the first person scores a 8. And, as the 2nd player, your very first roll is 3-3-2-2-2, you can simply stop right there and claim a win with a score of 6.
For a three, four, or more player game, only a single distinct winner can win the pot. So, say in a four player game, two people score 6, one guy scores 10, and another scores a 9. Since there is no distinct winner due to two best scores tying, there is no winner this round. All players ante up again, building the pot. All players are still in the game, and therefore, are eligible to win. You play until there is a single distinct winner, who scoops.
Is there an optimal strategy for this game? If so, what is it?