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?
Basic Strategy: Minimizing the Expected Score
A good place to start is to understand what score to expect when playing the game with $N$ dice. To begin, lets assume there is no score with which to beat so imagine you are going first. Given these assumptions, our objective is quite simple: minimize our expected score at every decision. Such an analysis is certainly best done in reverse.
Suppose we are playing with only one die where $D \in \{0,1,2,4,5,6\}$. We will assume that all dice are fair, so $P(X=x)=\frac{1}{6}$. The expected score of rolling one die, $i$, is
$E_{1}=\frac{1}{6}\sum_{i \in D}^{} i $
$E_{1}=\frac{18}{6}=3$
Suppose we are playing with two dice, $i$ and $j$; rearrange them so they are in ascending order and call them $d_{1} \leq d_{2}$. The rules require that we keep at least one, so clearly we keep $d_{1}$. The questions is whether to keep $d_{2}$. If we don’t keep it, our expected value from it will be 3 when we re-roll. Hence, we should keep $d_{2}$ if it is a 0, 1, or 2. Following this method, the expected value of our score when we play with two dice is expressible as
$E_{2}=\frac{1}{6^2}\sum_{i,j \in D}^{} d_{1}+min(d_{2}, E_{1})$
$E_{2}=\frac{158}{36}=4.3\overline{8}$
Suppose we are playing with three dice, $i$, $j$, and $k$; sort them and call them $d_{1} \leq d_{2} \leq d_{3}$. We must keep $d_{1}$, so we need to decide whether to keep either of the other two dice. If we keep both $d_{2}$ and $d_{3}$, our score is $d_{1}+d_{2}+d_{3}$. If we re-roll only $d_{3}$, then our expected score is $d_{1}+d_{2}+E_{1}$. If we re-roll both $d_{2}$ and $d_{3}$, then our expected score is $d_{1}+E_{2}$. Thus we want to choose the option with the lowest expected score. The expected value of our score when we roll three dice is expressible as
$E_{3}=\frac{1}{6^3}\sum_{i,j,k \in D}^{} d_{1}+min(d_{2}+d_{3}, d_{2}+E_{1}, E_{2})$
$E_{3}=\frac{2261}{432}=5.233796...$
Continuing on in this fashion, The expected value of our score when we roll four dice is
$E_{4}=\frac{1}{6^4}\sum_{i,j,k,l \in D}^{} d_{1}+min(d_{2}+d_{3}+d_{4},\ d_{2}+d_{3}+E_{1},\ d_{2}+E_{2},\ E_{3})$
$E_{4}=\frac{1663107}{279936}=5.833858...$
And five dice
$E_{5}= \frac{1}{6^5}\sum_{i,j,k,l,m \in D}^{} d_{1}+min(d_{2}+d_{3}+d_{4}+d_{5},\ d_{2}+d_{3}+d_{4}+E_{1}, d_{2}+d_{3}+E_{2}, d_{2}+E_{3},\ E_{4})$
$E_{5}=\frac{13613549985}{2176782336}=6.253979...$
Given the expected scores of playing with $N$ dice, we have encoded above a basic strategy which is optimal for a game of Threes where we have no score to beat. Unfortunately, it is a tricky strategy to paraphrase. Consider that if you roll 3-3-3-2-2 you should keep all the dice, but if you roll 3-3-2-2-2 you should re-roll the 2s since $6 > E_{3}$. So unfortunately there is not a simple statement summarizable to a "this die or more should always be kept on roll $r$" simplicity. However, a simplification to the strategies outlined above can be made by noting that 1) threes should always be kept and 2) the $E_{N}$ values that appear above can be replaced by their integer parts for easy memorization (for example a sum of integers is less than $4.3\overline{8}$ only if the sum is less than or equal to 4).
A simulation of 1000000 games using this strategy has the following distribution of scores
Given the expected scores of playing with $N$ dice, we have encoded above a basic strategy which is optimal for a game of Threes where we have no score to beat. Unfortunately, it is a tricky strategy to paraphrase. Consider that if you roll 3-3-3-2-2 you should keep all the dice, but if you roll 3-3-2-2-2 you should re-roll the 2s since $6 > E_{3}$. So unfortunately there is not a simple statement summarizable to a "this die or more should always be kept on roll $r$" simplicity. However, a simplification to the strategies outlined above can be made by noting that 1) threes should always be kept and 2) the $E_{N}$ values that appear above can be replaced by their integer parts for easy memorization (for example a sum of integers is less than $4.3\overline{8}$ only if the sum is less than or equal to 4).
A simulation of 1000000 games using this strategy has the following distribution of scores
------------------------------------------------------------------------------
X PDF CDF
------------------------------------------------------------------------------
0 0.011869 XX 0.011869
1 0.037897 XXXXXXXX 0.049766 X
2 0.076389 XXXXXXXXXXXXXXXX 0.126155 XXX
3 0.100920 XXXXXXXXXXXXXXXXXXXXX 0.227075 XXXXX
4 0.117054 XXXXXXXXXXXXXXXXXXXXXXXX 0.344129 XXXXXXXX
5 0.118088 XXXXXXXXXXXXXXXXXXXXXXXXX 0.462217 XXXXXXXXXXX
6 0.112624 XXXXXXXXXXXXXXXXXXXXXXX 0.574841 XXXXXXXXXXXXXX
7 0.098966 XXXXXXXXXXXXXXXXXXXX 0.673807 XXXXXXXXXXXXXXXX
8 0.082787 XXXXXXXXXXXXXXXXX 0.756594 XXXXXXXXXXXXXXXXXX
9 0.064524 XXXXXXXXXXXXX 0.821118 XXXXXXXXXXXXXXXXXXXX
10 0.053070 XXXXXXXXXXX 0.874188 XXXXXXXXXXXXXXXXXXXXX
11 0.041457 XXXXXXXX 0.915645 XXXXXXXXXXXXXXXXXXXXXX
12 0.030458 XXXXXX 0.946103 XXXXXXXXXXXXXXXXXXXXXXX
13 0.020815 XXXX 0.966918 XXXXXXXXXXXXXXXXXXXXXXXX
14 0.013893 XX 0.980811 XXXXXXXXXXXXXXXXXXXXXXXX
15 0.008638 X 0.989449 XXXXXXXXXXXXXXXXXXXXXXXX
16 0.005120 X 0.994569 XXXXXXXXXXXXXXXXXXXXXXXX
17 0.002887 0.997456 XXXXXXXXXXXXXXXXXXXXXXXX
18 0.001404 0.998860 XXXXXXXXXXXXXXXXXXXXXXXX
19 0.000647 0.999507 XXXXXXXXXXXXXXXXXXXXXXXX
20 0.000282 0.999789 XXXXXXXXXXXXXXXXXXXXXXXX
21 0.000144 0.999933 XXXXXXXXXXXXXXXXXXXXXXXX
22 0.000048 0.999981 XXXXXXXXXXXXXXXXXXXXXXXX
23 0.000011 0.999992 XXXXXXXXXXXXXXXXXXXXXXXX
24 0.000006 0.999998 XXXXXXXXXXXXXXXXXXXXXXXX
25 0.000002 1.000000 XXXXXXXXXXXXXXXXXXXXXXXXX
------------------------------------------------------------------------------
<X> = 6.28923
Fortunately, we do not always need to go first, so sometimes our objective is different than simply minimizing the expectation. Sometimes we have to take chances (and get lucky) because we have a score to beat. This is easy to see if we add a little more information to our example. Imagine you are rolling second with a score of 5 to beat and you have an opening roll of 3-3-2-2-2. Keeping all the die because $0+0+2+2+2 < E_{5}$ would guarantee you a loss.
So what is the best play? The answer is, it depends...
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