Threes

July 25, 2012 Leave a comment

Game Rule:

The player starts by rolling five standard dice. In the game, the threes count as zero, while the other faces count normally. The goal is to get as low a sum as possible. On each roll, at least one die must be kept, and any dice that are kept are added to the player’s sum. The game lasts at most five rolls, and the score can be anywhere from 0 to 30.

For example, a game might go like this. On the first roll the player rollss

2 3 3 4 6

The player decides to keep the 3s, and so has a score of zero. The other 3 dice are rolled, and the result is

1 5 5

Here the player keeps the 1, so their score is 1, and rerolls the other two dice. The result is

1 2

Here, the player decides to keep both dice, and their final score is 4.

Question: If a player plays optimally(using a strategy which minimizes the expected value of their score), what is the expected value of their score?

Let’s analysis from the basic situation: one dice.
So our expectation for rolling one die is given by following:

$\frac{1+2+4+5+6}{6}=3$

Now, suppose we roll 2 dice. The rules require that we keep at least one, so clearly we must keep the lower of the two. The question is whether to keep the other one. If we don’t keep it, our expected value from it will be 3 when we reroll. Hence, we should keep it if it is 3, or 1, or 2.

Following this method, the expected value with two dice is given by :

$E_2=\frac{1}{36} \sum_{i=1}^{6} \sum_{j=1}^{6} (\min(i,j)+ min(max(i,j),3))=\frac{94*2}{36}=4.4$

Suppose we roll three dice. We must keep the lowest die, so we need to decide whether to keep either of the other two dice. Obviously, if we keep only one of them, we would keep the lower one. Suppose the three dice :

$d_1\leq d_2 \leq d_3$

Then if we keep both $d_2$ and $d_3$ , our sum is $d_2+d_3$ . If we reroll only $d_3$, then our expected sum is $\latex d_2+3$. If we reroll both $\latex d_2$ and $d_3$ , then our expected sum is 4.4. Thus we want to choose the option so that our expected sum is