Say there is a dice with six faces, each face has a positive real number different from others. There is a chessman on the origin of the number axis. In each trial, the dice rolls infinite times. Every time the dice rolls, the chessman moves forward the distance as the dice rolls out. The incident for each point on the number axis happens when the chessman ever steps on the point in the trial. With infinite trials, we get the probability of the incident for every point. If a point, if exists, owns a highest chance over others on the axis, we call it **the Best Point** of the dice. If a dice, if exists, whose Best Point's chance is over other dice, then the corresponding six-number array is **the Best Array**, and its Best Point is called as **the Best Best Point**.

The question is to get the Best Array.

Supposing the six numbers are $0<x0<x1<x2<x3<x4<x5$, for any real number n, there exists:

$f(n) = \frac{1}{6}(\sum_{i=1}^6 f(n-x_i))$ $,(n>0)$

$f(n) = 1$ $,(n=0)$

$f(n) = 0$ $,(n<0)$

where $f(n)$ is the probability of point n.

If we define $\mathbf{x} = (x_1, x_2, x_3, x_4, x_5, x_6)$, the problem is equivalent to query $\mathbf{x} = argmax_\mathbf{x}(max_n(f(\mathbf{x}, n)))$.

According to the formula above, there are some easy conclusions I have got, based on the exist of the Best Array:

[1]: For any dice, the smallest best point should be in its six-number array;

[2]: The best best point should be $x_6$

[3]: For the best array, $x_6 = x_i + x_j (\exists i,j, 1<=i, j<=5)$

[4]: The numbers of the Best Array are in the same field, which means there exists an integer version of the Best Array, in consideration of conclusion [5].

[5]: Multiple the numbers of the Best Array by some number produces a new Best Array, which means there exists the most simplified integer version.

Using greedy algorithm or some program, I pretty sure the most simplified integer best array should be [1, 2, 3, 4, 5, 8]. However, I have no clue about proving it.

**Update:**

You can describe this question with the notation of this paper with $S$ of size 6 and $f(s)=1/6$. Certainly you can promote the question by replacing the constant 6 with a variable. The generating function can be found as $[x^n]\frac{1}{1-\sum_{s\in S}f(s)x^s}$ at the discussion part of the paper mentioned above.