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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.

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    $\begingroup$ It is not clear what probability space you consider. You say the chessman walks several times --- is the space all finite sequences of walks? If so, are each such walk equally likely? $\endgroup$ – Per Alexandersson Oct 1 '18 at 10:40
  • $\begingroup$ The dice is fair. In one trial, the chessman walks infinite steps, each step walks the distance the dice rolls out, and the incident for each point happens if the chessman steps on the point ever in this trial. For example, 6 numbers are [2,3,4,5,6,7], the chessman could never step on point 1, so the point 1's chance is 0. For point 4, the chessman has 2 method to step on: one is rolls 4 in one time, the other is rolls 2 twice in succession. So the possibility of point 2 in $1/6+(1/6)^2$. I hope the above explanation may make things clear. $\endgroup$ – rube wang Oct 2 '18 at 2:30
  • $\begingroup$ I might be a bit confused still, but if you speak of probability, the total sum of all probabilities must be 1. I cannot just add up the probabilities for walking on integer k, since this wont add up to 1, if I am not mistaken. This is the source of my confusion... $\endgroup$ – Per Alexandersson Oct 2 '18 at 4:26
  • $\begingroup$ For each point, the probability of incident of chessman ever stepped on it plus the probability of incident of chessman did not step on it equals 1. And we focus on the former. The total sum of the probability of incident of chessman step on each real number makes no sense. $\endgroup$ – rube wang Oct 2 '18 at 8:17
  • $\begingroup$ I have changed the main description. In this question, for each trail, there are multiple outcomes. This is a approximate metaphor. There is a bag of balls of 1000 different colors, and every trail I randomly picked up 100 balls in the bag. And I do it infinite trails. Then we get to know the chance of each ball been picked up (10%= the times of the trails the ball being picked up/ all trails' times) and not been picked up (90%= the times of the trails the ball not being picked up/ all trails' times). You adding all chance for all balls up will result to 100(=10%*1000) but not 1. $\endgroup$ – rube wang Oct 2 '18 at 8:40
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Ok, so I believe one can rephrase it a bit. Suppose that $a_i$ is the vector of numbers on the sides. Then you are interested in $$ f(x)=\prod_{i=1}^6 \frac{1}{1-\frac{x^{a_i}}{6}}. $$ Note that if $a = \{1,2,3,4,5,8\}$, then the Taylor expansion is $$ f(x) = 1 + \frac{x}{6}+\frac{7x^2}{36}+\dotsb $$ where the $7/36$ represents the sum $1/6$ chance of reaching $2$ with one step, and $1/36$ chance of reaching it in two steps (both of length 1). I believe this is what you are looking for, and you seek to maximize the coefficient of some $x^k$.

The above example, gives the largest coefficient about $0.245 x^8$. However, the array $1,2,\dotsc,6$ gives $0.2655x^6$, which is a bit bigger. Allowing for entries to be repeated in the array would give even higher coefficients.

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  • $\begingroup$ Not exactly. Your formula lacks of combinatorial number, which means walking to point 3 by rolling out 1 first and 2 later or 2 first and 1 later counts 1 time altogether, but it should count 2 times. The problem is set to different entries cause it would be trivial if the entries of the array are all the same. As a reference, for [1,2,3,4,5,6], x=6 gets the chance of 0.3602323388203018; for [1,2,3,4,5,8], x=8 gets the chance of 0.3653162389498552. $\endgroup$ – rube wang Oct 2 '18 at 12:40
  • $\begingroup$ I think $f(x)=\sum_{i=0}^{inf} (1/6*\sum_{j=1}^6 (x^{a_j}))^i$ $\endgroup$ – rube wang Oct 2 '18 at 12:52
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    $\begingroup$ The above formula is supported by MATLAB. Then $f(x)=\frac{6}{6-\sum_{i=1}^6 x^{a_i} }$. $\endgroup$ – rube wang Oct 2 '18 at 13:24

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