Problem 825

Chasing Game

Two cars are on a circular track of total length $2 n$ , facing the same direction, initially distance $n$ apart.

They move in turn. At each turn, the moving car will advance a distance of $1$ , $2$ or $3$ , with equal probabilities.

The chase ends when the moving car reaches or goes beyond the position of the other car. The moving car is declared the winner.

Let $S (n)$ be the difference between the winning probabilities of the two cars.

For example, when $n = 2$ , the winning probabilities of the two cars are $\frac{9}{11}$ and $\frac{2}{11}$ , and thus $S (2) = \frac{7}{11}$ .

Let $T (N) = \sum_{n = 2}^{N} S (n)$ .

You are given that $T (10) = 2.38235282$ rounded to $8$ digits after the decimal point.

Find $T (10^{14})$ , rounded to $8$ digits after the decimal point.

追逐游戏

在长度为 $2 n$ 的环形赛道上有两辆汽车沿相同方向前进，初始距离为 $n$ 。

两辆汽车轮流行动，每次行动时会以相等概率随机向前移动距离 $1$ 、 $2$ 或 $3$ 。

当其中一辆汽车追上或超过另一辆汽车时，游戏结束，前者获胜。

记 $S (n)$ 为两车获胜概率之差。

例如，当 $n = 2$ 时，两车获胜的概率分别是 $\frac{9}{11}$ 和 $\frac{2}{11}$ ，因此 $S (2) = \frac{7}{11}$ 。

记 $T (N) = \sum_{n = 2}^{N} S (n)$ 。

已知 $T (10) = 2.38235282$ ，四舍五入至小数点后 $8$ 位。

求 $T (10^{14})$ ，并将你的答案四舍五入至小数点后 $8$ 位。