Problem 589
Poohsticks Marathon
Christopher Robin and Pooh Bear love the game of Poohsticks so much that they invented a new version which allows them to play for longer before one of them wins and they have to go home for tea. The game starts as normal with both dropping a stick simultaneously on the upstream side of a bridge. But rather than the game ending when one of the sticks emerges on the downstream side, instead they fish their sticks out of the water, and drop them back in again on the upstream side. The game only ends when one of the sticks emerges from under the bridge ahead of the other one having also ‘lapped’ the other stick - that is, having made one additional journey under the bridge compared to the other stick.
On a particular day when playing this game, the time taken for a stick to travel under the bridge varies between a minimum of 30 seconds, and a maximum of 60 seconds. The time taken to fish a stick out of the water and drop it back in again on the other side is 5 seconds. The current under the bridge has the unusual property that the sticks’ journey time is always an integral number of seconds, and it is equally likely to emerge at any of the possible times between 30 and 60 seconds (inclusive). It turns out that under these circumstances, the expected time for playing a single game is 1036.15 seconds (rounded to 2 decimal places). This time is measured from the point of dropping the sticks for the first time, to the point where the winning stick emerges from under the bridge having lapped the other.
The stream flows at different rates each day, but maintains the property that the journey time in seconds is equally distributed amongst the integers from a minimum, $n$, to a maximum, $m$, inclusive. Let the expected time of play in seconds be $E(m,n)$. Hence $E(60,30)=1036.15…$
Let $S(k)=\sum_{m=2}^k\sum_{n=1}^{m-1}E(m,n)$.
For example $S(5)=7722.82$ rounded to 2 decimal places.
Find $S(100)$ and give your answer rounded to 2 decimal places.
维尼木棍马拉松
克里斯托佛·罗宾和小熊维尼特别喜欢玩维尼木棍这个游戏，因此他们发明了一种新的版本，这样他们就不用在一方获胜后早早回家喝茶而可以玩得更久。和一般的“维尼木棍”游戏一样，开始时，两人同时在桥的上游扔下一根木棍，但在其中一根木棍在桥的下游一边出现时，游戏并不结束，而是从水里捞起来然后再次扔到桥的上游。游戏只在当其中一根木棍从桥的下游出现时，另一根还未完成其上一轮从桥的下游出现——也就是说，套了另一根木棍整整一圈。
在某一天玩这个游戏时，一根木棍从桥的上游到达下游所用的时间最少为30秒，最多为60秒。将木棍捞起来并重新扔到桥的另一边所用的时间为5秒。桥下的水流有一种神奇的性质，使得木棍所用的时间总是整数秒，而且任意30秒至60秒之间（含）的时长都是等可能发生的。在这种情况下，进行一整轮游戏的期望时间为1036.15秒（保留2位小数），这个时间是从第一次将木棍扔入水中开始计算，直到获胜的木棍最后一次浮出水面为止。
每天水流的速度都不同，但是都保持着这个性质：木棍每一趟旅程的时间等可能地分布在最小值$n$和最大值$m$间（含）的所有整数上。记以秒计的每一轮的期望游戏时间为$E(m,n)$。因此$E(60,30)=1036.15…$。
令$S(k)=\sum_{m=2}^k\sum_{n=1}^{m-1}E(m,n)$。
例如$S(5)=7722.82$保留2位小数。
求$S(100)$并将你的答案保留2位小数。